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# 9.7: Solving Systems with Inverses

Solving Systems with Inverses
In this section, you will:
• Find the inverse of a matrix.
• Solve a system of linear equations using an inverse matrix.

Nancy plans to invest 10,500 into two different bonds to spread out her risk. The first bond has an annual return of 10%, and the second bond has an annual return of 6%. In order to receive an 8.5% return from the two bonds, how much should Nancy invest in each bond? What is the best method to solve this problem? There are several ways we can solve this problem. As we have seen in previous sections, systems of equations and matrices are useful in solving real-world problems involving finance. After studying this section, we will have the tools to solve the bond problem using the inverse of a matrix. # Finding the Inverse of a Matrix We know that the multiplicative inverse of a real number<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] a −1 , and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><msup/></mrow></annotation-xml></semantics>[/itex] a −1 = a −1 a=( 1 a )a=1. For example,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] 2 −1 = 1 2 and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 2 )2=1. The multiplicative inverse of a matrix is similar in concept, except that the product of matrix<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and its inverse<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] A −1 equals the identity matrix. The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. We identify identity matrices by<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] I n where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]represents the dimension of the matrix. [link] and [link] are the identity matrices for a<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mtext/><mo>×</mo><mtext/><mn>2</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]matrix and a<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>3</mn><mtext/><mo>×</mo><mtext/><mn>3</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]matrix, respectively. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mi>I</mi></msub></mrow></annotation-xml></semantics>[/itex] 2 =[ 1 0 0 1 ] <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mi>I</mi></msub></mrow></annotation-xml></semantics>[/itex] 3 =[ 1 0 0 0 1 0 0 0 1 ] The identity matrix acts as a 1 in matrix algebra. For example,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mi>I</mi><mo>=</mo><mi>I</mi><mi>A</mi><mo>=</mo><mi>A</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex] A matrix that has a multiplicative inverse has the properties <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtable columnalign="left"><mtr><mtd><mi>A</mi><msup/></mtd></mtr></mtable></annotation-xml></semantics>[/itex] A −1 =I A −1 A=I A matrix that has a multiplicative inverse is called an invertible matrix. Only a square matrix may have a multiplicative inverse, as the reversibility,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><msup/></mrow></annotation-xml></semantics>[/itex] A −1 = A −1 A=I, is a requirement. Not all square matrices have an inverse, but if<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is invertible, then<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] A −1 is unique. We will look at two methods for finding the inverse of a<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mtext> </mtext><mo>×</mo><mtext> </mtext><mn>2</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]matrix and a third method that can be used on both<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mtext> </mtext><mo>×</mo><mtext> </mtext><mn>2</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>3</mn><mtext> </mtext><mo>×</mo><mtext> </mtext><mn>3</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]matrices. The Identity Matrix and Multiplicative Inverse The identity matrix,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] I n , is a square matrix containing ones down the main diagonal and zeros everywhere else. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow/></mtd></mtr><mtr columnalign="left"><mtd columnalign="left"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow/></mtd></mtr><mtr columnalign="left"><mtd columnalign="left"><mrow><msub><mi>I</mi></msub></mrow></mtd></mtr></mtable></mrow></mtd></mtr></mtable></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 2 =[ 1 0 0 1 ] I 3 =[ 1 0 0 0 1 0 0 0 1 ] 2 × 2 3 × 3 If<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is an<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>n</mi><mtext> </mtext><mo>×</mo><mtext> </mtext><mi>n</mi><mtext> </mtext></annotation-xml></semantics>[/itex] matrix and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>B</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is an<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtext> </mtext><mi>n</mi><mtext> </mtext><mo>×</mo><mtext> </mtext><mi>n</mi><mtext> </mtext></annotation-xml></semantics>[/itex] matrix such that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mi>B</mi><mo>=</mo><mi>B</mi><mi>A</mi><mo>=</mo><msub/></mrow></annotation-xml></semantics>[/itex] I n , then<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>B</mi><mo>=</mo><msup/></mrow></annotation-xml></semantics>[/itex] A −1 , the multiplicative inverse of a matrix<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex] Showing That the Identity Matrix Acts as a 1 Given matrix A, show that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mi>I</mi><mo>=</mo><mi>I</mi><mi>A</mi><mo>=</mo><mi>A</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex] <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3 4 −2 5 ] Use matrix multiplication to show that the product of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and the identity is equal to the product of the identity and A. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mi>I</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3 4 −2 5 ] [ 1 0 0 1 ]=[ 3⋅1+4⋅0 3⋅0+4⋅1 −2⋅1+5⋅0 −2⋅0+5⋅1 ]=[ 3 4 −2 5 ] <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mi>I</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 0 0 1 ] [ 3 4 −2 5 ]=[ 1⋅3+0⋅(−2) 1⋅4+0⋅5 0⋅3+1⋅(−2) 0⋅4+1⋅5 ]=[ 3 4 −2 5 ] Given two matrices, show that one is the multiplicative inverse of the other. 1. Given matrix<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]of order<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mtext> </mtext><mo>×</mo><mtext> </mtext><mi>n</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and matrix<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>B</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]of order<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mtext> </mtext><mo>×</mo><mtext> </mtext><mi>n</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]multiply<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mi>B</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex] 2. If<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mi>B</mi><mo>=</mo><mi>I</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]then find the product<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>B</mi><mi>A</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]If<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>B</mi><mi>A</mi><mo>=</mo><mi>I</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]then<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>B</mi><mo>=</mo><msup/></mrow></annotation-xml></semantics>[/itex] A −1 and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>=</mo><msup/></mrow></annotation-xml></semantics>[/itex] B −1 . Showing That Matrix A Is the Multiplicative Inverse of Matrix B Show that the given matrices are multiplicative inverses of each other. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 5 −2 −9 ],B=[ −9 −5 2 1 ] Multiply<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mi>B</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>B</mi><mi>A</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]If both products equal the identity, then the two matrices are inverses of each other. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>A</mi><mi>B</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 1 5 −2 −9 ]·[ −9 −5 2 1 ] =[ 1(−9)+5(2) 1(−5)+5(1) −2(−9)−9(2) −2(−5)−9(1) ] =[ 1 0 0 1 ] <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>B</mi><mi>A</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] −9 −5 2 1 ]·[ 1 5 −2 −9 ] =[ −9(1)−5(−2) −9(5)−5(−9) 2(1)+1(−2) 2(−5)+1(−9) ] =[ 1 0 0 1 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mi>B</mi></annotation-xml></semantics>[/itex]are inverses of each other. Show that the following two matrices are inverses of each other. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 4 −1 −3 ],B=[ −3 −4 1 1 ] <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>A</mi><mi>B</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 1 4 −1 −3 ] [ −3 −4 1 1 ]=[ 1(−3)+4(1) 1(−4)+4(1) −1(−3)+−3(1) −1(−4)+−3(1) ]=[ 1 0 0 1 ] BA=[ −3 −4 1 1 ] [ 1 4 −1−3 ]=[ −3(1)+−4(−1) −3(4)+−4(−3) 1(1)+1(−1) 1(4)+1(−3) ]=[ 1 0 0 1 ] ## Finding the Multiplicative Inverse Using Matrix Multiplication We can now determine whether two matrices are inverses, but how would we find the inverse of a given matrix? Since we know that the product of a matrix and its inverse is the identity matrix, we can find the inverse of a matrix by setting up an equation using matrix multiplication. Finding the Multiplicative Inverse Using Matrix Multiplication Use matrix multiplication to find the inverse of the given matrix. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 −2 2 −3 ] For this method, we multiply<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]by a matrix containing unknown constants and set it equal to the identity. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 −2 2 −3 ] [ a b c d ]=[ 1 0 0 1 ] Find the product of the two matrices on the left side of the equal sign. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 −2 2 −3 ] [ a b c d ]=[ 1a−2c 1b−2d 2a−3c 2b−3d ] Next, set up a system of equations with the entry in row 1, column 1 of the new matrix equal to the first entry of the identity, 1. Set the entry in row 2, column 1 of the new matrix equal to the corresponding entry of the identity, which is 0. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtable><mtr><mtd><mn>1</mn><mi>a</mi><mn>−2</mn><mi>c</mi><mo>=</mo><mn>1</mn><mtext> </mtext><mtext> </mtext><msub/></mtd></mtr></mtable></annotation-xml></semantics>[/itex] R 1 2a−3c=0 R 2 Using row operations, multiply and add as follows:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">(</mo><mn>−2</mn><mo stretchy="false">)</mo><msub/></mrow></annotation-xml></semantics>[/itex] R 1 + R 2 → R 2 . Add the equations, and solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>c</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex] <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="right"><mtr columnalign="right"><mtd columnalign="right"><mrow><mn>1</mn><mi>a</mi><mo>−</mo><mn>2</mn><mi>c</mi><mo>=</mo><mn>1</mn><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 0+1c=−2 c=−2 Back-substitute to solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex] <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="right"><mtr columnalign="right"><mtd columnalign="right"><mrow><mi>a</mi><mn>−2</mn><mo stretchy="false">(</mo><mn>−2</mn><mo stretchy="false">)</mo><mo>=</mo><mn>1</mn><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] a+4=1 a=−3 Write another system of equations setting the entry in row 1, column 2 of the new matrix equal to the corresponding entry of the identity, 0. Set the entry in row 2, column 2 equal to the corresponding entry of the identity. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="right"><mtr columnalign="right"><mtd columnalign="right"><mrow><mn>1</mn><mi>b</mi><mn>−2</mn><mi>d</mi><mo>=</mo><mn>0</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] R 1 2b−3d=1 R 2 Using row operations, multiply and add as follows:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −2 ) R 1 + R 2 = R 2 . Add the two equations and solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>d</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex] <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="right"><mtr columnalign="right"><mtd columnalign="right"><mrow><mn>1</mn><mi>b</mi><mn>−2</mn><mi>d</mi><mo>=</mo><mn>0</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 0+1d=1 d=1 Once more, back-substitute and solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>b</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex] <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="right"><mtr columnalign="right"><mtd columnalign="right"><mrow><mi>b</mi><mn>−2</mn><mo stretchy="false">(</mo><mn>1</mn><mo stretchy="false">)</mo><mo>=</mo><mn>0</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] b−2=0 b=2 <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>A</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 =[ −3 2 −2 1 ] ## Finding the Multiplicative Inverse by Augmenting with the Identity Another way to find the multiplicative inverse is by augmenting with the identity. When matrix<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is transformed into<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>I</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]the augmented matrix<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>I</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]transforms into<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] A −1 . For example, given <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2 1 5 3 ] augment<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]with the identity <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2 1 5 3 | 1 0 0 1 ] Perform row operations with the goal of turning<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]into the identity. 1. Switch row 1 and row 2. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 5 3 2 1 | 0 1 1 0 ] 2. Multiply row 2 by<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>−2</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and add to row 1. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 1 2 1 | −2 1 1 0 ] 3. Multiply row 1 by<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>−2</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and add to row 2. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 1 0 −1 | −2 1 5 −2 ] 4. Add row 2 to row 1. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 0 0 −1 | 3 −1 5 −2 ] 5. Multiply row 2 by<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>−1.</mn></mrow></annotation-xml></semantics>[/itex] <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 0 0 1 | 3 −1 −5 2 ] The matrix we have found is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] A −1 . <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>A</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 =[ 3 −1 −5 2 ] ## Finding the Multiplicative Inverse of 2×2 Matrices Using a Formula When we need to find the multiplicative inverse of a<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mtext> </mtext><mo>×</mo><mtext> </mtext><mn>2</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]matrix, we can use a special formula instead of using matrix multiplication or augmenting with the identity. If<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is a<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mo>×</mo><mn>2</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]matrix, such as <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] a b c d ] the multiplicative inverse of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is given by the formula <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>A</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 = 1 ad−bc [ d −b −c a ] where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mi>d</mi><mo>−</mo><mi>b</mi><mi>c</mi><mo>≠</mo><mn>0.</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]If<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mi>d</mi><mo>−</mo><mi>b</mi><mi>c</mi><mo>=</mo><mn>0</mn><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]then<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]has no inverse. Using the Formula to Find the Multiplicative Inverse of Matrix A Use the formula to find the multiplicative inverse of <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 −2 2 −3 ] Using the formula, we have <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><msup><mi>A</mi></msup></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] −1 = 1 (1)(−3)−(−2)(2) [ −3 2 −2 1 ] = 1 −3+4 [ −3 2 −2 1 ] =[ −3 2 −2 1 ] Analysis We can check that our formula works by using one of the other methods to calculate the inverse. Let’s augment<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]with the identity. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 −2 2 −3 | 1 0 0 1 ] Perform row operations with the goal of turning<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]into the identity. 1. Multiply row 1 by<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>−2</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and add to row 2. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 −2 0 1 | 1 0 −2 1 ] 2. Multiply row 1 by 2 and add to row 1. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 0 0 1 | −3 2 −2 1 ] So, we have verified our original solution. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>A</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 =[ −3 2 −2 1 ] Use the formula to find the inverse of matrix<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Verify your answer by augmenting with the identity matrix. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 −1 2 3 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>A</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 =[ 3 5 1 5 − 2 5 1 5 ] Finding the Inverse of the Matrix, If It Exists Find the inverse, if it exists, of the given matrix. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3 6 1 2 ] We will use the method of augmenting with the identity. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3 6 1 3 | 1 0 0 1 ] 1. Switch row 1 and row 2. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 3 3 6 ​ ​ ​​| 0 1 1 0 ] 2. Multiply row 1 by −3 and add it to row 2. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 2 0 0 | 1 0 −3 1 ] 3. There is nothing further we can do. The zeros in row 2 indicate that this matrix has no inverse. ## Finding the Multiplicative Inverse of 3×3 Matrices Unfortunately, we do not have a formula similar to the one for a<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mtext/><mo>×</mo><mtext/><mn>2</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]matrix to find the inverse of a<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>3</mn><mtext/><mo>×</mo><mtext/><mn>3</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]matrix. Instead, we will augment the original matrix with the identity matrix and use row operations to obtain the inverse. Given a<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>3</mn><mtext> </mtext><mo>×</mo><mtext> </mtext><mn>3</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] matrix <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2 3 1 3 3 1 2 4 1 ] augment<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]with the identity matrix <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mo>|</mo><mi>I</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2 3 1 3 3 1 2 4 1 | 1 0 0 0 1 0 0 0 1 ] To begin, we write the augmented matrix with the identity on the right and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]on the left. Performing elementary row operations so that the identity matrix appears on the left, we will obtain the inverse matrix on the right. We will find the inverse of this matrix in the next example. Given a<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>3</mn><mtext> </mtext><mo>×</mo><mtext> </mtext><mn>3</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]matrix, find the inverse 1. Write the original matrix augmented with the identity matrix on the right. 2. Use elementary row operations so that the identity appears on the left. 3. What is obtained on the right is the inverse of the original matrix. 4. Use matrix multiplication to show that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><msup/></mrow></annotation-xml></semantics>[/itex] A −1 =I and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] A −1 A=I. Finding the Inverse of a 3 × 3 Matrix Given the<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>3</mn><mtext> </mtext><mo>×</mo><mtext> </mtext><mn>3</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]matrix<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]find the inverse. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2 3 1 3 3 1 2 4 1 ] Augment<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]with the identity matrix, and then begin row operations until the identity matrix replaces<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]The matrix on the right will be the inverse of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2 3 1 3 3 1 2 4 1 | 1 0 0 0 1 0 0 0 1 ] → Interchange R 2 and R 1 [ 3 3 1 2 3 1 2 4 1 | 0 1 0 1 0 0 0 0 1 ] <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo>−</mo><msub/></mrow></annotation-xml></semantics>[/itex] R 2 + R 1 = R 1 →[ 1 0 0 2 3 1 2 4 1 | −1 1 0 1 0 0 0 0 1 ] <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo>−</mo><msub/></mrow></annotation-xml></semantics>[/itex] R 2 + R 3 = R 3 →[ 1 0 0 2 3 1 0 1 0 | −1 1 0 1 0 0 −1 0 1 ] <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mi>R</mi></msub></mrow></annotation-xml></semantics>[/itex] 3 ↔ R 2 →[ 1 0 0 0 1 0 2 3 1 | −1 1 0 −1 0 1 1 0 0 ] <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>−2</mn><msub/></mrow></annotation-xml></semantics>[/itex] R 1 + R 3 = R 3 →[ 1 0 0 0 1 0 0 3 1 | −1 1 0 −1 0 1 3 −2 0 ] <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>−3</mn><msub/></mrow></annotation-xml></semantics>[/itex] R 2 + R 3 = R 3 →[ 1 0 0 0 1 0 0 0 1 | −1 1 0 −1 0 1 6 −2 −3 ] Thus, <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>A</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 =B=[ −1 1 0 −1 0 1 6 −2 −3 ] Analysis To prove that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>B</mi><mo>=</mo><msup/></mrow></annotation-xml></semantics>[/itex] A −1 , let’s multiply the two matrices together to see if the product equals the identity, if<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><msup/></mrow></annotation-xml></semantics>[/itex] A −1 =I and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] A −1 A=I. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow/></mtd></mtr><mtr columnalign="left"><mtd columnalign="left"><mrow/></mtd></mtr></mtable></mrow></mtd></mtr><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>A</mi><msup/></mrow></mtd></mtr></mtable></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] A −1 =[ 2 3 1 3 3 1 2 4 1 ] [ −1 1 0 −1 0 1 6 −2 −3 ] =[ 2(−1)+3(−1)+1(6) 2(1)+3(0)+1(−2) 2(0)+3(1)+1(−3)3(−1)+3(−1)+1(6) 3(1)+3(0)+1(−2) 3(0)+3(1)+1(−3) 2(−1)+4(−1)+1(6) 2(1)+4(0)+1(−2) 2(0)+4(1)+1(−3) ] =[ 1 00 0 1 0 0 0 1 ] <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow/></mtd></mtr><mtr columnalign="left"><mtd columnalign="left"><mrow/></mtd></mtr></mtable></mrow></mtd></mtr><mtr columnalign="left"><mtd columnalign="left"><mrow><msup><mi>A</mi></msup></mrow></mtd></mtr></mtable></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] −1 A=[ −1 1 0 −1 0 1 6 −2 −3 ] [ 2 3 1 3 3 1 2 4 1 ] =[ −1(2)+1(3)+0(2) −1(3)+1(3)+0(4) −1(1)+1(1)+0(1)−1(2)+0(3)+1(2) −1(3)+0(3)+1(4) −1(1)+0(1)+1(1) 6(2)+−2(3)+−3(2) 6(3)+−2(3)+−3(4) 6(1)+−2(1)+−3(1) ] =[ 10 0 0 1 0 0 0 1 ] Find the inverse of the<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>3</mn><mo>×</mo><mn>3</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]matrix. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2 −17 11 −1 11 −7 0 3 −2 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>A</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 =[ 1 1 2 2 4 −3 3 6 −5 ] # Solving a System of Linear Equations Using the Inverse of a Matrix Solving a system of linear equations using the inverse of a matrix requires the definition of two new matrices:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>X</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is the matrix representing the variables of the system, and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>B</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is the matrix representing the constants. Using matrix multiplication, we may define a system of equations with the same number of equations as variables as <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mi>X</mi><mo>=</mo><mi>B</mi></mrow></annotation-xml></semantics>[/itex] To solve a system of linear equations using an inverse matrix, let<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]be the coefficient matrix, let<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>X</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]be the variable matrix, and let<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>B</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]be the constant matrix. Thus, we want to solve a system<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mi>X</mi><mo>=</mo><mi>B</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]For example, look at the following system of equations. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable><mtr><mtd><mrow><msub><mi>a</mi></msub></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 1 x+ b 1 y= c 1 a 2 x+ b 2 y= c 2 From this system, the coefficient matrix is <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] a 1 b 1 a 2 b 2 ] The variable matrix is <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>X</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] x y ] And the constant matrix is <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>B</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] c 1 c 2 ] Then<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mi>X</mi><mo>=</mo><mi>B</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]looks like <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] a 1 b 1 a 2 b 2 ] [ x y ]=[ c 1 c 2 ] Recall the discussion earlier in this section regarding multiplying a real number by its inverse,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">(</mo><msup/></mrow></annotation-xml></semantics>[/itex] 2 −1 ) 2=( 1 2 ) 2=1. To solve a single linear equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mi>x</mi><mo>=</mo><mi>b</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]we would simply multiply both sides of the equation by the multiplicative inverse (reciprocal) of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Thus, <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable><mtr><mtd><mrow><mtext> </mtext><mi>a</mi><mi>x</mi><mo>=</mo><mi>b</mi></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] ( 1 a )ax=( 1 a )b ( a −1 )ax=( a −1 )b [( a −1 )a]x=( a −1 )b 1x=( a −1 )b x=( a −1 )b The only difference between a solving a linear equation and a system of equations written in matrix form is that finding the inverse of a matrix is more complicated, and matrix multiplication is a longer process. However, the goal is the same—to isolate the variable. We will investigate this idea in detail, but it is helpful to begin with a<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mtext> </mtext><mo>×</mo><mtext> </mtext><mn>2</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]system and then move on to a<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>3</mn><mtext> </mtext><mo>×</mo><mtext> </mtext><mn>3</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]system. Solving a System of Equations Using the Inverse of a Matrix Given a system of equations, write the coefficient matrix<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]the variable matrix<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>X</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and the constant matrix<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>B</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Then <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mi>X</mi><mo>=</mo><mi>B</mi></mrow></annotation-xml></semantics>[/itex] Multiply both sides by the inverse of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]to obtain the solution. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="right"><mtr columnalign="right"><mtd columnalign="right"><mrow><mrow><mo>(</mo></mrow></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] A −1 )AX=( A −1 )B [ ( A −1 )A ]X=( A −1 )B IX=( A −1 )B X=( A −1 )B If the coefficient matrix does not have an inverse, does that mean the system has no solution? No, if the coefficient matrix is not invertible, the system could be inconsistent and have no solution, or be dependent and have infinitely many solutions. Solving a 2 × 2 System Using the Inverse of a Matrix Solve the given system of equations using the inverse of a matrix. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="right"><mtr columnalign="right"><mtd columnalign="right"><mrow><mn>3</mn><mi>x</mi><mo>+</mo><mn>8</mn><mi>y</mi><mo>=</mo><mn>5</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 4x+11y=7 Write the system in terms of a coefficient matrix, a variable matrix, and a constant matrix. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mo>=</mo><mo stretchy="true">[</mo><mtable/></mrow></annotation-xml></semantics>[/itex] 3 8 4 11 ],X=[ x y ],B=[ 5 7 ] Then <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3 8 4 11 ] [ x y ]=[ 5 7 ] First, we need to calculate<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] A −1 . Using the formula to calculate the inverse of a 2 by 2 matrix, we have: <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><msup><mi>A</mi></msup></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] −1 = 1 ad−bc [ d −b −c a ] = 1 3(11)−8(4) [ 11 −8 −4 3 ] = 1 1 [ 11 −8 −4 3 ] So, <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>A</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 =[ 11 −8 −4 ​​ 3 ] Now we are ready to solve. Multiply both sides of the equation by<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] A −1 . <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mo stretchy="false">(</mo><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] A −1 )AX=( A −1 )B [ 11 −8 −4 3 ] [ 3 8 4 11 ] [ x y ]=[ 11 −8 −4 3 ] [ 5 7 ] [ 1 0 0 1] [ x y ]=[ 11(5)+(−8)7 −4(5)+3(7) ] [ x y ]=[ −1 1 ] The solution is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −1,1 ). Can we solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>X</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]by finding the product<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>B</mi><msup/></mrow></annotation-xml></semantics>[/itex] A −1 ? No, recall that matrix multiplication is not commutative, so<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] A −1 B≠B A −1 . Consider our steps for solving the matrix equation. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="right"><mtr columnalign="right"><mtd columnalign="right"><mrow><mrow><mo>(</mo></mrow></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] A −1 )AX=( A −1 )B [ ( A −1 )A ]X=( A −1 )B IX=( A −1 )B X=( A −1 )B Notice in the first step we multiplied both sides of the equation by<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] A −1 , but the<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] A −1 was to the left of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]on the left side and to the left of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>B</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]on the right side. Because matrix multiplication is not commutative, order matters. Solving a 3 × 3 System Using the Inverse of a Matrix Solve the following system using the inverse of a matrix. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="right"><mtr columnalign="right"><mtd columnalign="right"><mrow><mn>5</mn><mi>x</mi><mo>+</mo><mn>15</mn><mi>y</mi><mo>+</mo><mn>56</mn><mi>z</mi><mo>=</mo><mn>35</mn><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] −4x−11y−41z=−26 −x−3y−11z=−7 Write the equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mi>X</mi><mo>=</mo><mi>B</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 5 15 56 −4 −11 −41 −1 −3 −11 ] [ x y z ]=[ 35 −26 −7 ] First, we will find the inverse of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]by augmenting with the identity. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 5 15 56 −4 −11 −41 −1 −3 −11 | 1 0 0 0 1 0 0 0 1 ] Multiply row 1 by<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] 1 5 . <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 3 56 5 −4 −11 −41 −1 −3 −11 | 1 5 0 0 0 1 0 0 0 1 ] Multiply row 1 by 4 and add to row 2. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 3 56 5 0 1 19 5 −1 −3 −11 | 1 5 0 0 4 5 1 0 0 0 1 ] Add row 1 to row 3. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 3 56 5 0 1 19 5 0 0 1 5 | 1 5 0 0 4 5 1 0 1 5 0 1 ] Multiply row 2 by −3 and add to row 1. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 0 − 1 5 0 1 19 5 0 0 1 5 | − 11 5 −3 0 4 5 1 0 1 5 0 1 ] Multiply row 3 by 5. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 0 − 1 5 0 1 19 5 0 0 1 | − 11 5 −3 0 4 5 1 0 1 0 5 ] Multiply row 3 by<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] 1 5 and add to row 1. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 0 0 0 1 19 5 0 0 1 | −2 −3 1 4 5 1 0 1 0 5 ] Multiply row 3 by<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo>−</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 19 5 and add to row 2. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 0 0 0 1 0 0 0 1 | −2 −3 1 −3 1 −19 1 0 5 ] So, <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>A</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 =[ −2 −3 1 −3 1 −19 1 0 5 ] Multiply both sides of the equation by<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] A −1 . We want<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] A −1 AX= A −1 B: <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] −2 −3 1 −3 1 −19 1 0 5 ] [ 5 15 56 −4 −11 −41 −1 −3 −11 ] [ x y z ]=[ −2 −3 1 −3 1 −19 1 0 5 ] [ 35 −26 −7 ] Thus, <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>A</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 B=[ −70+78−7 −105−26+133 35+0−35 ]=[ 1 2 0 ] The solution is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1,2,0 ). Solve the system using the inverse of the coefficient matrix. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext> </mtext><mn>2</mn><mi>x</mi><mo>−</mo><mn>17</mn><mi>y</mi><mo>+</mo><mn>11</mn><mi>z</mi><mo>=</mo><mn>0</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] −x+11y−7z=8 3y−2z=−2 <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>X</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 4 38 58 ] Given a system of equations, solve with matrix inverses using a calculator. 1. Save the coefficient matrix and the constant matrix as matrix variables<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] A ] and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] B ]. 2. Enter the multiplication into the calculator, calling up each matrix variable as needed. 3. If the coefficient matrix is invertible, the calculator will present the solution matrix; if the coefficient matrix is not invertible, the calculator will present an error message. Using a Calculator to Solve a System of Equations with Matrix Inverses Solve the system of equations with matrix inverses using a calculator <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mn>2</mn><mi>x</mi><mo>+</mo><mn>3</mn><mi>y</mi><mo>+</mo><mi>z</mi><mo>=</mo><mn>32</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 3x+3y+z=−27 2x+4y+z=−2 On the matrix page of the calculator, enter the coefficient matrix as the matrix variable<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] A ], and enter the constant matrix as the matrix variable<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] B ]. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo stretchy="false">[</mo><mi>A</mi><mo stretchy="false">]</mo><mo>=</mo><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2 3 1 3 3 1 2 4 1 ], [B]=[ 32 −27 −2 ] On the home screen of the calculator, type in the multiplication to solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>X</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]calling up each matrix variable as needed. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mo stretchy="false">[</mo><mi>A</mi><mo stretchy="false">]</mo></mrow></msup></mrow></annotation-xml></semantics>[/itex] −1 ×[B] Evaluate the expression. <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] −59 −34 252 ] Access these online resources for additional instruction and practice with solving systems with inverses. # Key Equations  Identity matrix for a2 × 2[/itex]matrix I[/itex] 2 =[ 1 0 0 1 ] Identity matrix for a3 × 3[/itex]matrix I[/itex] 3 =[ 1 0 0 0 1 0 0 0 1 ] Multiplicative inverse of a2 × 2[/itex]matrix A[/itex] −1 = 1 ad−bc [ d −b −c a ], where ad−bc≠0 # Key Concepts • An identity matrix has the property<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mi>I</mi><mo>=</mo><mi>I</mi><mi>A</mi><mo>=</mo><mi>A</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]See [link]. • An invertible matrix has the property<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><msup/></mrow></annotation-xml></semantics>[/itex] A −1 = A −1 A=I. See [link]. • Use matrix multiplication and the identity to find the inverse of a<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mo>×</mo><mn>2</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]matrix. See [link]. • The multiplicative inverse can be found using a formula. See [link]. • Another method of finding the inverse is by augmenting with the identity. See [link]. • We can augment a<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>3</mn><mo>×</mo><mn>3</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]matrix with the identity on the right and use row operations to turn the original matrix into the identity, and the matrix on the right becomes the inverse. See [link]. • Write the system of equations as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mi>X</mi><mo>=</mo><mi>B</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and multiply both sides by the inverse of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>:</mo><msup/></mrow></annotation-xml></semantics>[/itex] A −1 AX= A −1 B. See [link] and[link]. • We can also use a calculator to solve a system of equations with matrix inverses. See [link]. # Section Exercises ## Verbal In a previous section, we showed that matrix multiplication is not commutative, that is,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mi>B</mi><mo>≠</mo><mi>B</mi><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]in most cases. Can you explain why matrix multiplication is commutative for matrix inverses, that is,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] A −1 A=A A −1 ? If<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] A −1 is the inverse of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]then<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><msup/></mrow></annotation-xml></semantics>[/itex] A −1 =I, the identity matrix. Since<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is also the inverse of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] A −1 , A −1 A=I. You can also check by proving this for a<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mo>×</mo><mn>2</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]matrix. Does every<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mo>×</mo><mn>2</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]matrix have an inverse? Explain why or why not. Explain what condition is necessary for an inverse to exist. Can you explain whether a<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mo>×</mo><mn>2</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]matrix with an entire row of zeros can have an inverse? No, because<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mi>d</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>b</mi><mi>c</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]are both 0, so<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mi>d</mi><mo>−</mo><mi>b</mi><mi>c</mi><mo>=</mo><mn>0</mn><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]which requires us to divide by 0 in the formula. Can a matrix with an entire column of zeros have an inverse? Explain why or why not. Can a matrix with zeros on the diagonal have an inverse? If so, find an example. If not, prove why not. For simplicity, assume a<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mo>×</mo><mn>2</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]matrix. Yes. Consider the matrix<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0 1 1 0 ]. The inverse is found with the following calculation:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] A −1 = 1 0(0)−1(1) [ 0 −1 −1 0 ]=[ 0 1 1 0 ]. ## Algebraic In the following exercises, show that matrix<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is the inverse of matrix<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>B</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 0 −1 1 ], B=[ 1 0 1 1 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 2 3 4 ], B=[ −2 1 3 2 − 1 2 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mi>B</mi><mo>=</mo><mi>B</mi><mi>A</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 0 0 1 ]=I <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 4 5 7 0 ], B=[ 0 1 7 1 5 − 4 35 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] −2 1 2 3 −1 ], B=[ −2 −1 −6 −4 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mi>B</mi><mo>=</mo><mi>B</mi><mi>A</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 0 0 1 ]=I <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 0 1 0 1 −1 0 1 1 ], B= 1 2 [ 2 1 −1 0 1 1 0 −1 1 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 2 3 4 0 2 1 6 9 ], B= 1 4 [ 6 0 −2 17 −3 −5 −12 2 4 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mi>B</mi><mo>=</mo><mi>B</mi><mi>A</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 0 0 0 1 0 0 0 1 ]=I <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mo>=</mo><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3 8 2 1 1 1 5 6 12 ], B= 1 36 [ −6 84 −6 7 −26 1 −1 −22 5 ] For the following exercises, find the multiplicative inverse of each matrix, if it exists. <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3 −2 1 9 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>1</mn></mfrac></mrow></annotation-xml></semantics>[/itex] 29 [ 9 2 −1 3 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] −2 2 3 1 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] −3 7 9 2 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>1</mn></mfrac></mrow></annotation-xml></semantics>[/itex] 69 [ −2 7 9 3 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] −4 −3 −5 8 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 1 2 2 ] There is no inverse <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0 1 1 0 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0.5 1.5 1 −0.5 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>4</mn></mfrac></mrow></annotation-xml></semantics>[/itex] 7 [ 0.5 1.5 1 −0.5 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 0 6 −2 1 7 3 0 2 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0 1 −3 4 1 0 1 0 5 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>1</mn></mfrac></mrow></annotation-xml></semantics>[/itex] 17 [ −5 5 −3 20 −3 12 1 −1 4 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 2 −1 −3 4 1 −2 −4 −5 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 9 −3 2 5 6 4 −2 7 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>1</mn></mfrac></mrow></annotation-xml></semantics>[/itex] 209 [ 47 −57 69 10 19 −12 −24 38 −13 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 −2 3 −4 8 −12 1 4 2 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 2 1 2 1 2 1 3 1 4 1 5 1 6 1 7 1 8 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 18 60 −168 −56 −140 448 40 80 −280 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 2 3 4 5 6 7 8 9 ] For the following exercises, solve the system using the inverse of a<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mtext> </mtext><mo>×</mo><mtext> </mtext><mn>2</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]matrix. <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext> </mtext><mn>5</mn><mi>x</mi><mo>−</mo><mn>6</mn><mi>y</mi><mo>=</mo><mo>−</mo><mn>61</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 4x+3y=−2 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −5,6 ) <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtable columnalign="left"><mtr><mtd><mn>8</mn><mi>x</mi><mo>+</mo><mn>4</mn><mi>y</mi><mo>=</mo><mn>−100</mn></mtd></mtr></mtable></annotation-xml></semantics>[/itex] 3x−4y=1 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext> </mtext><mn>3</mn><mi>x</mi><mn>−2</mn><mi>y</mi><mo>=</mo><mn>6</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] −x+5y=−2 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2,0 ) <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mn>5</mn><mi>x</mi><mn>−4</mn><mi>y</mi><mo>=</mo><mn>−5</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 4x+y=2.3 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mn>−3</mn><mi>x</mi><mn>−4</mn><mi>y</mi><mo>=</mo><mn>9</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 12x+4y=−6 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 3 ,− 5 2 ) <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mn>−2</mn><mi>x</mi><mo>+</mo><mn>3</mn><mi>y</mi><mo>=</mo><mfrac/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 3 10 −x+5y= 1 2 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mfrac/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 8 5 x− 4 5 y= 2 5 − 8 5 x+ 1 5 y= 7 10 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] − 2 3 ,− 11 6 ) <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtable columnalign="left"><mtr><mtd><mfrac><mn>1</mn></mfrac></mtd></mtr></mtable></annotation-xml></semantics>[/itex] 2 x+ 1 5 y=− 1 4 1 2 x− 3 5 y=− 9 4 For the following exercises, solve a system using the inverse of a<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>3</mn><mtext/><mo>×</mo><mtext/><mn>3</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] matrix. <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mn>3</mn><mi>x</mi><mn>−2</mn><mi>y</mi><mo>+</mo><mn>5</mn><mi>z</mi><mo>=</mo><mn>21</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 5x+4y=37 x−2y−5z=5 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 7, 1 2 , 1 5 ) <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext> </mtext><mn>4</mn><mi>x</mi><mo>+</mo><mn>4</mn><mi>y</mi><mo>+</mo><mn>4</mn><mi>z</mi><mo>=</mo><mn>40</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 2x−3y+4z=−12 −x+3y+4z=9 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext> </mtext><mn>6</mn><mi>x</mi><mo>−</mo><mn>5</mn><mi>y</mi><mo>−</mo><mi>z</mi><mo>=</mo><mn>31</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] −x+2y+z=−6 3x+3y+2z=13 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 5,0,−1 ) <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mn>6</mn><mi>x</mi><mn>−5</mn><mi>y</mi><mo>+</mo><mn>2</mn><mi>z</mi><mo>=</mo><mn>−4</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 2x+5y−z=12 2x+5y+z=12 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mn>4</mn><mi>x</mi><mn>−2</mn><mi>y</mi><mo>+</mo><mn>3</mn><mi>z</mi><mo>=</mo><mn>−12</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 2x+2y−9z=33 6y−4z=1 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>1</mn></mfrac></mrow></annotation-xml></semantics>[/itex] 34 ( −35,−97,−154 ) <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtable columnalign="left"><mtr><mtd><mfrac><mn>1</mn></mfrac></mtd></mtr></mtable></annotation-xml></semantics>[/itex] 10 x− 1 5 y+4z= −41 2 1 5 x−20y+ 2 5 z=−101 3 10 x+4y− 3 10 z=23 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mfrac/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 1 2 x− 1 5 y+ 1 5 z= 31 100 − 3 4 x− 1 4 y+ 1 2 z= 7 40 − 4 5 x− 1 2 y+ 3 2 z= 1 4 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>1</mn></mfrac></mrow></annotation-xml></semantics>[/itex] 690 ( 65,−1136,−229 ) <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mn>0.1</mn><mi>x</mi><mo>+</mo><mn>0.2</mn><mi>y</mi><mo>+</mo><mn>0.3</mn><mi>z</mi><mo>=</mo><mn>−1.4</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 0.1x−0.2y+0.3z=0.6 0.4y+0.9z=−2 ## Technology For the following exercises, use a calculator to solve the system of equations with matrix inverses. <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mn>2</mn><mi>x</mi><mo>−</mo><mi>y</mi><mo>=</mo><mn>−3</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] −x+2y=2.3 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] − 37 30 , 8 15 ) <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mo>−</mo><mfrac/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 1 2 x− 3 2 y=− 43 20 5 2 x+ 11 5 y= 31 4 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mn>12.3</mn><mi>x</mi><mn>−2</mn><mi>y</mi><mn>−2.5</mn><mi>z</mi><mo>=</mo><mn>2</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 36.9x+7y−7.5z=−7 8y−5z=−10 <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 10 123 ,−1, 2 5 ) <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mn>0.5</mn><mi>x</mi><mn>−3</mn><mi>y</mi><mo>+</mo><mn>6</mn><mi>z</mi><mo>=</mo><mn>−0.8</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 0.7x−2y=−0.06 0.5x+4y+5z=0 ## Extensions For the following exercises, find the inverse of the given matrix. <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 0 1 0 0 1 0 1 0 1 1 0 0 0 1 1 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>1</mn></mfrac></mrow></annotation-xml></semantics>[/itex] 2 [ 2 1 −1 −1 0 1 1 −1 0 −1 1 1 0 1 −1 1 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] −1 0 2 5 0 0 0 2 0 2 −1 0 1 −3 0 1 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 −2 3 0 0 1 0 2 1 4 −2 3 −5 0 1 1 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>1</mn></mfrac></mrow></annotation-xml></semantics>[/itex] 39 [ 3 2 1 −7 18 −53 32 10 24 −36 21 9 −9 46 −16 −5 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 2 0 2 3 0 2 1 0 0 0 0 3 0 1 0 2 0 0 1 0 0 1 2 0 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 1 1 1 1 1 ] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>[</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 −1 −1 −1 −1 −1 1 ] ## Real-World Applications For the following exercises, write a system of equations that represents the situation. Then, solve the system using the inverse of a matrix. 2,400 tickets were sold for a basketball game. If the prices for floor 1 and floor 2 were different, and the total amount of money brought in is64,000, how much was the price of each ticket?

In the previous exercise, if you were told there were 400 more tickets sold for floor 2 than floor 1, how much was the price of each ticket?

Infinite solutions.

A food drive collected two different types of canned goods, green beans and kidney beans. The total number of collected cans was 350 and the total weight of all donated food was 348 lb, 12 oz. If the green bean cans weigh 2 oz less than the kidney bean cans, how many of each can was donated?

Students were asked to bring their favorite fruit to class. 95% of the fruits consisted of banana, apple, and oranges. If oranges were twice as popular as bananas, and apples were 5% less popular than bananas, what are the percentages of each individual fruit?

50% oranges, 25% bananas, 20% apples

A sorority held a bake sale to raise money and sold brownies and chocolate chip cookies. They priced the brownies at $1 and the chocolate chip cookies at$0.75. They raised $700 and sold 850 items. How many brownies and how many cookies were sold? A clothing store needs to order new inventory. It has three different types of hats for sale: straw hats, beanies, and cowboy hats. The straw hat is priced at$13.99, the beanie at $7.99, and the cowboy hat at$14.49. If 100 hats were sold this past quarter, $1,119 was taken in by sales, and the amount of beanies sold was 10 more than cowboy hats, how many of each should the clothing store order to replace those already sold? 10 straw hats, 50 beanies, 40 cowboy hats Anna, Ashley, and Andrea weigh a combined 370 lb. If Andrea weighs 20 lb more than Ashley, and Anna weighs 1.5 times as much as Ashley, how much does each girl weigh? Three roommates shared a package of 12 ice cream bars, but no one remembers who ate how many. If Tom ate twice as many ice cream bars as Joe, and Albert ate three less than Tom, how many ice cream bars did each roommate eat? Tom ate 6, Joe ate 3, and Albert ate 3. A farmer constructed a chicken coop out of chicken wire, wood, and plywood. The chicken wire cost$2 per square foot, the wood $10 per square foot, and the plywood$5 per square foot. The farmer spent a total of \$51, and the total amount of materials used was<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>14</mn><msup/></mrow></annotation-xml></semantics>[/itex]  ft 2 . He used<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] 3 ft 2  more chicken wire than plywood. How much of each material in did the farmer use?

Jay has lemon, orange, and pomegranate trees in his backyard. An orange weighs 8 oz, a lemon 5 oz, and a pomegranate 11 oz. Jay picked 142 pieces of fruit weighing a total of 70 lb, 10 oz. He picked 15.5 times more oranges than pomegranates. How many of each fruit did Jay pick?

124 oranges, 10 lemons, 8 pomegranates

## Glossary

identity matrix
a square matrix containing ones down the main diagonal and zeros everywhere else; it acts as a 1 in matrix algebra
multiplicative inverse of a matrix
a matrix that, when multiplied by the original, equals the identity matrix