11.7: Solving Systems with Inverses
- Page ID
- 3102
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- 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 \(6%\). 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 \(a\) is \(a^{−1}\), so
\[aa^{−1}=a^{−1}a=\left(\dfrac{1}{a}\right)a=1 \label{eq0}\]
For example, consider the scalar multiplication situation
\[2^{−1}=\dfrac{1}{2} \nonumber\]
therefore from Equation \ref{eq0}
\[\left(\dfrac{1}{2}\right)2=1. \nonumber\]
The multiplicative inverse of a matrix is similar in concept, except that the product of matrix \(A\) and its inverse \(A^{−1}\) 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 \(I_n\) where \(n\) represents the dimension of the matrix. Equations \ref{eq1} and \ref{eq2} are the identity matrices for a \(2×2\) matrix and a \(3×3\) matrix, respectively:
\[I_2=\begin{bmatrix}1&0 \\ 0&1 \end{bmatrix} \label{eq1}\]
\[I_3=\begin{bmatrix}1&0&0 \\ 0&1&0 \\ 0&0&1\end{bmatrix} \label{eq2}\]
The identity matrix acts as a \(1\) in matrix algebra. For example,
\[AI=IA=A\nonumber\]
A matrix that has a multiplicative inverse has the properties
\[AA^{−1}=I\]
\[A^{−1}A=I\]
A matrix that has a multiplicative inverse is called an invertible matrix. Only a square matrix may have a multiplicative inverse, as the reversibility,
\[AA^{−1}=A^{−1}A=I\]
is a requirement. Not all square matrices have an inverse, but if \(A\) is invertible, then \(A^{−1}\) is unique. We will look at two methods for finding the inverse of a \(2 × 2\) matrix and a third method that can be used on both \(2 × 2\) and \(3 × 3\) matrices.
The identity matrix, \(I_n\), is a square matrix containing ones down the main diagonal and zeros everywhere else.
\[I_2=\begin{bmatrix}1&0 \nonumber \\ 0&1\end{bmatrix}\]
as for the \(2 × 2\) identity matrix
\[I_3=\begin{bmatrix}1&0&0 \nonumber \\ 0&1&0 \nonumber \\ 0&0&1\end{bmatrix}\]
as for the \(3 × 3\) identity matrix
If \(A\) is an \(n × n\) matrix and \(B\) is an \(n × n\) matrix such that \(AB=BA=I_n\), then \(B=A−1\), the multiplicative inverse of a matrix \(A\).
Given matrix \(A\), show that \(AI=IA=A\).
\[A=\begin{bmatrix}3&4 \nonumber \\ −2&5\end{bmatrix}\]
Solution
Use matrix multiplication to show that the product of \(A\) and the identity matric is equal to the product of the identity matrix and \(A\).
\[\begin{align*} AI&=\begin{bmatrix}3&4 \nonumber \\ −2&5\end{bmatrix}\begin{bmatrix}1&0 \nonumber \\ 0&1\end{bmatrix} \nonumber \\[4pt] &=\begin{bmatrix}3⋅1+4⋅0&3⋅0+4⋅1 \nonumber \\ −2⋅1+5⋅0&−2⋅0+5⋅1\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}3&4 \nonumber \\ −2&5\end{bmatrix} \end{align*}\]
\[\begin{align*} AI&=\begin{bmatrix}1&0 \nonumber \\ 0&1\end{bmatrix}\begin{bmatrix}3&4 \nonumber \\ −2&5\end{bmatrix} \nonumber \\[4pt] &=\begin{bmatrix}1⋅3+0⋅(−2)&1⋅4+0⋅5 \nonumber \\ 0⋅3+1⋅(−2)&0⋅4+1⋅5\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}3&4 \nonumber \\ −2&5\end{bmatrix} \end{align*}\]
- Given matrix \(A\) of order \(n × n\) and matrix \(B\) of order \(n × n\) multiply \(AB\).
- If \(AB=I\), then find the product \(BA\). If \(BA=I\), then \(B=A^{−1}\) and \(A=B^{−1}\).
Show that the given matrices are multiplicative inverses of each other.
\[A=\begin{bmatrix}1&5 \nonumber \\ −2&−9\end{bmatrix}\]
and
\[B=\begin{bmatrix}−9&−5 \nonumber \\ 2&1\end{bmatrix}\]
Solution
Multiply \(AB\) and \(BA\). If both products equal the identity, then the two matrices are inverses of each other.
\[\begin{align*} AB &= \begin{bmatrix}1&5 \nonumber \\ −2&−9\end{bmatrix}·\begin{bmatrix}−9&−5 \nonumber \\ 2&1\end{bmatrix} \nonumber \\[4pt] &=\begin{bmatrix}1(−9)+5(2)&1(−5)+5(1) \nonumber \\ −2(−9)−9(2)&−2(−5)−9(1)\end{bmatrix} \nonumber \\[4pt] &=\begin{bmatrix}1&0 \nonumber \\ 0&1\end{bmatrix} \end{align*}\]
and
\[\begin{align*} BA &= \begin{bmatrix}−9&−5 \nonumber \\ 2&1\end{bmatrix}·\begin{bmatrix}1&5 \nonumber \\ −2&−9\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}−9(1)−5(−2)&−9(5)−5(−9) \nonumber \\ 2(1)+1(−2)&2(−5)+1(−9)\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}1&0 \nonumber \\0&1\end{bmatrix} \end{align*}\]
\(A\) and \(B\) are inverses of each other.
Show that the following two matrices are inverses of each other.
\[A=\begin{bmatrix}1&4 \nonumber \\[4pt] −1&−3\end{bmatrix}\]
and
\[B=\begin{bmatrix}−3&−4 \nonumber \\[4pt] 1&1\end{bmatrix}\]
- Answer
-
\(\begin{align*} AB&=\begin{bmatrix}1&4 \nonumber \\[4pt] −1&−3\end{bmatrix}\begin{bmatrix}−3&−4 \nonumber \\[4pt] 1&1\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}1(−3)+4(1)&1(−4)+4(1) \nonumber \\[4pt] −1(−3)+−3(1)&−1(−4)+−3(1)\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}1&0 \nonumber \\[4pt] 0&1\end{bmatrix} \end{align*}\)
\(\begin{align*} BA&=\begin{bmatrix}−3&−4 \nonumber \\[4pt] 1&1\end{bmatrix}\begin{bmatrix}1&4 \nonumber \\[4pt] −1&−3\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}−3(1)+−4(−1)&−3(4)+−4(−3) \nonumber \\[4pt] 1(1)+1(−1)&1(4)+1(−3)\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}1&0 \nonumber \\[4pt] 0&1\end{bmatrix} \end{align*}\)
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.
Use matrix multiplication to find the inverse of the given matrix.
\[A=\begin{bmatrix}1&−2 \nonumber \\[4pt] 2&−3\end{bmatrix}\]
Solution
For this method, we multiply \(A\) by a matrix containing unknown constants and set it equal to the identity.
\(\begin{bmatrix}1&−2 \nonumber \\[4pt] 2&−3\end{bmatrix}\begin{bmatrix}a&b \nonumber \\[4pt] c&d\end{bmatrix}=\begin{bmatrix}1&0 \nonumber \\[4pt] 0&1\end{bmatrix}\)
Find the product of the two matrices on the left side of the equal sign.
\[\begin{bmatrix}1&−2 \nonumber \\[4pt] 2&−3\end{bmatrix}\begin{bmatrix}a&b \nonumber \\[4pt] c&d\end{bmatrix}=\begin{bmatrix}1a−2c&1b−2d \nonumber \\[4pt] 2a−3c&2b−3d\end{bmatrix}\]
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 \(0\).
\(1a−2c=1\space R_1\)
\(2a−3c=0\space R_2\)
Using row operations, multiply and add as follows: \((−2)R_1+R_2\rightarrow R_2\). Add the equations, and solve for \(c\).
\[ \begin{align*} 1a−2c &=1 \nonumber \\[4pt] 0+1c &=−2 \nonumber \\[4pt] c=−2 \nonumber \end{align*} \nonumber\]
Back-substitute to solve for \(a\).
\[ \begin{align*} a−2(−2)&=1 \nonumber \\[4pt] a+4&=1 \nonumber \\[4pt] a&=−3 \nonumber\end{align*} \nonumber\]
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.
\(1b−2d=0\space R_1\)
\(2b−3d=1\space R_2\)
Using row operations, multiply and add as follows: \((−2)R_1+R_2=R_2\). Add the two equations and solve for \(d\).
\[ \begin{align*} 1b−2d&=0 \nonumber \\[4pt] 0+1d&=1 \nonumber \\[4pt] d&=1 \nonumber \end{align*} \nonumber\]
Once more, back-substitute and solve for \(b\).
\[ \begin{align*} b−2(1)&=0 \nonumber \\[4pt] b&−2=0 \nonumber \\[4pt] b &=2 \nonumber \end{align*} \nonumber\]
\[A^{−1}=\begin{bmatrix}−3&2 \nonumber \\[4pt] −2&1\end{bmatrix}\]
Finding the Multiplicative Inverse by Augmenting with the Identity
Another way to find the multiplicative inverse is by augmenting with the identity. When matrix \(A\) is transformed into \(I\), the augmented matrix \(I\) transforms into \(A^{−1}\).
For example, given
\(A=\begin{bmatrix}2&1 \nonumber \\[4pt] 5&3\end{bmatrix}\)
augment \(A\) with the identity
\(\left[ \begin{array}{cc|cc} 2&1&1&0 \\ 5&3&0&1\end{array} \right]\)
Perform row operations with the goal of turning A into the identity.
- Switch row 1 and row 2.
\(\left[ \begin{array}{cc|cc} 5&3&0&1 \nonumber \\[4pt] 2&1&1&0\end{array} \right]\)
- Multiply row 2 by −2 and add to row 1.
\(\left[ \begin{array}{cc|cc} 1&1&-2&1 \nonumber \\[4pt] 2&1&1&0\end{array} \right]\)
- Multiply row 1 by −2 and add to row 2.
\(\left[ \begin{array}{cc|cc} 1&1&-2&1 \nonumber \\[4pt] 0&-1&5&-2\end{array} \right]\)
- Add row 2 to row 1.
\(\left[ \begin{array}{cc|cc} 1&0&3&-1 \nonumber \\[4pt] 0&-1&5&-2\end{array} \right]\)
- Multiply row 2 by−1. −1.
\(\left[ \begin{array}{cc|cc} 1&0&3&-1 \nonumber \\[4pt] 0&1&-5&2\end{array} \right]\)
The matrix we have found is \(A^{−1}\).
\(A^{−1}=\begin{bmatrix}3&−1 \nonumber \\[4pt] −5&2\end{bmatrix}\)
Finding the Multiplicative Inverse of \(2×2\) Matrices Using a Formula
When we need to find the multiplicative inverse of a \(2 × 2\) matrix, we can use a special formula instead of using matrix multiplication or augmenting with the identity.
If \(A\) is a \(2×2\) matrix, such as
\(A=\begin{bmatrix}a&b \nonumber \\[4pt] c&d\end{bmatrix}\)
the multiplicative inverse of \(A\) is given by the formula
\(A^{−1}=\dfrac{1}{ad−bc}\begin{bmatrix}d&−b \nonumber \\[4pt] −c&a\end{bmatrix}\)
where \(ad−bc≠0\). If \(ad−bc=0\), then \(A\) has no inverse.
Use the formula to find the multiplicative inverse of
\[A=\begin{bmatrix}1&−2 \nonumber \\[4pt] 2&−3\end{bmatrix}\]
Solution
We can check that our formula works by using one of the other methods to calculate the inverse. Let’s augment \(A\) with the identity.
\(\left[ \begin{array}{cc|cc} 1&-2&1&0 \nonumber \\[4pt] 2&-3&0&1\end{array}\right]\)
Perform row operations with the goal of turning \(A\) into the identity.
- Multiply row 1 by \(−2\) and add to row 2.
\(\left[ \begin{array}{cc|cc} 1&-2&1&0 \nonumber \\[4pt] 0&1&-2&1\end{array} \right]\)
- Multiply row 1 by \(2\) and add to row 1.
\(\left[ \begin{array}{cc|cc} 1&0&-3&2 \nonumber \\[4pt] 0&1&-2&1\end{array} \right]\)
So, we have verified our original solution.
\(A^{−1}=\begin{bmatrix}−3&2 \nonumber \\[4pt] −2&1\end{bmatrix}\)
Use the formula to find the inverse of matrix \(A\). Verify your answer by augmenting with the identity matrix.
\(A=\begin{bmatrix}1&−1 \nonumber \\[4pt] 2&3\end{bmatrix}\)
- Answer
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\(A^{−1}=\begin{bmatrix}\dfrac{3}{5}&\dfrac{1}{5} \nonumber \\[4pt] −\dfrac{2}{5}&\dfrac{1}{5}\end{bmatrix}\)
Find the inverse, if it exists, of the given matrix.
\(A=\begin{bmatrix}3&6 \nonumber \\[4pt] 1&2\end{bmatrix}\)
Solution
We will use the method of augmenting with the identity.
\(\left[ \begin{array}{cc|cc} 3&6&1&0 \nonumber \\[4pt] 1&3&0&1\end{array} \right]\)
- Switch row 1 and row 2.
\(\left[ \begin{array}{cc|cc} 1&3&0&1 \nonumber \\[4pt] 3&6&1&0\end{array} \right]\)
- Multiply row 1 by −3 and add it to row 2.
\(\left[ \begin{array}{cc|cc} 1&2&1&0 \nonumber \\[4pt] 0&0&-3&1\end{array} \right]\)
- 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 \(2×2\) matrix to find the inverse of a \(3×3\) matrix. Instead, we will augment the original matrix with the identity matrix and use row operations to obtain the inverse.
Given a \(3 × 3\) matrix
\[A=\begin{bmatrix}2&3&1 \nonumber \\[4pt] 3&3&1 \nonumber \\[4pt] 2&4&1\end{bmatrix}\]
augment \(A\) with the identity matrix
\[\begin{array}{c|c}A&I\end{array}=\left[ \begin{array}{ccc|ccc}2&3&1&1&0&0 \nonumber \\[4pt] 3&3&1&0&1&0 \nonumber \\[4pt] 2&4&1&0&0&1\end{array} \right]\]
To begin, we write the augmented matrix with the identity on the right and \(A\) 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.
- Write the original matrix augmented with the identity matrix on the right.
- Use elementary row operations so that the identity appears on the left.
- What is obtained on the right is the inverse of the original matrix.
- Use matrix multiplication to show that \(AA^{−1}=I\) and \(A^{−1}A=I\).
Given the \(3 × 3\) matrix \(A\), find the inverse.
\(A=\begin{bmatrix}2&3&1 \nonumber \\[4pt] 3&3&1 \nonumber \\[4pt] 2&4&1\end{bmatrix}\)
Solution
Augment \(A\) with the identity matrix, and then begin row operations until the identity matrix replaces \(A\). The matrix on the right will be the inverse of \(A\).
\(\left[ \begin{array}{ccc|ccc}2&3&1&1&0&0 \nonumber \\[4pt] 3&3&1&0&1&0 \nonumber \\[4pt] 2&4&1&0&0&1 \end{array} \right] \xrightarrow{Interchange\space R_2\space and\space R_1} \left[ \begin{array}{ccc|ccc}3&3&1&0&1&0 \nonumber \\[4pt] 2&3&1&1&0&0 \nonumber \\[4pt] 2&4&1&0&0&1 \end{array} \right]\)
\(−R_2+R_1=R_1\rightarrow \left[ \begin{array}{ccc|ccc}1&0&0&-1&1&0 \nonumber \\[4pt] 2&3&1&1&0&0 \nonumber \\[4pt] 2&4&1&0&0&1\end{array} \right]\)
\(−R_2+R_3=R_3\rightarrow \left[ \begin{array}{ccc|ccc}1&0&0&-1&1&0 \nonumber \\[4pt] 2&3&1&1&0&0 \nonumber \\[4pt] 0&1&0&-1&0&1\end{array} \right]\)
\(R_2\leftrightarrow R_3\rightarrow \left[ \begin{array}{ccc|ccc}1&0&0&-1&1&0 \nonumber \\[4pt] 0&1&0&-1&0&1 \nonumber \\[4pt] 2&3&1&1&0&0\end{array} \right]\)
\(−2R_1+R_3=R_3\rightarrow \left[ \begin{array}{ccc|ccc}1&0&0&-1&1&0 \nonumber \\[4pt] 0&1&0&-1&0&1 \nonumber \\[4pt] 0&3&1&3&-2&0\end{array} \right]\)
\(−3R_2+R_3=R_3\rightarrow \left[ \begin{array}{ccc|ccc}1&0&0&-1&1&0 \nonumber \\[4pt] 0&1&0&-1&0&1 \nonumber \\[4pt] 0&0&1&6&-2&-3\end{array} \right]\)
Thus,
\(A^{−1}=B=\begin{bmatrix}−1&1&0 \nonumber \\[4pt] −1&0&1 \nonumber \\[4pt] 6&−2&−3\end{bmatrix}\)
Analysis
To prove that \(B=A^{−1}\), let’s multiply the two matrices together to see if the product equals the identity, if \(AA^{−1}=I\) and \(A^{−1}A=I\).
\[\begin{align*} AA^{−1} & =\begin{bmatrix}2&3&1 \nonumber \\[4pt] 3&3&1 \nonumber \\[4pt] 2&4&1\end{bmatrix}\begin{bmatrix}−1&1&0 \nonumber \\[4pt] −1&0&1 \nonumber \\[4pt] 6&−2&−3\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}2(−1)+3(−1)+1(6)&2(1)+3(0)+1(−2)&2(0)+3(1)+1(−3) \nonumber \\[4pt] 3(−1)+3(−1)+1(6)& 3(1)+3(0)+1(−2)& 3(0)+3(1)+1(−3) \nonumber \\[4pt] 2(−1)+4(−1)+1(6)& 2(1)+4(0)+1(−2)& 2(0)+4(1)+1(−3)\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}1&0&0&0&1&0 \nonumber \\[4pt] 0&0&1\end{bmatrix} \nonumber \\[4pt] A^{−1}A &= \begin{bmatrix}−1&1&0 \nonumber \\[4pt] −1&0&1 \nonumber \\[4pt] 6&−2&−3\end{bmatrix}\begin{bmatrix}&2&31 \nonumber \\[4pt] 3&3&1 \nonumber \\[4pt] 2&4&1\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}−1(2)+1(3)+0(2)& −1(3)+1(3)+0(4)& −1(1)+1(1)+0(1) \nonumber \\[4pt] −1(2)+0(3)+1(2)& −1(3)+0(3)+1(4)& −1(1)+0(1)+1(1) \nonumber \\[4pt] 6(2)+−2(3)+−3(2)& 6(3)+−2(3)+−3(4)& 6(1)+−2(1)+−3(1)\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}1&0&0 \nonumber \\[4pt] 0&1&0 \nonumber \\[4pt] 0&0&1\end{bmatrix} \end{align*}\]
Find the inverse of the \(3×3\) matrix.
\(A=\begin{bmatrix}2&−17&11 \nonumber \\[4pt] −1&11&−7 \nonumber \\[4pt] 0&3&−2\end{bmatrix}\)
- Answer
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\(A^{−1}=\begin{bmatrix}1&1&2 \nonumber \\[4pt] 2&4&−3 \nonumber \\[4pt] 3&6&−5\end{bmatrix}\)
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: \(X\) is the matrix representing the variables of the system, and \(B\) 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
\(AX=B\)
To solve a system of linear equations using an inverse matrix, let \(A\) be the coefficient matrix, let \(X\) be the variable matrix, and let \(B\) be the constant matrix. Thus, we want to solve a system \(AX=B\). For example, look at the following system of equations.
\(a_1x+b_1y=c_1\)
\(a_2x+b_2y=c_2\)
From this system, the coefficient matrix is
\(A=\begin{bmatrix}a_1&b_1 \nonumber \\[4pt] a_2&b_2\end{bmatrix}\)
The variable matrix is
\(X=\begin{bmatrix}x \nonumber \\[4pt] y\end{bmatrix}\)
And the constant matrix is
\(B=\begin{bmatrix}c_1 \nonumber \\[4pt] c_2\end{bmatrix}\)
Then \(AX=B\) looks like
\(\begin{bmatrix}a_1&b_1 \nonumber \\[4pt] a_2&b_2\end{bmatrix}\begin{bmatrix}x \nonumber \\[4pt] y\end{bmatrix}=\begin{bmatrix}c_1 \nonumber \\[4pt] c_2\end{bmatrix}\)
Recall the discussion earlier in this section regarding multiplying a real number by its inverse, \((2^{−1}) 2=\left(\dfrac{1}{2}\right) 2=1\). To solve a single linear equation \(ax=b\) for \(x\), we would simply multiply both sides of the equation by the multiplicative inverse (reciprocal) of \(a\). Thus,
\[\begin{align*} ax&= b\\ \left(\dfrac{1}{a}\right)ax&= \left(\dfrac{1}{a}\right)b\\ \left(a^{-1}\right)ax&= \left(a^{-1}\right)b\\ \left[\left(a^{-1}\right)a\right]x&= \left(a^{-1}\right)b\\ 1x&= \left(a^{-1}\right)b\\ x&= \left(a^{-1}\right)b \end{align*}\]
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 \(2 × 2\) system and then move on to a \(3 × 3\) system.
Given a system of equations, write the coefficient matrix \(A\), the variable matrix \(X\), and the constant matrix \(B\). Then
\(AX=B\)
Multiply both sides by the inverse of \(A\) to obtain the solution.
\[\begin{align*} \left(A^{-1}\right)AX&= \left(A^{-1}\right)B\\ \left[\left(A^{-1}\right)A \right]X&= \left(A^{-1}\right)B\\ IX&= \left(A^{-1}\right)B\\ X&= \left(A^{-1}\right)B \end{align*}\]
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.
Solve the given system of equations using the inverse of a matrix.
\[\begin{align*} 3x+8y&= 5\\ 4x+11y&= 7 \end{align*}\]
Solution
Write the system in terms of a coefficient matrix, a variable matrix, and a constant matrix.
\(A=\begin{bmatrix}3&8 \nonumber \\[4pt] 4&11\end{bmatrix}\), \(X=\begin{bmatrix}x \nonumber \\[4pt] y\end{bmatrix}\), \(B=\begin{bmatrix}5 \nonumber \\[4pt] 7\end{bmatrix}\)
Then
\(\begin{bmatrix}3&8 \nonumber \\[4pt] 4&11\end{bmatrix}\begin{bmatrix}x \nonumber \\[4pt] y\end{bmatrix}=\begin{bmatrix}5 \nonumber \\[4pt] 7\end{bmatrix}\)
First, we need to calculate \(A^{−1}\). Using the formula to calculate the inverse of a \(2\) by \(2\) matrix, we have:
\[\begin{align*} A^{−1} &= \dfrac{1}{ad−bc}\begin{bmatrix}d&−b \nonumber \\[4pt] −c&a\end{bmatrix} \\ &= \dfrac{1}{3(11)−8(4)}\begin{bmatrix}11&−8 \nonumber \\[4pt] −4&3\end{bmatrix} \\ &=\dfrac{1}{1}\begin{bmatrix}11&−8 \nonumber \\[4pt] −4&3\end{bmatrix} \end{align*}\]
So,
\(A^{−1}=\begin{bmatrix}11&−8 \nonumber \\[4pt] −4 &3\end{bmatrix}\)
Now we are ready to solve. Multiply both sides of the equation by \(A^{−1}\).
\[\begin{align*} \left(A^{−1}\right)AX&=\left(A^{−1}\right)B \\[4pt] \begin{bmatrix}11&−8 \nonumber \\[4pt] −4&3\end{bmatrix}\begin{bmatrix}3&8 \nonumber \\[4pt] 4&11\end{bmatrix}\begin{bmatrix}x \nonumber \\[4pt] y\end{bmatrix}&=\begin{bmatrix}11&−8 \nonumber \\[4pt] −4&3\end{bmatrix}\begin{bmatrix}5 \nonumber \\[4pt] 7\end{bmatrix} \\[4pt] \begin{bmatrix}1&0 \nonumber \\[4pt] 0&1\end{bmatrix}\begin{bmatrix}x \nonumber \\[4pt] y\end{bmatrix}&=\begin{bmatrix}11(5)+(−8)7 \nonumber \\[4pt] −4(5)+3(7)\end{bmatrix} \\[4pt] \begin{bmatrix}x \nonumber \\[4pt] y\end{bmatrix}&=\begin{bmatrix}−1 \nonumber \\[4pt] 1\end{bmatrix} \end{align*}\]
The solution is \((−1,1)\).
No, recall that matrix multiplication is not commutative, so \(A^{−1}B≠BA^{−1}\). Consider our steps for solving the matrix equation.
\[\begin{align*} \left(A^{-1}\right)AX&= \left(A^{-1}\right)B\\ \left[ \left(A^{-1}\right)A \right]X&= \left(A^{-1}\right)B\\ IX&= \left(A^{-1}\right)B\\ X&= \left(A^{-1}\right)B \end{align*}\]
Notice in the first step we multiplied both sides of the equation by \(A^{−1}\), but the \(A^{−1}\) was to the left of \(A\) on the left side and to the left of \(B\) on the right side. Because matrix multiplication is not commutative, order matters.
Solve the following system using the inverse of a matrix.
\[\begin{align*} 5x+15y+56z&= 35\\ -4x-11y-41z&= -26\\ -x-3y-11z&= -7 \end{align*}\]
Solution
Write the equation \(AX=B\).
\(\begin{bmatrix}5&15&56 \nonumber \\[4pt] −4&−11&−41 \nonumber \\[4pt] −1&−3&−11\end{bmatrix}\begin{bmatrix}x \nonumber \\[4pt] y \nonumber \\[4pt] z\end{bmatrix}=\begin{bmatrix}35 \nonumber \\[4pt] −26 \nonumber \\[4pt] −7\end{bmatrix}\)
First, we will find the inverse of \(A\) by augmenting with the identity.
\(\left[ \begin{array}{ccc|ccc}5&15&56&1&0&0 \nonumber \\[4pt] −4&−11&−41&0&1&0 \nonumber \\[4pt] −1&−3&−11&0&0&1\end{array} \right]\)
Multiply row 1 by \(\dfrac{1}{5}\).
\(\left[ \begin{array}{ccc|ccc}1&3&\dfrac{56}{5}&\dfrac{1}{5}&0&0 \nonumber \\[4pt] −4&−11&−41&0&1&0 \nonumber \\[4pt] −1&−3&−11&0&0&1\end{array} \right]\)
Multiply row 1 by \(4\) and add to row 2.
\(\left[ \begin{array}{ccc|ccc}1&3&\dfrac{56}{5}&\dfrac{1}{5}&0&0 \nonumber \\[4pt] 0&1&\dfrac{19}{5}&\dfrac{4}{5}&1&0 \nonumber \\[4pt] −1&−3&−11&0&0&1\end{array} \right]\)
Add row 1 to row 3.
\(\left[ \begin{array}{ccc|ccc}1&3&\dfrac{56}{5}&\dfrac{1}{5}&0&0 \nonumber \\[4pt] 0&1&\dfrac{19}{5}&\dfrac{4}{5}&1&0 \nonumber \\[4pt] 0&0&\dfrac{1}{5}&\dfrac{1}{5}&0&1\end{array} \right]\)
Multiply row 2 by \(−3\) and add to row 1.
\(\left[ \begin{array}{ccc|ccc}1&0&-\dfrac{1}{5}&-\dfrac{11}{5}&-3&0 \nonumber \\[4pt] 0&1&\dfrac{19}{5}&\dfrac{4}{5}&1&0 \nonumber \\[4pt] 0&0&\dfrac{1}{5}&\dfrac{1}{5}&0&1\end{array} \right]\)
Multiply row 3 by \(5\).
\(\left[ \begin{array}{ccc|ccc}1&0&-\dfrac{1}{5}&-\dfrac{11}{5}&-3&0 \nonumber \\[4pt] 0&1&\dfrac{19}{5}&\dfrac{4}{5}&1&0 \nonumber \\[4pt] 0&0&1&1&0&5\end{array} \right]\)
Multiply row 3 by \(\dfrac{1}{5}\) and add to row 1.
\(\left[ \begin{array}{ccc|ccc}1&0&0&-2&-3&1 \nonumber \\[4pt] 0&1&\dfrac{19}{5}&\dfrac{4}{5}&1&0 \nonumber \\[4pt] 0&0&1&1&0&5\end{array} \right]\)
Multiply row 3 by \(−\dfrac{19}{5}\) and add to row 2.
\(\left[ \begin{array}{ccc|ccc}1&0&0&-2&-3&1 \nonumber \\[4pt] 0&1&0&-3&1&-19 \nonumber \\[4pt] 0&0&1&1&0&5\end{array} \right]\)
So,
\(A^{−1}=\begin{bmatrix}−2&−3&1 \nonumber \\[4pt] −3&1&−19 \nonumber \\[4pt] 1&0&5\end{bmatrix}\)
Multiply both sides of the equation by \(A^{−1}\). We want \(A^{−1}AX=A^{−1}B\):
\(\begin{bmatrix}−2&−3&1 \nonumber \\[4pt] −3&1&−19 \nonumber \\[4pt] 1&0&5\end{bmatrix}\begin{bmatrix}5&15&56 \nonumber \\[4pt] −4&−11&−41 \nonumber \\[4pt] −1&−3&−11\end{bmatrix}\begin{bmatrix}x \nonumber \\[4pt] y \nonumber \\[4pt] z\end{bmatrix}=\begin{bmatrix}−2&−3&1 \nonumber \\[4pt] −3&1&−19 \nonumber \\[4pt] 1&0&5\end{bmatrix}\begin{bmatrix}35 \nonumber \\[4pt] −26 \nonumber \\[4pt] −7\end{bmatrix}\)
Thus,
\(A^{−1}B=\begin{bmatrix}−70+78−7 \nonumber \\[4pt] −105−26+133 \nonumber \\[4pt] 35+0−35\end{bmatrix}=\begin{bmatrix}1 \nonumber \\[4pt] 2 \nonumber \\[4pt] 0\end{bmatrix}\)
The solution is \((1,2,0)\).
Solve the system using the inverse of the coefficient matrix.
\[\begin{align*} 2x-17y+11z&= 0\\ -x+11y-7z&= 8\\ 3y-2z&= -2 \end{align*}\]
- Answer
-
\(X=\begin{bmatrix}4 \nonumber \\[4pt] 38 \nonumber \\[4pt] 58\end{bmatrix}\)
- Save the coefficient matrix and the constant matrix as matrix variables \([ A ]\) and \([ B ]\).
- Enter the multiplication into the calculator, calling up each matrix variable as needed.
- 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.
Solve the system of equations with matrix inverses using a calculator
\[\begin{align*} 2x+3y+z&= 32\\ 3x+3y+z&= -27\\ 2x+4y+z&= -2 \end{align*}\]
Solution
On the matrix page of the calculator, enter the coefficient matrix as the matrix variable \([ A ]\), and enter the constant matrix as the matrix variable \([ B ]\).
\([A]=\begin{bmatrix}2&3&1 \nonumber \\[4pt] 3&3&1 \nonumber \\[4pt] 2&4&1\end{bmatrix}\), \([B]=\begin{bmatrix}32 \nonumber \\[4pt] −27 \nonumber \\[4pt] −2\end{bmatrix}\)
On the home screen of the calculator, type in the multiplication to solve for \(X\), calling up each matrix variable as needed.
\([A]^{−1}×[B]\)
Evaluate the expression.
\(\begin{bmatrix}−59 \nonumber \\[4pt] −34 \nonumber \\[4pt] 252\end{bmatrix}\)
Access these online resources for additional instruction and practice with solving systems with inverses.
Key Equations
| Identity matrix for a \(2 × 2\) matrix | \(I_2=\begin{bmatrix}1&0 \nonumber \\[4pt] 0&1\end{bmatrix}\) |
| Identity matrix for a \(3 × 3\) matrix | \(I_3=\begin{bmatrix}1&0&0 \nonumber \\[4pt] 0&1&0 \nonumber \\[4pt] 0&0&1\end{bmatrix}\) |
| Multiplicative inverse of a \(2 × 2\) matrix | \(A^{−1}=\dfrac{1}{ad−bc}\begin{bmatrix}d&−b \nonumber \\[4pt] −c&a\end{bmatrix}\), where \(ad−bc≠0\) |
Key Concepts
- An identity matrix has the property \(AI=IA=A\). See Example \(\PageIndex{1}\).
- An invertible matrix has the property \(AA^{−1}=A^{−1}A=I\). See Example \(\PageIndex{2}\).
- Use matrix multiplication and the identity to find the inverse of a \(2×2\) matrix. See Example \(\PageIndex{3}\).
- The multiplicative inverse can be found using a formula. See Example \(\PageIndex{4}\).
- Another method of finding the inverse is by augmenting with the identity. See Example \(\PageIndex{5}\).
- We can augment a \(3×3\) 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 Example \(\PageIndex{6}\).
- Write the system of equations as \(AX=B\), and multiply both sides by the inverse of \(A\): \(A^{−1}AX=A^{−1}B\). See Example \(\PageIndex{7}\) and Example \(\PageIndex{8}\).
- We can also use a calculator to solve a system of equations with matrix inverses. See Example \(\PageIndex{9}\).