3.4: Solving Systems with Inverses

Learning Objectives

• Find the inverse of a matrix.
• Solve a system of linear equations using an inverse matrix

Prerequisite Skills

Before you get started, take this prerequisite quiz.

1. Use your calculator to find \(A \cdot B\) if \(A=\begin{bmatrix} 2 & 1 \\5 & 3 \end{bmatrix} \), \(B=\begin{bmatrix} 3 & -1\\-5 & 2 \end{bmatrix} \).

Click here to check your answer

\(\begin{bmatrix} 1 & 0\\0 & 1 \end{bmatrix} \)

If you missed any part of this problem, review Section 3.2. (Note that this will open in a new window.)

 

2. Simplify \(5 \cdot \dfrac{1}{5}\) without using a calculator.

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\(1\)

If you missed this problem, review here. (Note that this will open a different textbook in a new window.)

 

3. Simplify \(\dfrac{3}{2} \cdot \dfrac{2}{3}\) without using a calculator.

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\(1\)

If you missed this problem, review here. (Note that this will open a different textbook in a new window.)

 

4. Solve \(-\dfrac{3}{2}x=\dfrac{5}{8}\) by isolating the \(x\) first.  (Do not multiply both sides by the LCD.)

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\(x=-\dfrac{5}{12}\) 

If you missed this problem, review here. (Note that this will open a different textbook in a new window.)

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.

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. The identity matrices for a \(2×2\) matrix and a \(3×3\) matrix are, respectively:

\[I_2=\begin{bmatrix}1&0 \\ 0&1 \end{bmatrix} \nonumber \]

\[I_3=\begin{bmatrix}1&0&0 \\ 0&1&0 \\ 0&0&1\end{bmatrix} \nonumber \]

The identity matrix acts as a \(1\) in matrix algebra. For example,

\[AI=IA=A\nonumber\]

Definition: THE IDENTITY MATRIX

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

Example \(\PageIndex{1}\): Showing That the Identity Matrix Acts as a 1

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 \(A\).

\[\begin{align*} AI&=\begin{bmatrix}3&4 \nonumber \\ −2&5\end{bmatrix}\begin{bmatrix}1&0 \nonumber \\ 0&1\end{bmatrix} \nonumber \\[4pt] &=\begin{bmatrix}3⋅1+4⋅0&3⋅0+4⋅1 \nonumber \\ −2⋅1+5⋅0&−2⋅0+5⋅1\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}3&4 \nonumber \\ −2&5\end{bmatrix} \end{align*}\]

\[\begin{align*} AI&=\begin{bmatrix}1&0 \nonumber \\ 0&1\end{bmatrix}\begin{bmatrix}3&4 \nonumber \\ −2&5\end{bmatrix} \nonumber \\[4pt] &=\begin{bmatrix}1⋅3+0⋅(−2)&1⋅4+0⋅5 \nonumber \\ 0⋅3+1⋅(−2)&0⋅4+1⋅5\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}3&4 \nonumber \\ −2&5\end{bmatrix} \end{align*}\]

We know that the multiplicative inverse of a real number is that which multiplies to create the multiplicative identity, which is 1.  Remember that \(a^{-1}=\dfrac{1}{a}\) and that \((a)(\dfrac{1}{a})\)=1.  Since \((a)(a^{−1})\)=1 and \((a^{−1})(a)\)=1, \(a\) and \(a^{−1}\) are multiplicative inverses of each other.

\[aa^{−1}=a^{−1}a=\left(\dfrac{1}{a}\right)a=1 \nonumber \]

For example, consider the example

\[2^{−1}=\dfrac{1}{2} \nonumber\]

As a multiplication problem,

\[2\left(\dfrac{1}{2}\right)=\left(\dfrac{1}{2}\right)2=1 \nonumber\]

therefore 2 and \(\left(\dfrac{1}{2}\right)\) are multiplicative inverses.

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.

A matrix that has a multiplicative inverse has the properties

\[AA^{−1}=I\nonumber \]

\[A^{−1}A=I\nonumber \]

A matrix that has a multiplicative inverse is called an invertible matrix. Only a square matrix may have a multiplicative inverse, as the reversibility,

\[AA^{−1}=A^{−1}A=I\nonumber \]

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.

 

Definition: THE MULTIPLICATIVE INVERSE

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 \(A\).

How to: Given two matrices, show that one is the multiplicative inverse of the other

• 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 \(A=B^{−1}\).

Example \(\PageIndex{2}\): Showing That Matrix \(A\) Is the Multiplicative Inverse of Matrix \(B\)

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

\[\begin{align*} BA &= \begin{bmatrix}−9&−5 \nonumber \\ 2&1\end{bmatrix}·\begin{bmatrix}1&5 \nonumber \\ −2&−9\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}−9(1)−5(−2)&−9(5)−5(−9) \nonumber \\ 2(1)+1(−2)&2(−5)+1(−9)\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}1&0 \nonumber \\0&1\end{bmatrix} \end{align*}\]

Since both \(AB\) and \(BA\) equal the identity matrix, \(A\) and \(B\) are inverses of each other.

Exercise \(\PageIndex{1}\)

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 by Augmenting with the Identity

We can now determine whether two matrices are inverses, but how would we find the inverse of a given matrix? One 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.

1. Switch row 1 and row 2.

   \(\left[ \begin{array}{cc|cc} 5&3&0&1 \nonumber \\[4pt] 2&1&1&0\end{array} \right]\)

2. Multiply row 2 by −2 and add to row 1.

   \(\left[ \begin{array}{cc|cc} 1&1&-2&1 \nonumber \\[4pt] 2&1&1&0\end{array} \right]\)

3. Multiply row 1 by −2 and add to row 2.

   \(\left[ \begin{array}{cc|cc} 1&1&-2&1 \nonumber \\[4pt] 0&-1&5&-2\end{array} \right]\)

4. Add row 2 to row 1.

   \(\left[ \begin{array}{cc|cc} 1&0&3&-1 \nonumber \\[4pt] 0&-1&5&-2\end{array} \right]\)

5. 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.

Example \(\PageIndex{4}\): Using the Formula to Find the Multiplicative Inverse of Matrix \(A\)

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.

1. Multiply row 1 by \(−2\) and add to row 2.

   \(\left[ \begin{array}{cc|cc} 1&-2&1&0 \nonumber \\[4pt] 0&1&-2&1\end{array} \right]\)

2. 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.

\(A^{−1}=\begin{bmatrix}−3&2 \nonumber \\[4pt] −2&1\end{bmatrix}\)

Exercise \(\PageIndex{2}\)

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

\(A^{−1}=\begin{bmatrix}\dfrac{3}{5}&\dfrac{1}{5} \nonumber \\[4pt] −\dfrac{2}{5}&\dfrac{1}{5}\end{bmatrix}\)

Finding the Multiplicative Inverse Using a Calculator

For the purposes of this course, we can use the calculator to find a matrix inverse.

How to: Find the Matrix Inverse Using a Calculator

1. Enter the original matrix into the calculator using the MATRIX EDIT screen.
2. Press 2^nd MATRIX and use down arrow key to select the correct matrix name from the NAMES menu.
3. Press the x^-1 key (for inverse). Press ENTER.
4. The screen shows the command such as [A]^-1. Press ENTER.
5. If the matrix has an inverse, the calculator will display it.  If the matrix does not have an inverse, the calculator will display an error.

 

Exercise \(\PageIndex{3}\)

Use a calculator to 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

\(A^{−1}=\begin{bmatrix}1&1&2 \nonumber \\[4pt] 2&4&−3 \nonumber \\[4pt] 3&6&−5\end{bmatrix}\)

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 new matrices: let \(A\) be the coefficient matrix, let \(X\) be the variable matrix, and let \(B\) be the constant matrix.  Provided that the system of equations has the same number of equations as it has variables, matrix multiplication allows us to define the system of equations as a new matrix equation,

\(AX=B\)

Then we will multiply both sides of the equation by the inverse of \(A\) to obtain the solution for \(X\).

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.

SOLVING A SYSTEM OF EQUATIONS USING THE INVERSE OF A MATRIX

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.

\[\begin{align*} \left(A^{-1}\right)AX&= \left(A^{-1}\right)B\\ \left[\left(A^{-1}\right)A \right]X&= \left(A^{-1}\right)B\\ IX&= \left(A^{-1}\right)B\\ X&= \left(A^{-1}\right)B \end{align*}\]

Q&A: 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.

Example \(\PageIndex{7}\): Solving a \(2 × 2\) System Using the Inverse of a Matrix

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

\(\begin{bmatrix}3&8 \nonumber \\[4pt] 4&11\end{bmatrix}\begin{bmatrix}x \nonumber \\[4pt] y\end{bmatrix}=\begin{bmatrix}5 \nonumber \\[4pt] 7\end{bmatrix}\)

or

\([A][X]=[B]\)

First, we need to calculate \(A^{−1}\).  Entering both [A] and [B] into the calculator and calling \([A]^{-1}\) finds that 

\(A^{−1}=\begin{bmatrix}11&−8 \nonumber \\[4pt] −4 &3\end{bmatrix}\)

Now we are ready to solve. Multiply both sides of the equation by \(A^{−1}\).

\[\begin{align*} \left(A^{-1}\right)AX&= \left(A^{-1}\right)B\\  X&= \left(A^{-1}\right)B \end{align*}\]

Entering \([A]^{-1}[B]\) into the calculator gives the product of

\begin{bmatrix}−1 \nonumber \\[4pt] 1\end{bmatrix} 

The solution is \((−1,1)\).

Q&A: Can we solve for \(X\) by finding the product \(BA^{−1}\)?

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.

Exercise \(\PageIndex{4}\)

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

\(X=\begin{bmatrix}4 \nonumber \\[4pt] 38 \nonumber \\[4pt] 58\end{bmatrix}\)

Media

Access these online resources for additional instruction and practice with solving systems with inverses.

• The Identity Matrix
• Determining Inverse Matrices
• Using a Matrix Equation to Solve a System of Equations

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 \(\PageIndex{9}\).

Contributors and Attributions

• Jay Abramson (Arizona State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at https://openstax.org/details/books/precalculus.