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Study Guides > College Algebra

Find the Inverse of a Matrix

Learning Objectives

  • Verify that multiplying a matrix by it's inverse results in 1
  • Use matrix multiplication to find the inverse of a matrix
  • Finding an inverse by augmenting with an identity matrix
We know that the multiplicative inverse of a real number [latex]a[/latex] is [latex]{a}^{-1}[/latex], and [latex]a{a}^{-1}={a}^{-1}a=\left(\frac{1}{a}\right)a=1[/latex]. For example, [latex]{2}^{-1}=\frac{1}{2}[/latex] and [latex]\left(\frac{1}{2}\right)2=1[/latex]. The multiplicative inverse of a matrix is similar in concept, except that the product of matrix [latex]A[/latex] and its inverse [latex]{A}^{-1}[/latex] 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 [latex]{I}_{n}[/latex] where [latex]n[/latex] represents the dimension of the matrix. The equations below are the identity matrices for a [latex]2\text{}\times \text{}2[/latex] matrix and a [latex]3\text{}\times \text{}3[/latex] matrix, respectively.

[latex]{I}_{2}=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 0\\ \hfill 0& \hfill & \hfill 1\end{array}\right][/latex]

[latex]{I}_{3}=\left[\begin{array}{rrrrr}\hfill 1& \hfill & \hfill 0& \hfill & \hfill 0\\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill 0\\ \hfill 0& \hfill & \hfill 0& \hfill & \hfill 1\end{array}\right][/latex]

The identity matrix acts as a 1 in matrix algebra. For example, [latex]AI=IA=A[/latex]. A matrix that has a multiplicative inverse has the properties

[latex]\begin{array}{l}A{A}^{-1}=I\\ {A}^{-1}A=I\end{array}[/latex]

A matrix that has a multiplicative inverse is called an invertible matrix. Only a square matrix may have a multiplicative inverse, as the reversibility, [latex]A{A}^{-1}={A}^{-1}A=I[/latex], is a requirement. Not all square matrices have an inverse, but if [latex]A[/latex] is invertible, then [latex]{A}^{-1}[/latex] is unique. We will look at two methods for finding the inverse of a [latex]2\text{}\times \text{}2[/latex] matrix and a third method that can be used on both [latex]2\text{}\times \text{}2[/latex] and [latex]3\text{}\times \text{}3[/latex] matrices.

A General Note: The Identity Matrix and Multiplicative Inverse

The identity matrix, [latex]{I}_{n}[/latex], is a square matrix containing ones down the main diagonal and zeros everywhere else.

[latex]\begin{array}{l}\hfill\begin{array}{l}\begin{array}{l}\hfill \\ {I}_{2}=\left[\begin{array}{rr}\hfill 1& \hfill 0\\ \hfill 0& \hfill 1\end{array}\right]\begin{array}{cccc}& & & \end{array}{I}_{3}=\left[\begin{array}{rrr}\hfill 1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 1& \hfill 0\\ \hfill 0& \hfill 0& \hfill 1\end{array}\right]\hfill \end{array}\hfill \\ \text{ }2\times 2\text{ 3}\times 3\hfill \end{array}\hfill \end{array}[/latex]

If [latex]A[/latex] is an [latex]n\times n[/latex] matrix and [latex]B[/latex] is an [latex]n\times n[/latex] matrix such that [latex]AB=BA={I}_{n}[/latex], then [latex]B={A}^{-1}[/latex], the multiplicative inverse of a matrix [latex]A[/latex].

Example: Showing That the Identity Matrix Acts as a 1

Given matrix A, show that [latex]AI=IA=A[/latex]. [latex-display]A=\left[\begin{array}{cc}3& 4\\ -2& 5\end{array}\right][/latex-display]

Answer: Use matrix multiplication to show that the product of [latex]A[/latex] and the identity is equal to the product of the identity and A.

[latex]AI=\left[\begin{array}{rrr}\hfill 3& \hfill & \hfill 4\\ \hfill -2& \hfill & \hfill 5\end{array}\right]\begin{array}{r}\hfill \end{array}\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 0\\ \hfill 0& \hfill & \hfill 1\end{array}\right]=\left[\begin{array}{rrrr}\hfill 3\cdot 1+4\cdot 0& \hfill & \hfill & \hfill 3\cdot 0+4\cdot 1\\ \hfill -2\cdot 1+5\cdot 0& \hfill & \hfill & \hfill -2\cdot 0+5\cdot 1\end{array}\right]=\left[\begin{array}{rrr}\hfill 3& \hfill & \hfill 4\\ \hfill -2& \hfill & \hfill 5\end{array}\right][/latex]

[latex]AI=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 0\\ \hfill 0& \hfill & \hfill 1\end{array}\right]\begin{array}{r}\hfill \end{array}\left[\begin{array}{rrr}\hfill 3& \hfill & \hfill 4\\ \hfill -2& \hfill & \hfill 5\end{array}\right]=\left[\begin{array}{rrrr}\hfill 1\cdot 3+0\cdot \left(-2\right)& \hfill & \hfill & \hfill 1\cdot 4+0\cdot 5\\ \hfill 0\cdot 3+1\cdot \left(-2\right)& \hfill & \hfill & \hfill 0\cdot 4+1\cdot 5\end{array}\right]=\left[\begin{array}{rrr}\hfill 3& \hfill & \hfill 4\\ \hfill -2& \hfill & \hfill 5\end{array}\right][/latex]

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

  1. Given matrix [latex]A[/latex] of order [latex]n\times n[/latex] and matrix [latex]B[/latex] of order [latex]n\times n[/latex] multiply [latex]AB[/latex].
  2. If [latex]AB=I[/latex], then find the product [latex]BA[/latex]. If [latex]BA=I[/latex], then [latex]B={A}^{-1}[/latex] and [latex]A={B}^{-1}[/latex].

Example: Showing That Matrix A Is the Multiplicative Inverse of Matrix B

Show that the given matrices are multiplicative inverses of each other.

[latex]A=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 5\\ \hfill -2& \hfill & \hfill -9\end{array}\right],B=\left[\begin{array}{rrr}\hfill -9& \hfill & \hfill -5\\ \hfill 2& \hfill & \hfill 1\end{array}\right][/latex]

Answer: Multiply [latex]AB[/latex] and [latex]BA[/latex]. If both products equal the identity, then the two matrices are inverses of each other.

[latex]\begin{array}{l}AB=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 5\\ \hfill -2& \hfill & \hfill -9\end{array}\right]\cdot \left[\begin{array}{rrr}\hfill -9& \hfill & \hfill -5\\ \hfill 2& \hfill & \hfill 1\end{array}\right]\hfill \\ =\left[\begin{array}{rrr}\hfill 1\left(-9\right)+5\left(2\right)& \hfill & \hfill 1\left(-5\right)+5\left(1\right)\\ \hfill -2\left(-9\right)-9\left(2\right)& \hfill & \hfill -2\left(-5\right)-9\left(1\right)\end{array}\right]\hfill \\ =\left[\begin{array}{ccc}1& & 0\\ 0& & 1\end{array}\right]\hfill \end{array}[/latex]

[latex]\begin{array}{l}BA=\left[\begin{array}{rrr}\hfill -9& \hfill & \hfill -5\\ \hfill 2& \hfill & \hfill 1\end{array}\right]\cdot \left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 5\\ \hfill -2& \hfill & \hfill -9\end{array}\right]\hfill \\ =\left[\begin{array}{rrr}\hfill -9\left(1\right)-5\left(-2\right)& \hfill & \hfill -9\left(5\right)-5\left(-9\right)\\ \hfill 2\left(1\right)+1\left(-2\right)& \hfill & \hfill 2\left(-5\right)+1\left(-9\right)\end{array}\right]\hfill \\ =\left[\begin{array}{ccc}1& & 0\\ 0& & 1\end{array}\right]\hfill \end{array}[/latex]

[latex]A[/latex] and [latex]B[/latex] are inverses of each other.

Try It

Show that the following two matrices are inverses of each other.

[latex]A=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 4\\ \hfill -1& \hfill & \hfill -3\end{array}\right],B=\left[\begin{array}{rrr}\hfill -3& \hfill & \hfill -4\\ \hfill 1& \hfill & \hfill 1\end{array}\right][/latex]

Answer:

[latex]\begin{array}{l}AB=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 4\\ \hfill -1& \hfill & \hfill -3\end{array}\right]\begin{array}{r}\hfill \end{array}\left[\begin{array}{rrr}\hfill -3& \hfill & \hfill -4\\ \hfill 1& \hfill & \hfill 1\end{array}\right]=\left[\begin{array}{rrr}\hfill 1\left(-3\right)+4\left(1\right)& \hfill & \hfill 1\left(-4\right)+4\left(1\right)\\ \hfill -1\left(-3\right)+-3\left(1\right)& \hfill & \hfill -1\left(-4\right)+-3\left(1\right)\end{array}\right]=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 0\\ \hfill 0& \hfill & \hfill 1\end{array}\right]\hfill \\ BA=\left[\begin{array}{rrr}\hfill -3& \hfill & \hfill -4\\ \hfill 1& \hfill & \hfill 1\end{array}\right]\begin{array}{r}\hfill \end{array}\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 4\\ \hfill -1& \hfill & \hfill -3\end{array}\right]=\left[\begin{array}{rrr}\hfill -3\left(1\right)+-4\left(-1\right)& \hfill & \hfill -3\left(4\right)+-4\left(-3\right)\\ \hfill 1\left(1\right)+1\left(-1\right)& \hfill & \hfill 1\left(4\right)+1\left(-3\right)\end{array}\right]=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 0\\ \hfill 0& \hfill & \hfill 1\end{array}\right]\hfill \end{array}[/latex]

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.

Example: Finding the Multiplicative Inverse Using Matrix Multiplication

Use matrix multiplication to find the inverse of the given matrix.

[latex]A=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill -2\\ \hfill 2& \hfill & \hfill -3\end{array}\right][/latex]

Answer: For this method, we multiply [latex]A[/latex] by a matrix containing unknown constants and set it equal to the identity.

[latex]\left[\begin{array}{rr}\hfill 1& \hfill -2\\ \hfill 2& \hfill -3\end{array}\right]\text{ }\left[\begin{array}{rr}\hfill a& \hfill b\\ \hfill c& \hfill d\end{array}\right]=\left[\begin{array}{rr}\hfill 1& \hfill 0\\ \hfill 0& \hfill 1\end{array}\right][/latex]

Find the product of the two matrices on the left side of the equal sign.

[latex]\left[\begin{array}{rr}\hfill 1& \hfill -2\\ \hfill 2& \hfill -3\end{array}\right]\text{ }\left[\begin{array}{rr}\hfill a& \hfill b\\ \hfill c& \hfill d\end{array}\right]=\left[\begin{array}{rr}\hfill 1a - 2c& \hfill 1b - 2d\\ \hfill 2a - 3c& \hfill 2b - 3d\end{array}\right][/latex]

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.

[latex]\begin{array}{c}1a - 2c=1\text{ }{R}_{1}\\ 2a - 3c=0\text{ }{R}_{2}\end{array}[/latex]

Using row operations, multiply and add as follows: [latex]\left(-2\right){R}_{1}+{R}_{2}\to {R}_{2}[/latex]. Add the equations, and solve for [latex]c[/latex].

[latex]\begin{array}{r}\hfill 1a - 2c=1\\ \hfill 0+1c=-2\\ \hfill c=-2\end{array}[/latex]

Back-substitute to solve for [latex]a[/latex].

[latex]\begin{array}{r}\hfill a - 2\left(-2\right)=1\\ \hfill a+4=1\\ \hfill a=-3\end{array}[/latex]

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.

[latex]\begin{array}{rr}\hfill 1b - 2d=0& \hfill {R}_{1}\\ \hfill 2b - 3d=1& \hfill {R}_{2}\end{array}[/latex]

Using row operations, multiply and add as follows: [latex]\left(-2\right){R}_{1}+{R}_{2}={R}_{2}[/latex]. Add the two equations and solve for [latex]d[/latex].

[latex]\begin{array}{r}\hfill 1b - 2d=0\\ \hfill \frac{0+1d=1}{d=1}\\ \hfill \end{array}[/latex]

Once more, back-substitute and solve for [latex]b[/latex].

[latex]\begin{array}{r}\hfill b - 2\left(1\right)=0\\ \hfill b - 2=0\\ \hfill b=2\end{array}[/latex]

[latex]{A}^{-1}=\left[\begin{array}{rrr}\hfill -3& \hfill & \hfill 2\\ \hfill -2& \hfill & \hfill 1\end{array}\right][/latex]

Finding the Multiplicative Inverse by Augmenting with the Identity

Another way to find the multiplicative inverse is by augmenting with the identity. When matrix [latex]A[/latex] is transformed into [latex]I[/latex], the augmented matrix [latex]I[/latex] transforms into [latex]{A}^{-1}[/latex]. For example, given

[latex]A=\left[\begin{array}{rrr}\hfill 2& \hfill & \hfill 1\\ \hfill 5& \hfill & \hfill 3\end{array}\right][/latex]

augment [latex]A[/latex] with the identity

[latex]\left[\begin{array}{rr}\hfill 2& \hfill 1\\ \hfill 5& \hfill 3\end{array}\text{ }|\text{ }\begin{array}{rr}\hfill 1& \hfill 0\\ \hfill 0& \hfill 1\end{array}\right][/latex]

Perform row operations with the goal of turning [latex]A[/latex] into the identity.
  1. Switch row 1 and row 2. [latex]\left[\begin{array}{rr}\hfill 5& \hfill 3\\ \hfill 2& \hfill 1\end{array}\text{ }|\text{ }\begin{array}{rr}\hfill 0& \hfill 1\\ \hfill 1& \hfill 0\end{array}\right][/latex]
  2. Multiply row 2 by [latex]-2[/latex] and add to row 1. [latex]\left[\begin{array}{rr}\hfill 1& \hfill 1\\ \hfill 2& \hfill 1\end{array}\text{ }|\text{ }\begin{array}{rr}\hfill -2& \hfill 1\\ \hfill 1& \hfill 0\end{array}\right][/latex]
  3. Multiply row 1 by [latex]-2[/latex] and add to row 2. [latex]\left[\begin{array}{rr}\hfill 1& \hfill 1\\ \hfill 0& \hfill -1\end{array}\text{ }|\text{ }\begin{array}{rr}\hfill -2& \hfill 1\\ \hfill 5& \hfill -2\end{array}\right][/latex]
  4. Add row 2 to row 1. [latex]\left[\begin{array}{rr}\hfill 1& \hfill 0\\ \hfill 0& \hfill -1\end{array}\text{ }|\text{ }\begin{array}{rr}\hfill 3& \hfill -1\\ \hfill 5& \hfill -2\end{array}\right][/latex]
  5. Multiply row 2 by [latex]-1[/latex]. [latex]\left[\begin{array}{rr}\hfill 1& \hfill 0\\ \hfill 0& \hfill 1\end{array}\text{ }|\text{ }\begin{array}{rr}\hfill 3& \hfill -1\\ \hfill -5& \hfill 2\end{array}\right][/latex]
The matrix we have found is [latex]{A}^{-1}[/latex].

[latex]{A}^{-1}=\left[\begin{array}{rrr}\hfill 3& \hfill & \hfill -1\\ \hfill -5& \hfill & \hfill 2\end{array}\right][/latex]

Finding the Multiplicative Inverse of 2×2 Matrices Using a Formula

When we need to find the multiplicative inverse of a [latex]2\times 2[/latex] matrix, we can use a special formula instead of using matrix multiplication or augmenting with the identity. If [latex]A[/latex] is a [latex]2\times 2[/latex] matrix, such as

[latex]A=\left[\begin{array}{rrr}\hfill a& \hfill & \hfill b\\ \hfill c& \hfill & \hfill d\end{array}\right][/latex]

the multiplicative inverse of [latex]A[/latex] is given by the formula

[latex]{A}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{rrr}\hfill d& \hfill & \hfill -b\\ \hfill -c& \hfill & \hfill a\end{array}\right][/latex]

where [latex]ad-bc\ne 0[/latex]. If [latex]ad-bc=0[/latex], then [latex]A[/latex] has no inverse.

Example: Using the Formula to Find the Multiplicative Inverse of Matrix A

Use the formula to find the multiplicative inverse of

[latex]A=\left[\begin{array}{cc}1& -2\\ 2& -3\end{array}\right][/latex]

Answer: Using the formula, we have

[latex]\begin{array}{l}{A}^{-1}=\frac{1}{\left(1\right)\left(-3\right)-\left(-2\right)\left(2\right)}\left[\begin{array}{cc}-3& 2\\ -2& 1\end{array}\right]\hfill \\ =\frac{1}{-3+4}\left[\begin{array}{cc}-3& 2\\ -2& 1\end{array}\right]\hfill \\ =\left[\begin{array}{cc}-3& 2\\ -2& 1\end{array}\right]\hfill \end{array}[/latex]

Analysis of the Solution

We can check that our formula works by using one of the other methods to calculate the inverse. Let’s augment [latex]A[/latex] with the identity.

[latex]\left[\begin{array}{cc}1& -2\\ 2& -3\end{array}|\begin{array}{cc}1& 0\\ 0& 1\end{array}\right][/latex]

Perform row operations with the goal of turning [latex]A[/latex] into the identity.
  1. Multiply row 1 by [latex]-2[/latex] and add to row 2. [latex]\left[\begin{array}{cc}1& -2\\ 0& 1\end{array}|\begin{array}{cc}1& 0\\ -2& 1\end{array}\right][/latex]
  2. Multiply row 1 by 2 and add to row 1. [latex]\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}|\begin{array}{cc}-3& 2\\ -2& 1\end{array}\right][/latex]
So, we have verified our original solution.

[latex]{A}^{-1}=\left[\begin{array}{cc}-3& 2\\ -2& 1\end{array}\right][/latex]

Try It

Use the formula to find the inverse of matrix [latex]A[/latex]. Verify your answer by augmenting with the identity matrix.

[latex]A=\left[\begin{array}{cc}1& -1\\ 2& 3\end{array}\right][/latex]

Answer: [latex]{A}^{-1}=\left[\begin{array}{cc}\frac{3}{5}& \frac{1}{5}\\ -\frac{2}{5}& \frac{1}{5}\end{array}\right][/latex]

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