Solving Systems with Inverses: Learn It 2

Lumen Learning, OpenStax

Solving Systems with Inverses: Learn It 2

Finding the Multiplicative Inverse

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

There are several methods to accomplish this, each with its own advantages and applications. Understanding these techniques will provide you with a solid foundation for solving systems of linear equations and other matrix-related problems. Let’s dive into the different approaches to finding the inverse of a matrix!

Matrix Multiplication

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.

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

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]

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}{cc|cc}\hfill 2& \hfill 1& \hfill 1& \hfill 0\\ \hfill 5& \hfill 3& \hfill 0& \hfill 1\\ \end{array}\right][/latex]

Perform row operations with the goal of turning [latex]A[/latex] into the identity.

Switch row 1 and row 2.
[latex]\left[\begin{array}{cc|cc}\hfill 5& \hfill 3& \hfill 0& \hfill 1\\ \hfill 2& \hfill 1& \hfill 1& \hfill 0\\ \end{array}\right][/latex]
Multiply row 2 by [latex]-2[/latex] and add to row 1.
[latex]\left[\begin{array}{cc|cc}\hfill 1& \hfill 1& \hfill -2& \hfill 1\\ \hfill 2& \hfill 1& \hfill 1& \hfill 0\\ \end{array}\right][/latex]
Multiply row 1 by [latex]-2[/latex] and add to row 2.
[latex]\left[\begin{array}{cc|cc}\hfill 1& \hfill 1& \hfill -2& \hfill 1\\ \hfill 0& \hfill -1& \hfill 5& \hfill -2\\ \end{array}\right][/latex]
Add row 2 to row 1.
[latex]\left[\begin{array}{cc|cc}\hfill 1& \hfill 0& \hfill 3& \hfill -1\\ \hfill 0& \hfill -1& \hfill 5& \hfill -2\\ \end{array}\right][/latex]
Multiply row 2 by [latex]-1[/latex].
[latex]\left[\begin{array}{cc|cc}\hfill 1& \hfill 0& \hfill 3& \hfill -1\\ \hfill 0& \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]

Find the inverse of the given matrix by augmenting with the identity.

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