Example 3 from Algebra and Trigonometry, 11.7 Solving Systems with Inverses

Finding the Multiplicative Inverse Using Matrix Multiplication

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

A = [\begin{array}{r} 1 & −2 \\ 2 & −3 \end{array}]

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
For this method, we multiply $A$ by a matrix containing unknown constants and set it equal to the identity.

$[\begin{array}{r} 1 & −2 \\ 2 & −3 \end{array}] [\begin{array}{r} a & b \\ c & d \end{array}] = [\begin{array}{r} 1 & 0 \\ 0 & 1 \end{array}]$
Find the product of the two matrices on the left side of the equal sign.

$[\begin{array}{r} 1 & −2 \\ 2 & −3 \end{array}] [\begin{array}{r} a & b \\ c & d \end{array}] = [\begin{array}{r} 1 a −2 c & 1 b −2 d \\ 2 a −3 c & 2 b −3 d \end{array}]$
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.
$\begin{matrix} 1 a −2 c = 1 R_{1} \\ 2 a −3 c = 0 R_{2} \end{matrix}$

Using row operations, multiply and add as follows: $(−2) R_{1} + R_{2} \to R_{2} .$ Add the equations, and solve for $c .$
$\begin{array}{r} 1 a - 2 c = 1 \\ 0 + 1 c = - 2 \\ c = - 2 \end{array}$
Back-substitute to solve for $a .$

$\begin{array}{r} a −2 (−2) = 1 \\ a + 4 = 1 \\ a = −3 \end{array}$
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.

$\begin{array}{r} 1 b −2 d = 0 & R_{1} \\ 2 b −3 d = 1 & R_{2} \end{array}$
Using row operations, multiply and add as follows: $(−2) R_{1} + R_{2} = R_{2} .$ Add the two equations and solve for $d .$

$\begin{array}{r} 1 b −2 d = 0 \\ \frac{0 + 1 d = 1}{d = 1} \end{array}$
Once more, back-substitute and solve for $b .$

$\begin{array}{r} b −2 (1) = 0 \\ b −2 = 0 \\ b = 2 \end{array}$
$A^{−1} = [\begin{array}{r} −3 & 2 \\ −2 & 1 \end{array}]$

How did it go?

Solution

Solution