College Algebra, Chapter 7, 7.3, Section 7.3, Problem 12

Determine the inverse of the matrix $\left[ \begin{array}{cc}
-7 & 4 \\
8 & -5
\end{array} \right]$ if it exists.

First, let's add the identity matrix to the right of our matrix

$\left[ \begin{array}{cc|cc}
-7 & 4 & 1 & 0 \\
8 & -5 & 0 & 1
\end{array} \right]$

By using Gauss-Jordan Elimination

$\displaystyle \frac{-1}{7} R_1$

$\left[ \begin{array}{cc|cc}
1 & \displaystyle \frac{-4}{7} & \displaystyle \frac{-1}{7} & 0 \\
8 & -5 & 0 & 1
\end{array} \right]$

$\displaystyle R_2 - 8 R_1 \to R_2$

$\left[ \begin{array}{cc|cc}
1 & \displaystyle \frac{-4}{7} & \displaystyle \frac{-1}{7} & 0 \\
0 & \displaystyle \frac{-3}{7} & \displaystyle \frac{8}{7} & 1
\end{array} \right]$

$\displaystyle \frac{-7}{3} R_2$

$\left[ \begin{array}{cc|cc}
1 & \displaystyle \frac{-4}{7} & \displaystyle \frac{-1}{7} & 0 \\
0 & 1 & \displaystyle \frac{-8}{3} & \displaystyle \frac{-7}{3}
\end{array} \right]$

$\displaystyle R_1 + \frac{4}{7} R_2 \to R_1$

$\left[ \begin{array}{cc|cc}
1 & 0 & \displaystyle \frac{-5}{3} & \displaystyle \frac{-4}{3} \\
0 & 1 & \displaystyle \frac{-8}{3} & \displaystyle \frac{-7}{3}
\end{array} \right]$

The inverse matrix can now be found in the right half of our reduced row-echelon matrix. So the inverse matrix is

$\left[ \begin{array}{cc}
\displaystyle \frac{-5}{3} & \displaystyle \frac{-4}{3} \\
\displaystyle \frac{-8}{3} & \displaystyle \frac{-7}{3}
\end{array} \right]$