College Algebra, Chapter 7, 7.3, Section 7.3, Problem 18

Determine the inverse of the matrix $\left[ \begin{array}{ccc}
5 & 7 & 4 \\
3 & -1 & 3 \\
6 & 7 & 5
\end{array} \right]$ if it exists.

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

$\left[ \begin{array}{ccc|ccc}
5 & 7 & 4 & 1 & 0 & 0 \\
3 & -1 & 3 & 0 & 1 & 0 \\
6 & 7 & 5 & 0 & 0 & 1
\end{array} \right]$

By using Gauss-Jordan Elimination

$\displaystyle \frac{1}{5} R_1$

$\left[ \begin{array}{ccc|ccc}
1 & \displaystyle \frac{7}{5} & \displaystyle \frac{4}{5} & \displaystyle \frac{1}{5} & 0 & 0 \\
3 & -1 & 3 & 0 & 1 & 0 \\
6 & 7 & 5 & 0 & 0 & 1
\end{array} \right]$

$\displaystyle R_2 - 3 R_1 \to R_2$

$\left[ \begin{array}{ccc|ccc}
1 & \displaystyle \frac{7}{5} & \displaystyle \frac{4}{5} & \displaystyle \frac{1}{5} & 0 & 0 \\
0 & \displaystyle \frac{-26}{5} & \displaystyle \frac{3}{5} & \displaystyle \frac{-3}{5} & 1 & 0 \\
0 & \displaystyle \frac{-7}{5} & \displaystyle \frac{1}{5} & \displaystyle \frac{-6}{5} & 0 & 1
\end{array} \right]$

$\displaystyle \frac{-5}{26} R_2$

$\left[ \begin{array}{ccc|ccc}
1 & \displaystyle \frac{7}{5} & \displaystyle \frac{4}{5} & \displaystyle \frac{1}{5} & 0 & 0 \\
0 & 1 & \displaystyle \frac{-3}{26} & \displaystyle \frac{3}{26} & \displaystyle \frac{-5}{26} & 0 \\
0 & \displaystyle \frac{-7}{5} & \displaystyle \frac{1}{5} & \displaystyle \frac{-6}{5} & 0 & 1
\end{array} \right]$

$\displaystyle R_3 + \frac{7}{5} R_2 \to R_3$

$\left[ \begin{array}{ccc|ccc}
1 & \displaystyle \frac{7}{5} & \displaystyle \frac{4}{5} & \displaystyle \frac{1}{5} & 0 & 0 \\
0 & 1 & \displaystyle \frac{-3}{26} & \displaystyle \frac{3}{26} & \displaystyle \frac{-5}{26} & 0 \\
0 & 0 & \displaystyle \frac{1}{26} & \displaystyle \frac{-27}{26} & \displaystyle \frac{-7}{26} & 1
\end{array} \right]$


$\displaystyle 26 R_3$

$\left[ \begin{array}{ccc|ccc}
1 & \displaystyle \frac{7}{5} & \displaystyle \frac{4}{5} & \displaystyle \frac{1}{5} & 0 & 0 \\
0 & 1 & \displaystyle \frac{-3}{26} & \displaystyle \frac{3}{26} & \displaystyle \frac{-5}{26} & 0 \\
0 & 0 & 1 & -27 & -7 & 26
\end{array} \right]$


$\displaystyle R_2 + \frac{3}{26} R_3 \to R_2$

$\left[ \begin{array}{ccc|ccc}
1 & \displaystyle \frac{7}{5} & \displaystyle \frac{4}{5} & \displaystyle \frac{1}{5} & 0 & 0 \\
0 & 1 & 0 & -3 & -1 & 3 \\
0 & 0 & 1 & -27 & -7 & 26
\end{array} \right]$


$\displaystyle R_1 - \frac{4}{5} R_3 \to R_1$

$\left[ \begin{array}{ccc|ccc}
1 & \displaystyle \frac{7}{5} & 0 & \displaystyle \frac{109}{5} & \displaystyle \frac{28}{5} & \displaystyle \frac{-104}{5} \\
0 & 1 & 0 & -3 & -1 & 3 \\
0 & 0 & 1 & -27 & -7 & 26
\end{array} \right]$


$\displaystyle R_1 - \frac{7}{5} R_2 \to R_1$

$\left[ \begin{array}{ccc|ccc}
1 & 0 & 0 & 26 & 7 & -25 \\
0 & 1 & 0 & -3 & -1 & 3 \\
0 & 0 & 1 & -27 & -7 & 26
\end{array} \right]$


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

$\left[ \begin{array}{ccc}
26 & 7 & -25 \\
-3 & -1 & 3 \\
-27 & -7 & 26
\end{array} \right]$