College Algebra, Chapter 7, 7.3, Section 7.3, Problem 20

Determine the inverse of the matrix $\left[ \begin{array}{ccc}
2 & 1 & 0 \\
1 & 1 & 4 \\
2 & 1 & 2
\end{array} \right]$ if it exists.

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

$\left[ \begin{array}{ccc|ccc}
2 & 1 & 0 & 1 & 0 & 0 \\
1 & 1 & 4 & 0 & 1 & 0 \\
2 & 1 & 2 & 0 & 0 & 1
\end{array} \right]$

By using Gauss-Jordan Elimination

$\displaystyle \frac{1}{2} R_1 $

$\left[ \begin{array}{ccc|ccc}
1 & \displaystyle \frac{1}{2} & 0 & \displaystyle \frac{1}{2} & 0 & 0 \\
1 & 1 & 4 & 0 & 1 & 0 \\
2 & 1 & 2 & 0 & 0 & 1
\end{array} \right]$

$\displaystyle R_2 - R_1 \to R_2$

$\left[ \begin{array}{ccc|ccc}
1 & \displaystyle \frac{1}{2} & 0 & \displaystyle \frac{1}{2} & 0 & 0 \\
0 & \displaystyle \frac{1}{2} & 4 & \displaystyle \frac{-1}{2} & 1 & 0 \\
2 & 1 & 2 & 0 & 0 & 1
\end{array} \right]$

$\displaystyle R_3 - 2 R_1 \to R_3$

$\left[ \begin{array}{ccc|ccc}
1 & \displaystyle \frac{1}{2} & 0 & \displaystyle \frac{1}{2} & 0 & 0 \\
0 & \displaystyle \frac{1}{2} & 4 & \displaystyle \frac{-1}{2} & 1 & 0 \\
0 & 0 & 2 & -1 & 0 & 1
\end{array} \right]$

$\displaystyle 2 R_2$

$\left[ \begin{array}{ccc|ccc}
1 & \displaystyle \frac{1}{2} & 0 & \displaystyle \frac{1}{2} & 0 & 0 \\
0 & 1 & 8 & -1 & 2 & 0 \\
0 & 0 & 2 & -1 & 0 & 1
\end{array} \right]$

$\displaystyle \frac{1}{2} R_3$

$\left[ \begin{array}{ccc|ccc}
1 & \displaystyle \frac{1}{2} & 0 & \displaystyle \frac{1}{2} & 0 & 0 \\
0 & 1 & 8 & -1 & 2 & 0 \\
0 & 0 & 1 & \displaystyle \frac{-1}{2} & 0 & \displaystyle \frac{1}{2}
\end{array} \right]$

$\displaystyle R_2 - 8 R_3 \to R_2$

$\left[ \begin{array}{ccc|ccc}
1 & \displaystyle \frac{1}{2} & 0 & \displaystyle \frac{1}{2} & 0 & 0 \\
0 & 1 & 0 & 3 & 2 & -4 \\
0 & 0 & 1 & \displaystyle \frac{-1}{2} & 0 & \displaystyle \frac{1}{2}
\end{array} \right]$

$\displaystyle R_1 - \frac{1}{2} \to R_1$

$\left[ \begin{array}{ccc|ccc}
1 & 0 & 0 & -1 & -1 & 2 \\
0 & 1 & 0 & 3 & 2 & -4 \\
0 & 0 & 1 & \displaystyle \frac{-1}{2} & 0 & \displaystyle \frac{1}{2}
\end{array} \right]$

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

$\left[ \begin{array}{ccc}
-1 & -1 & 2 \\
3 & 2 & -4 \\
\displaystyle \frac{-1}{2} & 0 & \displaystyle \frac{1}{2}
\end{array} \right]$