College Algebra, Chapter 7, 7.3, Section 7.3, Problem 16

Determine the inverse of the matrix $\left[ \begin{array}{ccc}
4 & 2 & 3 \\
3 & 3 & 2 \\
1 & 0 & 1
\end{array} \right]$ if it exists.

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

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

By using Gauss-Jordan Elimination

$\displaystyle \frac{1}{4} R_1$

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

$\displaystyle R_2 - 3 R_1 \to R_2$

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

$\displaystyle R_3 - R_1 \to R_3$

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

$\displaystyle \frac{2}{3} R_2$

$\left[ \begin{array}{ccc|ccc}
1 & \displaystyle \frac{1}{2} & \displaystyle \frac{3}{4} & \displaystyle \frac{1}{4} & 0 & 0 \\
0 & 1 & \displaystyle \frac{-1}{6} & \displaystyle \frac{-1}{2} & \displaystyle \frac{2}{3} & 0 \\
0 & \displaystyle \frac{-1}{2} & \displaystyle \frac{1}{4} & \displaystyle \frac{-1}{4} & 0 & 1
\end{array} \right]$

$\displaystyle R_3 + \frac{1}{2} R_2 \to R_3$

$\left[ \begin{array}{ccc|ccc}
1 & \displaystyle \frac{1}{2} & \displaystyle \frac{3}{4} & \displaystyle \frac{1}{4} & 0 & 0 \\
0 & 1 & \displaystyle \frac{-1}{6} & \displaystyle \frac{-1}{2} & \displaystyle \frac{2}{3} & 0 \\
0 & 0 & \displaystyle \frac{1}{6} & \displaystyle \frac{-1}{2} & \displaystyle \frac{1}{3} & 1
\end{array} \right]$

$\displaystyle 6 R_3$

$\left[ \begin{array}{ccc|ccc}
1 & \displaystyle \frac{1}{2} & \displaystyle \frac{3}{4} & \displaystyle \frac{1}{4} & 0 & 0 \\
0 & 1 & \displaystyle \frac{-1}{6} & \displaystyle \frac{-1}{2} & \displaystyle \frac{2}{3} & 0 \\
0 & 0 & 1 & -3 & 2 & 6
\end{array} \right]$

$\displaystyle R_2 + \frac{1}{6} R_3 \to R_2$

$\left[ \begin{array}{ccc|ccc}
1 & \displaystyle \frac{1}{2} & \displaystyle \frac{3}{4} & \displaystyle \frac{1}{4} & 0 & 0 \\
0 & 1 & 0 & -1 & 1 & 1 \\
0 & 0 & 1 & -3 & 2 & 6
\end{array} \right]$

$\displaystyle R_1 - \frac{3}{4} R_3 \to R_1$

$\left[ \begin{array}{ccc|ccc}
1 & \displaystyle \frac{1}{2} & 0 & \displaystyle \frac{5}{2} & \displaystyle \frac{-3}{2} & \displaystyle \frac{-9}{2} \\
0 & 1 & 0 & -1 & 1 & 1 \\
0 & 0 & 1 & -3 & 2 & 6
\end{array} \right]$

$\displaystyle R_1 - \frac{1}{2} R_2 \to R_1$

$\left[ \begin{array}{ccc|ccc}
1 & 0 & 0 & 3 & -2 & -5 \\
0 & 1 & 0 & -1 & 1 & 1 \\
0 & 0 & 1 & -3 & 2 & 6
\end{array} \right]$


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

$\left[ \begin{array}{ccc}
3 & -2 & -5 \\
-1 & 1 & 1 \\
-3 & 2 & 6
\end{array} \right]$

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