College Algebra, Chapter 7, Review Exercises, Section Review Exercises, Problem 48
Determine the determinant of the matrix $\displaystyle A = \left[
\begin{array}{ccc}
1 & 2 & 3 \\
2 & 4 & 5 \\
2 & 5 & 6
\end{array}
\right]$ and if possible, the inverse of the matrix.
Using the formula
$\displaystyle |D| = \left[
\begin{array}{ccc}
1 & 2 & 3 \\
2 & 4 & 5 \\
2 & 5 & 6
\end{array}
\right] = 1 \left|
\begin{array}{cc}
4 & 5 \\
5 & 6
\end{array}
\right| -2 \left|
\begin{array}{cc}
2 & 5 \\
2 & 6
\end{array}
\right| + 3 \left|
\begin{array}{cc}
2 & 4 \\
2 & 5
\end{array}
\right| = 4 \cdot 6 - 5 \cdot 5 - 2 (2 \cdot 6 - 5 \cdot 2) + 3 (2 \cdot 5 - 4 \cdot 2) = -1-4+6 = 1$
The matrix has an inverse.
First, let's add the identity matrix to the right of our matrix.
$\displaystyle \left[ \begin{array}{ccc|ccc}
1 & 2 & 3 & 1 & 0 & 0 \\
2 & 4 & 5 & 0 & 1 & 0 \\
2 & 5 & 6 & 0 & 0 & 1
\end{array} \right]$
Using Gauss-Jordan Elimination
$R_2 - 2R_1 \to R_2$
$\displaystyle \left[ \begin{array}{ccc|ccc}
1 & 2 & 3 & 1 & 0 & 0 \\
0 & 0 & -1 & -2 & 1 & 0 \\
2 & 5 & 6 & 0 & 0 & 1
\end{array} \right]$
$R_3 - 2 R_1 \to R_3$
$\displaystyle \left[ \begin{array}{ccc|ccc}
1 & 2 & 3 & 1 & 0 & 0 \\
0 & 0 & -1 & -2 & 1 & 0 \\
0 & 1 & 0 & -2 & 0 & 1
\end{array} \right]$
$R_3 \longleftrightarrow R_2$
$\displaystyle \left[ \begin{array}{ccc|ccc}
1 & 2 & 3 & 1 & 0 & 0 \\
0 & 1 & 0 & -2 & 0 & 1 \\
0 & 0 & -1 & -2 & 1 & 0
\end{array} \right]$
$- R_3$
$\displaystyle \left[ \begin{array}{ccc|ccc}
1 & 2 & 3 & 1 & 0 & 0 \\
0 & 1 & 0 & -2 & 0 & 1 \\
0 & 0 & 1 & 2 & -1 & 0
\end{array} \right]$
$R_1 - 3 R_3 \to R_1$
$\displaystyle \left[ \begin{array}{ccc|ccc}
1 & 2 & 0 & -5 & 3 & 0 \\
0 & 1 & 0 & -2 & 0 & 1 \\
0 & 0 & 1 & 2 & -1 & 0
\end{array} \right]$
$R_1 - 2R_2 \to R_1$
$\displaystyle \left[ \begin{array}{ccc|ccc}
1 & 0 & 0 & -1 & 3 & -2 \\
0 & 1 & 0 & -2 & 0 & 1 \\
0 & 0 & 1 & 2 & -1 & 0
\end{array} \right]$
The inverse of matrix $A$ is
$\displaystyle A^{-1} = \left[ \begin{array}{ccc}
-1 & 3 & -2 \\
-2 & 0 & 1 \\
2 & -1 & 0
\end{array} \right]$
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