Answer
$|A|=-1$
$\begin{array} \\A^{-1}=\\ \end{array}
\begin{bmatrix} 3&2&-3\\ 2&1&-2\\-8&-6&9 \end{bmatrix}$
Work Step by Step
Step 1. List the matrix needed:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 3&0&1\\ 2&-3&0\\4&-2&1 \end{bmatrix}$
Step 2. Calculate the determinant use row-1 expansion (use the formula from section 10.6):
$|A|=3(-3\times1-0)-0+1(-2\times2+3\times4)=-9+8=-1$
Step 3. Derive $A^{-1}$ (use the procedures from section 10.5):
$\begin{array} \\AI=\\ \end{array}
\begin{bmatrix} 3&0&1&|&1&0&0\\ 2&-3&0&|&0&1&0\\4&-2&1&|&0&0&1 \end{bmatrix}
\begin{array} \\ \\2R_1-3R_2\to R_2\\4R_1-3R_3\to R_3 \end{array}$
Step 4. Peform the row operations shown on the right side of the matrix:
$\begin{array} \\AI=\\ \end{array}
\begin{bmatrix} 3&0&1&|&1&0&0\\ 0&9&2&|&2&-3&0\\0&6&1&|&4&0&-3 \end{bmatrix}
\begin{array} \\ \\ \\3R_3-2R_2\to R_3 \end{array}$
Step 5. Peform the row operations shown on the right side of the matrix:
$\begin{array} \\AI=\\ \end{array}
\begin{bmatrix} 3&0&1&|&1&0&0\\ 0&9&2&|&2&-3&0\\0&0&-1&|&8&6&-9 \end{bmatrix}
\begin{array} \\ \\R_1+R_3\to R_1 \\R_2-2R_3\to R_2 \\ \end{array}$
Step 6. Peform the row operations shown on the right side of the matrix:
$\begin{array} \\AI=\\ \end{array}
\begin{bmatrix} 3&0&0&|&9&6&-9\\ 0&9&0&|&18&9&-18\\0&0&-1&|&8&6&-9 \end{bmatrix}
\begin{array} \\R_1/3\to R_1 \\R_2/9\to R_2 \\ -R_3\to R_3\end{array}$
Step 7. Peform the row operations shown on the right side of the matrix:
$\begin{array} \\AI=\\ \end{array}
\begin{bmatrix} 1&0&0&|&3&2&-3\\ 0&1&0&|&2&1&-2\\0&0&1&|&-8&-6&9 \end{bmatrix}$
Step 8. Write the result as:
$\begin{array} \\A^{-1}=\\ \end{array}
\begin{bmatrix} 3&2&-3\\ 2&1&-2\\-8&-6&9 \end{bmatrix}$
