Find the inverse of a 3x3 matrix: \[ \begin{pmatrix} 1 & 3 & 0 \\ 0 & 3 & 1 \\ 2 & -1 & 2 \end{pmatrix} \]

To find the inverse of a 3x3 matrix, we can use the formula:

\[ A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) \]

where det(A) is the determinant of matrix A and adj(A) is the adjugate of matrix A.

First, let's find the determinant of the given matrix:

\[ \text{det}(A) = 1(3*2 - 1*1) - 3(0*2 - 1*2) + 0(0*1 - 3*2) \]
\[ \text{det}(A) = 1(6 - 1) - 3(0 - 2) + 0(0 - 6) \]
\[ \text{det}(A) = 5 + 6 + 0 \]
\[ \text{det}(A) = 11 \]

Next, let's find the adjugate of the matrix. The adjugate of a matrix is the transpose of the cofactor matrix. The cofactor matrix is obtained by taking the determinant of each 2x2 minor matrix and multiplying by -1 if the sum of the row and column indices is odd.

\[
\text{adj}(A) =
\begin{pmatrix}
(3*2 - 1*1) & -(0*2 - 1*2) & (0*1 - 3*2) \\
-3 & 1 & 3 \\
-3 & 0 & 1
\end{pmatrix}
\]

\[
\text{adj}(A) =
\begin{pmatrix}
5 & -2 & -6 \\
-3 & 1 & 3 \\
-3 & 0 & 1
\end{pmatrix}
\]

Now, we can find the inverse of the matrix:

\[ A^{-1} = \frac{1}{11}
\begin{pmatrix}
5 & -2 & -6 \\
-3 & 1 & 3 \\
-3 & 0 & 1
\end{pmatrix}
\]

\[ A^{-1} =
\begin{pmatrix}
\frac{5}{11} & \frac{-2}{11} & \frac{-6}{11} \\
\frac{-3}{11} & \frac{1}{11} & \frac{3}{11} \\
\frac{-3}{11} & 0 & \frac{1}{11}
\end{pmatrix}
\]

Therefore, the inverse of the given 3x3 matrix is:

\[
\begin{pmatrix}
\frac{5}{11} & \frac{-2}{11} & \frac{-6}{11} \\
\frac{-3}{11} & \frac{1}{11} & \frac{3}{11} \\
\frac{-3}{11} & 0 & \frac{1}{11}
\end{pmatrix}
\]

Step 1: Find the determinant of the matrix.

The determinant is a numerical value that tells us the area or volume of the parallelepiped formed by the three column vectors of the matrix. In this case, the determinant is:

$$ det \begin{pmatrix} 1 & 3 & 0 \\ 0 & 3 & 1 \\ 2 & -1 & 2 \end{pmatrix} = 1 \cdot 6 - 3 \cdot 0 + 0 \cdot (-2) = 6 $$

Step 2: Find the adjoint of the matrix.

The adjoint of a matrix is the transpose of the cofactor matrix. The cofactor matrix is found by replacing each element in the original matrix with the determinant of the 2x2 submatrix that remains when that element is removed. In this case, the cofactor matrix is:

$$ C = \begin{pmatrix} 6 & -1 & -2 \\ 0 & 2 & -2 \\ -6 & 3 & 6 \end{pmatrix} $$

and the adjoint is:

$$ adj(A) = C^T = \begin{pmatrix} 6 & 0 & -6 \\ -1 & 2 & 3 \\ -2 & -2 & 6 \end{pmatrix} $$

Step 3: Find the inverse of the matrix.

The inverse of a matrix is found by dividing the adjoint by the determinant. In this case, the inverse is:

$$ A^{-1} = \frac{1}{det(A)} adj(A) = \frac{1}{6} \begin{pmatrix} 6 & 0 & -6 \\ -1 & 2 & 3 \\ -2 & -2 & 6 \end{pmatrix} $$

$$ = \begin{pmatrix} 1 & 0 & -1 \\ -\frac{1}{6} & \frac{1}{3} & \frac{1}{2} \\ -\frac{1}{3} & -\frac{1}{3} & 1 \end{pmatrix} $$

Therefore, the inverse of the given matrix is:

$$ \begin{pmatrix} 1 & 0 & -1 \\ -\frac{1}{6} & \frac{1}{3} & \frac{1}{2} \\ -\frac{1}{3} & -\frac{1}{3} & 1 \end{pmatrix} $$