3.6 Let \( A=\left[\begin{array}{ccc}1 & 0 & 2 \\ 0 & -3 & -1 \\ 2 & -1 & -2\end{array}\right] \).Use the adjoint method to find the inverse of \( A \).

To find the inverse of the matrix \( A \) using the adjoint method, we follow these steps: 1. **Calculate the determinant of \( A \)**. 2. **Find the matrix of minors**. 3. **Find the matrix of cofactors**. 4. **Find the adjugate (adjoint) of \( A \)**. 5. **Use the formula for the inverse**: \[ A^{-1} = \frac{1}{\det(A)} \text{adj}(A) \] ### Step 1: Calculate the Determinant of \( A \) The determinant of a \( 3 \times 3 \) matrix \( A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \) is given by: \[ \det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \] For our matrix \( A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & -3 & -1 \\ 2 & -1 & -2 \end{bmatrix} \): - \( a = 1, b = 0, c = 2 \) - \( d = 0, e = -3, f = -1 \) - \( g = 2, h = -1, i = -2 \) Calculating the determinant: \[ \det(A) = 1((-3)(-2) - (-1)(-1)) - 0 + 2(0(-1) - (-3)(2)) \] \[ = 1(6 - 1) + 2(0 + 6) \] \[ = 1 \cdot 5 + 2 \cdot 6 = 5 + 12 = 17 \] ### Step 2: Find the Matrix of Minors The minor \( M_{ij} \) is the determinant of the \( 2 \times 2 \) matrix that remains after removing the \( i \)-th row and \( j \)-th column. Calculating the minors: - \( M_{11} = \det\begin{bmatrix} -3 & -1 \\ -1 & -2 \end{bmatrix} = (-3)(-2) - (-1)(-1) = 6 - 1 = 5 \) - \( M_{12} = \det\begin{bmatrix} 0 & -1 \\ 2 & -2 \end{bmatrix} = (0)(-2) - (-1)(2) = 0 + 2 = 2 \) - \( M_{13} = \det\begin{bmatrix} 0 & -3 \\ 2 & -1 \end{bmatrix} = (0)(-1) - (-3)(2) = 0 + 6 = 6 \) - \( M_{21} = \det\begin{bmatrix} 0 & 2 \\ -1 & -2 \end{bmatrix} = (0)(-2) - (2)(-1) = 0 + 2 = 2 \) - \( M_{22} = \det\begin{bmatrix} 1 & 2 \\ 2 & -2 \end{bmatrix} = (1)(-2) - (2)(2) = -2 - 4 = -6 \) - \( M_{23} = \det\begin{bmatrix} 1 & 0 \\ 2 & -1 \end{bmatrix} = (1)(-1) - (0)(2) = -1 \) - \( M_{31} = \det\begin{bmatrix} 0 & 2 \\ -3 & -1 \end{bmatrix} = (0)(-1) - (2)(-3) = 0 + 6 = 6 \) - \( M_{32} = \det\begin{bmatrix} 1 & 2 \\ 0 & -1 \end{bmatrix} = (1)(-1) - (2)(0) = -1 \) - \( M_{33} = \det\begin{bmatrix} 1 & 0 \\ 0 & -3 \end{bmatrix} = (1)(-3) - (0)(0) = -3 \) Thus, the matrix of minors is: \[ M = \begin{bmatrix} 5 & 2 & 6 \\ 2 & -6 & -1 \\ 6 & -1 & -3 \