Let's use the **adjoint method** to find the inverse of the matrix \( A \): \[ A = \begin{bmatrix} 2 & 3 & 1 \\ 0 & 4 & 1 \\ 3 & -3 & 2 \end{bmatrix} \] ### **Step 1: Find the Determinant of \( A \) and the Matrix of Cofactors** #### **Determinant of \( A \):** \[ |A| = 2 \begin{vmatrix} 4 & 1 \\ -3 & 2 \end{vmatrix} - 3 \begin{vmatrix} 0 & 1 \\ 3 & 2 \end{vmatrix} + 1 \begin{vmatrix} 0 & 4 \\ 3 & -3 \end{vmatrix} \] Calculate each minor: - \( \begin{vmatrix} 4 & 1 \\ -3 & 2 \end{vmatrix} = (4)(2) - (1)(-3) = 8 + 3 = 11 \) - \( \begin{vmatrix} 0 & 1 \\ 3 & 2 \end{vmatrix} = (0)(2) - (1)(3) = 0 - 3 = -3 \) - \( \begin{vmatrix} 0 & 4 \\ 3 & -3 \end{vmatrix} = (0)(-3) - (4)(3) = 0 - 12 = -12 \) Plug them back in: \[ |A| = 2(11) - 3(-3) + 1(-12) = 22 + 9 - 12 = 19 \] **Explanation:** The determinant is necessary because the formula for the inverse is \(\frac{1}{|A|} \text{adj}(A)\). If \(|A| = 0\), the matrix is not invertible. --- ### **Step 2: Find the Adjoint of \( A \) and Then the Inverse** #### **Cofactor Matrix:** Let's calculate each cofactor: - \(C_{11} = +\begin{vmatrix} 4 & 1 \\ -3 & 2 \end{vmatrix} = 11\) - \(C_{12} = -\begin{vmatrix} 0 & 1 \\ 3 & 2 \end{vmatrix} = -(-3) = 3\) - \(C_{13} = +\begin{vmatrix} 0 & 4 \\ 3 & -3 \end{vmatrix} = -12\) - \(C_{21} = -\begin{vmatrix} 3 & 1 \\ -3 & 2 \end{vmatrix} = -[(3)(2) - (1)(-3)] = -[6 + 3] = -9\) - \(C_{22} = +\begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix} = (2)(2) - (1)(3) = 4 - 3 = 1\) - \(C_{23} = -\begin{vmatrix} 2 & 3 \\ 3 & -3 \end{vmatrix} = -[(2)(-3) - (3)(3)] = -[-6 - 9] = 15\) - \(C_{31} = +\begin{vmatrix} 3 & 1 \\ 4 & 1 \end{vmatrix} = (3)(1) - (1)(4) = 3 - 4 = -1\) - \(C_{32} = -\begin{vmatrix} 2 & 1 \\ 0 & 1 \end{vmatrix} = -[(2)(1) - (1)(0)] = -[2 - 0] = -2\) - \(C_{33} = +\begin{vmatrix} 2 & 3 \\ 0 & 4 \end{vmatrix} = (2)(4) - (3)(0) = 8 - 0 = 8\) So the cofactor matrix is: \[ \begin{bmatrix} 11 & 3 & -12 \\ -9 & 1 & 15 \\ -1 & -2 & 8 \end{bmatrix} \] Take its transpose (adjugate/adjoint matrix): \[ \text{adj}(A) = \begin{bmatrix} 11 & -9 & -1 \\ 3 & 1 & -2 \\ -12 & 15 & 8 \end{bmatrix} \] **Explanation:** The adjoint (or adjugate) matrix is the transpose of the matrix of cofactors. --- ### **Final Answer: Inverse of \( A \)** \[ A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{19} \begin{bmatrix} 11 & -9 & -1 \\ 3 & 1 & -2 \\ -12 & 15 & 8 \end{bmatrix} \] Or explicitly: \[ A^{-1} = \begin{bmatrix} \frac{11}{19} & \frac{-9}{19} & \frac{-1}{19} \\ \frac{3}{19} & \frac{1}{19} & \frac{-2}{19} \\ \frac{-12}{19} & \frac{15}{19} & \frac{8}{19} \end{bmatrix} \]