Question

If \[ A = \begin{bmatrix} 3 & 1 & 1 \\ 15 & 6 & 5 \\ 5 & 2 & 2 \end{bmatrix} \] then find \( A^{-1} \).

Hint:

To find \( A^{-1} \), use \( A^{-1} = \frac{1}{\det(A)} \text{adj}(A) \), where adjoint is the transpose of the cofactor matrix.

Answer 1

The inverse of a \( 3 \times 3 \) matrix is given by: \[ A^{-1} = \frac{1}{\det(A)} \text{adj}(A) \] Computing \( \det(A) \): \[ \det(A) = 3(6 \cdot 2 - 5 \cdot 2) - 1(15 \cdot 2 - 5 \cdot 5) + 1(15 \cdot 2 - 6 \cdot 5) = 3(12 - 10) - (30 - 25) + (30 - 30) = 6 - 5 = 1. \] Since \( \det(A) = 1 \), we compute \( \text{adj}(A) \) and find: \[ A^{-1} = \begin{bmatrix} 4 & -1 & -1 \\ -5 & 2 & 1 \\ 5 & -2 & -1 \end{bmatrix}. \]