Question

If matrix \[ A = \begin{bmatrix} 1 & 1 & 3 \\ 1 & 3 & -3 \\ -2 & -4 & -4 \end{bmatrix}, \] then find \( A^{-1} \).

Hint:

The inverse of a matrix is given by \( A^{-1} = \frac{\text{adj}(A)}{\det(A)} \), where adj(A) is the adjugate matrix.

Answer 1

Step 1: Compute determinant of \( A \). \[ \det(A) = \begin{vmatrix} 1 & 1 & 3 \\ 1 & 3 & -3 \\ -2 & -4 & -4 \end{vmatrix} \] Expanding along first row: \[ = 1 \begin{vmatrix} 3 & -3 \\ -4 & -4 \end{vmatrix} - 1 \begin{vmatrix} 1 & -3 \\ -2 & -4 \end{vmatrix} + 3 \begin{vmatrix} 1 & 3 \\ -2 & -4 \end{vmatrix} \] \[ = 1(3(-4) - (-3)(-4)) - 1(1(-4) - (-3)(-2)) + 3(1(-4) - 3(-2)) \] \[ = 1(-12 - 12) - 1(-4 - 6) + 3(-4 + 6) \] \[ = 1(-24) - 1(-10) + 3(2) = -24 + 10 + 6 = -8 \] Step 2: Compute adjugate and inverse using \( A^{-1} = \frac{\text{adj}(A)}{\det(A)} \). After calculation, \[ A^{-1} = \frac{1}{-8} \begin{bmatrix} -12 & 6 & -2 \\ -2 & -4 & 2 \\ -4 & 2 & 2 \end{bmatrix} \] \[ = \begin{bmatrix} \frac{12}{8} & -\frac{6}{8} & \frac{2}{8} \\ \frac{2}{8} & \frac{4}{8} & -\frac{2}{8} \\ \frac{4}{8} & -\frac{2}{8} & -\frac{2}{8} \end{bmatrix} \] \[ = \begin{bmatrix} \frac{3}{2} & -\frac{3}{4} & \frac{1}{4} \\ \frac{1}{4} & \frac{1}{2} & -\frac{1}{4} \\ \frac{1}{2} & -\frac{1}{4} & -\frac{1}{4} \end{bmatrix} \]