Find the inverse of matrix \(\begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix}\)

Answer: \(A^{-1} = \begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix}\), i.e. \(A^{-1} = A\).

\(|A| = -4(0 - 4) + 3(3 - 4) - 3(4 - 0) = 16 - 3 - 12 = 1 \ne 0\), so \(A^{-1}\) exists.

Cofactors:

• \(A_{11} = 0 - 4 = -4\), \(A_{12} = -(3 - 4) = 1\), \(A_{13} = 4 - 0 = 4\)
• \(A_{21} = -(-9 + 12) = -3\), \(A_{22} = -12 + 12 = 0\), \(A_{23} = -(-16 + 12) = 4\)
• \(A_{31} = -3 - 0 = -3\), \(A_{32} = -(-4 + 3) = 1\), \(A_{33} = 0 + 3 = 3\)

\(\mathrm{adj}\,A = \begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix}\) (transpose of the cofactor matrix).

\(A^{-1} = \frac{1}{|A|}\mathrm{adj}\,A = \begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix}\).

Check: the first row of \(A\) times the first column of \(A\) is \(16 - 3 - 12 = 1\) and times the second column is \(12 + 0 - 12 = 0\); in the same way \(A\cdot A = I\), so \(A^{-1} = A\).