For the matrix $A = \begin{bmatrix} 4 & -4 & 5 \\ -2 & 3 & -3 \\ 3 & -3 & 4 \end{bmatrix}$ find $A^{-2}$.
Step-by-Step Solution
Key Concept: To find $A^{-2}$, first calculate the inverse matrix $A^{-1}$ using the formula $A^{-1} = \frac{1}{|A|} \text{adj } A$, and then compute the matrix product of $A^{-1}$ with itself.
<div>$A^{-2} = A^{-1} \times A^{-1}$<br>Given $A = \begin{bmatrix} 4 & -4 & 5 \\ -2 & 3 & -3 \\ 3 & -3 & 4 \end{bmatrix}$<br>Cofactor matrix $[C_{ij}] = \begin{bmatrix} 3 & -1 & -3 \\ 1 & 1 & 0 \\ -3 & 2 & 4 \end{bmatrix}$<br>$\text{adj } A = [C_{ij}]^T = \begin{bmatrix} 3 & 1 & -3 \\ -1 & 1 & 2 \\ -3 & 0 & 4 \end{bmatrix}$<br>$|A| = \begin{vmatrix} 4 & -4 & 5 \\ -2 & 3 & -3 \\ 3 & -3 & 4 \end{vmatrix} = 4(3) + 4 - 15 = 1$<br>$A^{-1} = \frac{\text{adj } A}{|A|} = \begin{bmatrix} 3 & 1 & -3 \\ -1 & 1 & 2 \\ -3 & 0 & 4 \end{bmatrix}$<br>$A^{-2} = A^{-1} A^{-1} = \begin{bmatrix} 3 & 1 & -3 \\ -1 & 1 & 2 \\ -3 & 0 & 4 \end{bmatrix} \begin{bmatrix} 3 & 1 & -3 \\ -1 & 1 & 2 \\ -3 & 0 & 4 \end{bmatrix} = \begin{bmatrix} 17 & 4 & -19 \\ -10 & 0 & 13 \\ -21 & -3 & 25 \end{bmatrix}$</div>
Correct Answer: $\begin{bmatrix} 17 & 4 & -19 \\ -10 & 0 & 13 \\ -21 & -3 & 25 \end{bmatrix}$