Let $A = \begin{bmatrix} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix}$. Verify that $[adj A]^{-1} = adj(A^{-1})$.

(N/A) Given $A = \begin{bmatrix} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix}$.
First,calculate the determinant $|A| = 1(15 - 1) + 2(-10 - 1) + 1(-2 - 3) = 14 - 22 - 5 = -13$.
The matrix of cofactors is calculated as:
$C_{11} = 14, C_{12} = 11, C_{13} = -5$
$C_{21} = 11, C_{22} = 4, C_{23} = -3$
$C_{31} = -5, C_{32} = -3, C_{33} = -1$
Thus,$adj A = \begin{bmatrix} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{bmatrix}$.
Now,$[adj A]^{-1} = \frac{1}{|adj A|} adj(adj A)$.
$|adj A| = 14(-4 - 9) - 11(-11 - 15) - 5(-33 + 20) = 14(-13) - 11(-26) - 5(-13) = -182 + 286 + 65 = 169$.
The adjoint of $adj A$ is $\begin{bmatrix} -13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65 \end{bmatrix}$.
So,$[adj A]^{-1} = \frac{1}{169} \begin{bmatrix} -13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65 \end{bmatrix} = \frac{1}{13} \begin{bmatrix} -1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5 \end{bmatrix}$.
Next,$A^{-1} = \frac{1}{|A|} adj A = -\frac{1}{13} \begin{bmatrix} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{bmatrix} = \frac{1}{13} \begin{bmatrix} -14 & -11 & 5 \\ -11 & -4 & 3 \\ 5 & 3 & 1 \end{bmatrix}$.
Calculating $adj(A^{-1})$ involves finding the cofactors of $A^{-1}$,which results in $\frac{1}{13} \begin{bmatrix} -1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5 \end{bmatrix}$.
Thus,$[adj A]^{-1} = adj(A^{-1})$ is verified.