Let $A=\left[\begin{array}{ccc}1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5\end{array}\right]$. Verify that $\left(A^{-1}\right)^{-1}=A$.

(N/A) Given $A=\left[\begin{array}{ccc}1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5\end{array}\right]$.
First,calculate the determinant $|A|$:
$|A| = 1(15-1) - (-2)(-10-1) + 1(-2-3) = 1(14) + 2(-11) + 1(-5) = 14 - 22 - 5 = -13$.
Next,find the matrix of cofactors $C_{ij}$:
$C_{11} = +(15-1) = 14, C_{12} = -( -10-1) = 11, C_{13} = +(-2-3) = -5$.
$C_{21} = -(-10-1) = 11, C_{22} = +(5-1) = 4, C_{23} = -(1+2) = -3$.
$C_{31} = +(-2-3) = -5, C_{32} = -(1+2) = -3, C_{33} = +(3-4) = -1$.
Thus,$adj(A) = \left[\begin{array}{ccc}14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1\end{array}\right]$.
$A^{-1} = \frac{1}{|A|} adj(A) = -\frac{1}{13} \left[\begin{array}{ccc}14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1\end{array}\right] = \frac{1}{13} \left[\begin{array}{ccc}-14 & -11 & 5 \\ -11 & -4 & 3 \\ 5 & 3 & 1\end{array}\right]$.
Now,to find $(A^{-1})^{-1}$,we find the determinant and adjoint of $A^{-1}$.
$|A^{-1}| = \frac{1}{|A|} = -\frac{1}{13}$.
The adjoint of $A^{-1}$ is given by $adj(A^{-1}) = \frac{1}{|A|^{n-1}} A = \frac{1}{(-13)^2} A = \frac{1}{169} \left[\begin{array}{ccc}1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5\end{array}\right]$.
Finally,$(A^{-1})^{-1} = \frac{adj(A^{-1})}{|A^{-1}|} = \frac{\frac{1}{169} A}{-\frac{1}{13}} = -\frac{13}{169} A = -\frac{1}{13} A = -\frac{1}{13} \left[\begin{array}{ccc}1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5\end{array}\right] \times (-13) = A$.
Hence,$\left(A^{-1}\right)^{-1}=A$ is verified.