Find inverse of matrix by using Cayley Hamilton Theorem.
A = \( ~ \begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 1\\
0 & -2 & 4\\
\end{bmatrix} \)

By Cayley Hamliton Theorem
We have ?³ – S₁ ?² + S₂ ? – |A| = 0 characteristics equation of matrix A
Also A satisfies above equation according to theorem.
? A³– S₁A²+ S₂A – 6 = 0……… (i) Where S₁=6, S₂=6 & |A|=6
? A³ – 6A² + 6A – 6 = 0
Multiplying equation (i) by A^-1
A^-1 (A³) – 6 A^-1 (A²) + A^-1 (A) – 6 A^-1 = 0
=> A² – 6A + 6I – 6 A^-1 = \(\frac{1}{6}\) A^-1
=> A^-1 = \(\frac{1}{6} \Bigg\{\begin{bmatrix}
1 & 0 & 0\\
0 & -1 & 5\\
0 & -10 & 14\\
\end{bmatrix} – \begin{bmatrix}
6 & 0 & 0\\
0 & 6 & 6\\
0 & -12 & -24\\
\end{bmatrix} + \begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{bmatrix} \Bigg\} \)
\(? A^{-1} = \frac{1}{6} \begin{bmatrix}
1 & 0 & 0\\
0 & -1 & -1\\
0 & 2 & -4\\
\end{bmatrix} \).