Question

A = [1 3 2 5]
Find A-1 using Cayley -Hamilton theorem.​

Answer 1

$\large\underline{\sf{Solution-}}$

Cayley Hamilton Theorem states that

• Every square matrix 'A' satisfy its characteristic equation where characteristic equation is given by

$\sf \: |A - \lambda \: I| = 0$

Now,

Given matrix,

$\rm :\longmapsto\:A = \: \begin{bmatrix} 1 & 3\\ 2 & 5\end{bmatrix}$

So,

$\rm :\longmapsto\:A - \lambda \:I = \: \begin{bmatrix} 1 & 3\\ 2 & 5\end{bmatrix} - \lambda \:\: \begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}$

$\rm :\longmapsto\:A - \lambda \:I = \: \begin{bmatrix} 1 & 3\\ 2 & 5\end{bmatrix} - \:\: \begin{bmatrix} \lambda \: & 0\\ 0 & \lambda \:\end{bmatrix}$

$\rm :\longmapsto\:A - \lambda \:I = \: \begin{bmatrix} 1 - \lambda \: & 3\\ 2 & 5 - \lambda \:\end{bmatrix}$

So,

$\rm :\longmapsto\:|A - \lambda \: I| = (1 - \lambda \:)(5 - \lambda \:) - 6$

$\rm :\longmapsto\:|A - \lambda \: I| =5 - \lambda \: - 5\lambda \: + {\lambda \:}^{2} - 6$

$\rm :\longmapsto\:|A - \lambda \: I| = {\lambda \:}^{2} - 6\lambda \: - 1$

Now, Cayley Hamilton Theorem states A must satisfy its characteristic equation,

So, we have to verify that,

$\rm :\longmapsto\: {A}^{2} - 6A - I = 0$

Now, Consider

$\rm :\longmapsto\: {A}^{2} = A \times A = \: \begin{bmatrix} 1 & 3\\ 2 & 5\end{bmatrix} \times \: \begin{bmatrix} 1 & 3\\ 2 & 5\end{bmatrix}$

$\sf \: \: \: = \: \: \: \: \begin{bmatrix} 1 + 6 & 3 + 15\\ 2 + 10 & 6 + 25\end{bmatrix}$

$\sf \: \: \: = \: \: \: \: \begin{bmatrix} 7 & 18\\ 12 & 31\end{bmatrix}$

Now,

Consider,

$\rm :\longmapsto\: {A}^{2} - 6A - I$

$\sf \: \: \: = \: \: \: \: \begin{bmatrix} 7 & 18\\ 12 & 31\end{bmatrix} - 6\: \begin{bmatrix} 1 & 3\\ 2 & 5\end{bmatrix} - \: \begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}$

$\sf \: \: \: = \: \: \: \: \begin{bmatrix} 7 & 18\\ 12 & 31\end{bmatrix} - \: \begin{bmatrix} 6 & 18\\ 12 & 30\end{bmatrix} - \: \begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}$

$\sf \: \: \: = \: \: \: \: \begin{bmatrix} 0 & 0\\ 0 & 0\end{bmatrix}$

$\bf\implies \: {A}^{2} - 6A - I = 0$

Now, to find inverse of matrix A,

$\boxed{ \bf \: Pre - multiply \: {A}^{ - 1} \: on \: both \: sides}$

$\rm :\longmapsto\: {A}^{ - 1} {A}^{2} - 6 {A}^{ - 1} A - {A}^{ - 1} I = 0$

$\rm :\longmapsto\:A - 6I - {A}^{ - 1} = 0$

$\rm :\longmapsto\: {A}^{ - 1} = A - 6I$

$\rm :\longmapsto\: {A}^{ - 1} = \: \begin{bmatrix} 1 & 3\\ 2 & 5\end{bmatrix} - 6\: \begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}$

$\rm :\longmapsto\: {A}^{ - 1} = \: \begin{bmatrix} 1 & 3\\ 2 & 5\end{bmatrix} - \: \begin{bmatrix} 6 & 0\\ 0 & 6\end{bmatrix}$

$\rm :\longmapsto\: {A}^{ - 1} = \: \begin{bmatrix} - 5 & 3\\ 2 & - 1\end{bmatrix}$

$\boxed{\bf \: Hence \: \: \: {A}^{ - 1} = \: \begin{bmatrix} - 5 & 3\\ 2 & - 1\end{bmatrix}}$