Question

Cayley Hamilton theorem for the matrix =
[3 1
1 4]

Answer 1

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

Given matrix is

$\red{\rm :\longmapsto\:A = \bigg[ \begin{matrix}3&1 \\ 1&4 \end{matrix} \bigg]}$

Cayley Hamilton Theorem states that Every square matrix A must satisfy its characteristic equation | A - kI |.

So, first find characteristic equation.

Consider,

$\rm :\longmapsto\: A - kI$

$\rm \:  =  \:  \: \bigg[ \begin{matrix}3&1 \\ 1&4 \end{matrix} \bigg] - k\bigg[ \begin{matrix}1&0 \\ 0&1 \end{matrix} \bigg]$

$\rm \:  =  \:  \: \bigg[ \begin{matrix}3&1 \\ 1&4 \end{matrix} \bigg] - \bigg[ \begin{matrix}k&0 \\ 0&k \end{matrix} \bigg]$

$\rm \:  =  \:  \: \bigg[ \begin{matrix}3 - k&1 \\ 1&4 - k \end{matrix} \bigg]$

So,

Characteristic equation is given by

$\rm :\longmapsto\: |A - kI| = 0$

$\rm :\longmapsto\:\begin{array}{|cc|}\sf 3 - k &\sf 1 \\ \sf 1 &\sf 4 - k \\\end{array} = 0$

$\rm :\longmapsto\:(3 - k)(4 - k) - 1 = 0$

$\rm :\longmapsto\:12 - 3k - 4k + {k}^{2} - 1 = 0$

$\rm :\longmapsto\: {k}^{2} - 7k + 11 = 0$

So, We have to show that A must satisfy

$\rm :\longmapsto\: {k}^{2} - 7k + 11 = 0$

That is

$\rm :\longmapsto\: {A}^{2} - 7A + 11I = 0$

So, Consider

$\rm :\longmapsto\: {A}^{2} - 7A + 11I$

$\rm\:=\bigg[ \begin{matrix}3&1 \\ 1&4 \end{matrix} \bigg]\bigg[ \begin{matrix}3&1 \\ 1&4 \end{matrix} \bigg] - 7\bigg[ \begin{matrix}3&1 \\ 1&4 \end{matrix} \bigg] + 11\bigg[ \begin{matrix}1&0 \\ 0&1 \end{matrix} \bigg]$

$\rm\:=\bigg[ \begin{matrix}9 + 1&3 + 4 \\ 3 + 4&1 + 16 \end{matrix} \bigg] - \bigg[ \begin{matrix}21&7 \\ 7&28 \end{matrix} \bigg] + \bigg[ \begin{matrix}11&0 \\ 0&11 \end{matrix} \bigg]$

$\rm\:=\bigg[ \begin{matrix}10&7 \\ 7&17 \end{matrix} \bigg] + \bigg[ \begin{matrix} - 10& - 7 \\ - 7& - 17 \end{matrix} \bigg]$

$\rm \:  =  \:  \: \bigg[ \begin{matrix}0&0 \\ 0&0 \end{matrix} \bigg]$

Hence, Verified