Solution:

$A=\begin{bmatrix} 1 & 4 \\ 2 & 3 \\ \end{bmatrix}$

characteristic equation A is given by,

$|A-\lambda I|=0$

$\left | \begin{matrix} 1-\lambda & 4 \\ 2 & 3-\lambda \\ \end{matrix} \right |$

$(1-\lambda)(3-\lambda)-8=0$

$3-\lambda -3\lambda+\lambda^2-8=0$

$\lambda^2-4\lambda-5=0-----(1)$

by cayley-hamilton Theorem, This equation is satisfied by A

$A^{2}-4A-5I=0----(2)$

$A^2 =A.A=\begin{bmatrix} 9 & 16 \\ 8 & 17 \\ \end{bmatrix}$----(3)

$I=\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}$----(4)

$A=\begin{bmatrix} 1 & 4 \\ 2 & 3 \\ \end{bmatrix}$----(5)

Put eqn (3),(4),(5) in (2)

$\begin{bmatrix} 9 & 16 \\ 8 & 17 \\ \end{bmatrix}-4\begin{bmatrix} 1 & 4 \\ 2 & 3 \\ \end{bmatrix}-5\begin{bmatrix} 1 & 0 \\ 0 & 1\\ \end{bmatrix}$

$\begin{bmatrix} 5 & 0 \\ 0 & 5 \\ \end{bmatrix}-\begin{bmatrix} 5 & 0 \\ 0 & 5 \\ \end{bmatrix}$

$\begin{bmatrix} 0 & 0 \\ 0 & 0 \\ \end{bmatrix}$

Hence cayley hamilton theorem verified

To find $A^{-1}$, Multiply eqn(2) by $A^{-1}$ we get

$A-4I-5A^{-1}=0$------(6)

$5A^{-1} = A-4I$

put eqn(4) & (5) in eqn(6)

$5A^{-1} = \begin{bmatrix} 1 & 4 \\ 2 & 3 \\ \end{bmatrix}-4\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}$

$5A^{-1}=\begin{bmatrix} -3 & 4 \\ 2 & -1 \\ \end{bmatrix}$

$5A^{-1}=\frac{1}{5}\begin{bmatrix} -3 & 4 \\ 2 & -1 \\ \end{bmatrix}$

Now to find ,

$A^{5} - 4A^{4} - 7A^{3} +11A^{2}-A-10I$

Convert above eqn in terms of $\lambda $

$\lambda^{5} - 4\lambda^{4} - 7\lambda^{3} +11\lambda^{2}-\lambda-10$

Now divide ,

$\lambda^{5} - 4\lambda^{4} - 7\lambda^{3} +11\lambda^{2}-\lambda-10$ by eqn (1)

$dividend = (divisor \times quotient )+Remainder $

$\lambda^5-4\lambda^4-7\lambda^3+11\lambda^2-\lambda-10=(\lambda^2-4\lambda-5)(\lambda^3+2\lambda+3)+(\lambda+5)$

by cayley hamilton theorem above equation is satisfied by A..

$A^5-4A^4-7A^3+11A^2-A-10I=(A^2-4A-5I)(A^3+2A+3I)+(A+5I)$

But from equation (2) value of $(A^2-4A-5I)=0$

$A^5-4A^4-7A^3+11A^2-A-10I=0(A^3+2A+3I)+(A+5I)$

$A^5-4A^4-7A^3+11A^2-A-10I=A+5I$