Let λ be eigen value of matrix A

Characteristic equation is [A- λ I ] =0

$ \therefore \left| \begin{array}{ccc} 1-λ & 4 \\ 2 & 3-λ \end{array}\right| \;=\; 0 \\ \; \\ \; \\ ∴ (1- λ) (3- λ)-8=0 \\ \; \\ \; \\ ∴ 3 -λ-3λ+ λ^2-8 = 0 \\ \; \\ \; \\ ∴ λ^2-4 λ-5=0--------(1) $

Cayley Hamilton theorem states that the characteristic equation is satisfied by A

∴ Replacing λ with A

∴ $A^2-4 A-5I=0$--------(2)

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

∴ Dividing Equation (3) by Equation (2)

$ \hspace{2.5 cm} A^3-2 A+3 \\ A^2-4 A-5) \overline{ \; A^5- 4A^4-7 A^3+11 A^2-A-10I }\\ \; \; \; \; \; \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ A^5-4 A^4-5A^3 \\ \hspace{2.2 cm} - \; \; + \; \; \; \; \; \; + \; \; \; \; \ \ \ \ \ \\ \hspace{2 cm} \overline{\hspace{2 cm} -2 A^3+11 A^2-A-10I } \\ \hspace{3.8 cm} -2 A^3+8 A^2+10A \\ \hspace{4 cm} + \ \ \ - \ \ \ \ \ \ \ \ - \\ \hspace{2 cm} \overline{\hspace{3 cm} 3 A^2-11A-10I} \\ \hspace{5 cm} 3 A^2-12A-15I \\ \hspace{5 cm} - \ \ \ + \ \ \ + \\ \hspace{2 cm} \overline{\hspace{4 cm} A + 5 \hspace{2 cm}} \\ \; \\ \; \\ $

∴ Dividend = Divisor * Quotient + Remainder

∴ $A^5- A^4-7 A^3+11 A^2-A-10 \\ =(A^2-4 A-5)*(A^3-2 A+3)+(A+5)\\ = 0 * (A^3-2 A+3)+(A+5) -------(from (2)) \\ = (A+5) \\ \; \\$ ∴ $A^5- 4A^4-7 A^3+11 A^2-A-10 I=A+5I
$