According to Cayley-Hamilton Theorem ,
"Every Square Matrix satisfies its own characteristic equation"
Ref :- https://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem
So , for the given matrix :-
characteristic equation is :- (1-λ)(3-λ) - 4*2 = 0 where λ = eigen value
So, λ2 - 4λ -5 = 0
Now , given square matrix will satisfy this characteristic equation.
So, A2 - 4A - 5I = 0 -------- equation (1)
now , here polynomial is :-
A5−4A4−7A3+11A2−2A+kI=0
A5 - 4A4 - 5A3 - 2A3 + 11A2 - 2A +kI = 0
A3(A2-4A - 5I) - 2A3 + 11A2 - 2A +kI = 0
-2A3 + 11A2 - 2A +kI = 0 (Since , according to equation (1) , A2 - 4A - 5I = 0 )
-2A3 + 8A2 +10A +3A2 -2A - 10A+kI = 0
-2A(A2 -4A - 5I) +3A2 -2A - 10A+kI = 0
3A2 -12A +kI = 0 (According to equation (1))
3A2 -12A -15I +15I +kI = 0
3(A2-4A - 5I) +15I +kI = 0
15I + kI = 0 (According to equation (1))
(15+k)I = 0
now put Identity matrix 'I' here and after solving it ,we will get , k = -15