Matrices and Determinants Part 2

Given matrix,
A=[

0	2
K	−1

]
Characteristic equation of A is
|A−λI|=0
⇒|

−λ	2	‌
K	−1	−λ

|=0
⇒λ(1+λ)−2k=0
⇒λ2+λ−2k=0
∵ Every square matrix satisfied its own characteristic equation.
∴ A2+A−2‌KI=0
⇒A2=2‌KI−A
⇒A4=4K2I+A2−2(2‌KI)(A)
⇒A4=4K2I+2‌KI−A−4‌KA
⇒A4=(4K2+2K)I−(1+4K)A ...(i)
Now, A(A3+3I)=2I
⇒A4=2I−3A ...(ii)
Comparing Eqs. (i) and (ii), we get
1+4K=3
⇒K=

1

2