UPSC NDA Math Model Paper 2

Let us find the characteristic polynomial p(λ) of the given matrix A=[

1	5
4	2

]
p(λ)=|λ==I−==A|
⇒p(λ)=|

λ−1	5
4	λ−2

|
⇒p(λ)=(λ−1)(λ−2)−(4)(5)
⇒p(λ)=λ2−2λ−λ+2−20
⇒p(λ)=λ2−3λ−18
According to the Cayley - Hamilton theorem, p(A)=0.
⇒A2−3A−18I=0
Now, the expression given to evaluate is A2−4A−12I, which can be written as:
(A2−3A−18I)−A+6I
=0−[

1	5
4	2

]+[

6	0
0	6

]
=[

5	−5
−4	4

]