AP EAPCET 19 May 2024 Shift 2 Solved Paper

To solve the given equation αA2+βA=2I where A=[

1	2
−2	−5

], we first need to compute A2 :
A2=[

1	2
−2	−5

][

1	2
−2	−5

]
Calculating the matrix multiplication:
A2=[

(1)(1)+(2)(−2)	(1)(2)+(2)(−5)
(−2)(1)+(−5)(−2)	(−2)(2)+(−5)(−5)

]=[

−3	−8
8	21

]
Next, substitute A2 into the equation αA2+βA=2I :
α[

−3	−8
8	21

]+β[

1	2
−2	−5

]=[

2	0
0	2

]
Break this down into:
[

−3α	−8α
8α	21α

]+[

β	2β
−2β	−5β

]=[

2	0
0	2

]
Equate the corresponding elements from both sides:
‌−3α+β=2
‌−8α+2β=0
‌8α−2β=0
‌21α−5β=2
From the second equation −8α+2β=0, we solve for β :
2β=8α⇒β=4α
Substitute β=4α into the first equation:
−3α+4α=2⇒α=2
Using β=4α, we find:
β=4(2)=8
Finally, calculate α+β :
α+β=2+8=10