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TS EAMCET 20-July-2022 Shift 1 Solved Paper
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© examsnet.com
Question : 5
Total: 160
Let
A
=
(
2
1
1
0
1
0
1
1
2
)
. If
A
−
1
=
α
A
2
+
β
A
+
γ
I
, where
α
,
β
,
γ
are real numbers and
I
is a
3
×
3
identity matrix, then
17
α
+
5
β
+
γ
=
−
1
−
1
3
2
3
3
Validate
Solution:
👈: Video Solution
Given,
A
=
[
2
1
1
0
1
0
1
1
2
]
∵
A
−
1
=
α
A
2
+
β
A
+
γ
I
Multiply by
A
both sides,
A
−
1
⋅
A
=
α
A
3
+
β
A
2
+
γ
I
⋅
A
I
=
α
A
3
+
β
A
2
+
γ
A
For matrix
A
det
(
A
−
λ
I
)
=
0
|
2
−
λ
1
1
0
1
−
λ
0
1
1
2
−
λ
|
=
0
Expanding along
R
1
(
2
−
λ
)
(
1
−
λ
)
(
2
−
λ
)
−
1
(
0
)
+
1
(
λ
−
1
)
=
0
⇒
(
1
−
λ
)
(
λ
2
−
4
λ
+
4
)
+
λ
−
1
=
0
⇒
λ
2
−
4
λ
+
4
−
λ
3
+
4
λ
2
−
4
λ
+
λ
−
1
=
0
⇒
−
λ
3
+
5
λ
2
−
7
λ
+
3
=
0
⇒
1
3
λ
3
−
5
3
λ
2
+
7
3
λ
=
1
Compare with
α
A
3
+
β
A
2
+
γ
A
=
I
,
So,
α
=
1
3
,
β
=
−
5
3
,
γ
=
7
3
∴
17
α
+
5
β
+
γ
=
17
3
−
25
3
+
7
3
=
−
1
3
© examsnet.com
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