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TS EAMCET 7 May 2018 Shift 1 Solved Paper
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© examsnet.com
Question : 3
Total: 160
If A =
[
1
1
0
1
]
and I =
[
1
0
0
1
]
, then for all n ∊ N
A
n
= nA
A
n
= nA + (n - 1) I
A
n
= (n - 1) A - nI
A
n
= nA - (n - 1) I
Validate
Solution:
The square of the matrix A,
A
2
=
A
.
A
=
[
1
1
0
1
]
.
[
1
1
0
1
]
=
2
[
1
1
0
1
]
−
[
1
0
0
1
]
=
2
A
−
(
2
−
1
)
I
The cube of the matrix A,
A
3
=
A
2
.
A
=
[
1
2
0
1
]
[
1
1
0
1
]
=
3
[
1
1
0
1
]
−
2
[
1
0
0
1
]
=
3
A
−
(
3
−
1
)
I
So,
A
n
=
n
A
−
(
n
−
1
)
I
© examsnet.com
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