If A is a square matrix that |A|=2 then for any positive integer n,|A^n|=
Answers
Answered by
1
Answer:
We have,
A
2
=A
∴(I+A)
2
=(I+A)(I+A)=I+2A+A
2
=I+3A
and (I+A)
3
=(I+A)
2
(I+A)
=(I+3A)(I+A) ....... [∵(I+A)
2
=I+3A]
=I+4A+3A
2
=I+7A [∵A
2
=A]
Thus, we have
(I+A)
2
=I+3A and (I+A)
3
=I+7A
⇒(I+A)
2
=I+(2
2
−1)A
and (I+A)
3
=I+(2
3
−1)A
Hence, (I+A)
n
=I+(2
n
−1)A
∴(I+A)
n
=I+λA
⇒λ=2
n
−1
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Answered by
1
Answer:
2^n will be the answer according to me
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