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If `A=[1a0 1]`, then `A^n`(where `n in N)`equals`[1n a0 1]`(b) `[1n^2a0 1]`(c) `[1n a0 0]`(d) `[nn a0n]`A. `{:A=[(1,na),(0,1)]:}`B. `{:A=[(1,n^2a),(0,1)]:}`C. `{:A=[(1,na),(0,0)]:}`D. `{:A=[(1,2a),(0,n)]:}` |
Answer» Correct Answer - B We have, `{:A=[(1,a),(0,1)]:}` `:.A^2={:[(1,a),(0,1)][(1,a),(0,1)]=[(1,2a),(0,1)]:}` It can proved by the principle of mathematical induction that `A^n={:[(1,na),(0,1)]:}` for all `n in N. |
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