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Question

# Let A=⎡⎢⎣100101010⎤⎥⎦ satisfies An=An−2+A2−I for n≥3. And trace of a square matrix X is equal to the sum of elements in its principal diagonal. Further consider a matrix ∪3×3 with its column as ∪1,∪2,∪3 such that A50 ∪1=⎡⎢⎣12525⎤⎥⎦,A50 ∪2=⎡⎢⎣010⎤⎥⎦, A50 ∪3=⎡⎢⎣001⎤⎥⎦ Then, Trace of A50 equals

A
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B
1
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C
2
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D
3
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Solution

## The correct option is D 3An−An−2=A2−I ⇒A50=A48+A2−I Further, A48=A46+A2−I A46=A44+A2−I ⋮ ⋮ ⋮ A4=A2+A2−I _________________ A50=25A2−24I Here, A2=⎡⎢⎣100101010⎤⎥⎦⎡⎢⎣100101010⎤⎥⎦ =⎡⎢⎣100110101⎤⎥⎦ ⇒A50=⎡⎢⎣25002525025025⎤⎥⎦−24⎡⎢⎣100010001⎤⎥⎦ =⎡⎢⎣10025102501⎤⎥⎦ ⇒tr(A50)=1+1+1=3.

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