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Matrices & Determinants: Consider Matrix Pmatrix Pmatrix Equals Unit Matrix Order

JEE Maths question with a full step-by-step solution.

Question
Consider a matrix A=(31−6−2)A = \begin{pmatrix}3 & 1\\ -6 & -2\end{pmatrix}, then (1+A)99(1+A)^{99} equals, where II is a unit matrix of order 2:
AI+298AI + 2^{98}A
BI+299AI + 2^{99}A
CI+(299+1)AI + (2^{99}+1)A
DI+(299−1)AI + (2^{99}-1)Acorrect
Solution
Step 1: Show that AA is idempotent
A2=(31−6−2)(31−6−2)=(9−63−2−18+12−6+4)=(31−6−2)=AA^2 = \begin{pmatrix}3 & 1\\ -6 & -2\end{pmatrix}\begin{pmatrix}3 & 1\\ -6 & -2\end{pmatrix} = \begin{pmatrix}9-6 & 3-2\\ -18+12 & -6+4\end{pmatrix} = \begin{pmatrix}3 & 1\\ -6 & -2\end{pmatrix} = A
 Since A2=AA^2 = A, we have An=AA^n = A for all n≥1n \geq 1. Step 2: Expand using the binomial theorem
(I+A)99=∑k=099(99k)Ak=I+∑k=199(99k)Ak=I+(∑k=199(99k))A(I+A)^{99} = \sum_{k=0}^{99}\binom{99}{k}A^k = I + \sum_{k=1}^{99}\binom{99}{k}A^k = I + \left(\sum_{k=1}^{99}\binom{99}{k}\right)A
=I+(299−1)A= I + (2^{99}-1)A
 Answer: (4)
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