Matrices & DeterminantsmediumFree

Matrices & Determinants — JEE Maths practice question

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

Question

Consider a matrix

A = \begin{pmatrix}3 & 1\\ -6 & -2\end{pmatrix}

, then

(1+A)^{99}

equals, where

I

is a unit matrix of order 2:

I + 2^{98}A

I + 2^{99}A

I + (2^{99}+1)A

I + (2^{99}-1)A

correct

Solution

Step 1: Show that

A

is idempotent

A^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

A^2 = A

, we have

A^n = A

for all

n \geq 1

. Step 2: Expand using the binomial theorem

(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 + (2^{99}-1)A

Answer: (4)

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