Let $\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]$ be a square matrix of order 3 such that $\mathrm{a}_{\mathrm{ij}}=2^{\mathrm{j}-\mathrm{i}}$, for all $\mathrm{i}, \mathrm{j}=1,2,3$. Then, the matrix $\mathrm{A}^{2}+\mathrm{A}^{3}+\ldots+\mathrm{A}^{10}$ is equal to :

(A) $\left(\frac{3^{10}-3}{2}\right) A$

(B) $\left(\frac{3^{10}-1}{2}\right) A$

(C) $\left(\frac{3^{10}+1}{2}\right) A$

(D) $\left(\frac{3^{10}+3}{2}\right) A$

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Matrices

Let

A = [a_{ij}]

be a square matrix of order 3 such that

a_{ij} = 2^{j - i}

, for all

i, j = 1, 2, 3

. Then, the matrix

A^{2} + A^{3} + \dots + A^{10}

is equal to :

Answer: (A)

The matrix is:

A = 1 1/2 1/4 21 1/2 421

By computation,

A^{2} = 3 A

. This means

A^{n} = 3^{n - 1} A

for all

n \geq 1

.

Therefore:

A^{2} + A^{3} + \dots + A^{10} = 3 A + 3^{2} A + \dots + 3^{9} A

= A (3 + 3^{2} + \dots + 3^{9})

= A \cdot \frac{3 ( 3 ^{9} - 1 )}{3 - 1}

= \frac{3 ^{10} - 3}{2} A

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Single correct option