JEE Main 2022 · 29 Jun · Shift 1 · Mathematics Q10

← JEE Main 2022 · 29 Jun · Shift 1

MathematicsSingle correct+4 / -1Question 10 of 90

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 :

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

\left(\dfrac{3^{10}-1}2\right)A

\left(\dfrac{3^{10}+1}2\right)A

\left(\dfrac{3^{10}+3}2\right)A

Answer and worked solution

Official answer: A

Rank one. $a_{ij}=2^j\cdot2^{-i}$ , so $A^2=(\operatorname{tr}A)A=3A$ and $A^n=3^{n-1}A$ .

Sum. $(3+3^2+\dots+3^9)A=\dfrac{3^{10}-3}2A$ .

(A)

Topics:Matrix algebra Geometric progression

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JEE Main 2022 · 29 Jun · Shift 1, Mathematics question 10