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← JEE Main 2022 · 26 Jul · Shift 2
MathematicsSingle correct+4 / -1Question 3 of 90

JEE Main 2022 · 26 Jul · Shift 2, Mathematics question 3

Let A=[111]A=\begin{bmatrix}1\\1\\1\end{bmatrix} and B=[92−102112122132−142−152162172]B=\begin{bmatrix}9^2&-10^2&11^2\\12^2&13^2&-14^2\\-15^2&16^2&17^2\end{bmatrix}, then the value of A′BAA'BA is:

Choose one option
Answer and worked solution

Official answer: D

A′BAA'BA is the sum of all entries of B: 81−100+121+144+169−196−225+256+289=53981-100+121+144+169-196-225+256+289=539.

(D)