Let A= 1 1 1 and B= 9^2 -10^2 11^2 12^2 13^2 -14^2 -15^2 16^2 17^2 , then the value of A^(')BA is;

Correct answer: (D) 539. Given A= 1 1 1 and B= 9^2 -10^2 11^2 12^2 13^2 -14^2 -15^2 16^2 17^2 then A^(')= 1 1 1 So A^(')B= 1 1 1 9^2 -10^2 11^2 12^2 13^2 -14^2 -15^2 16^2 17^2 = 9^2+12^2-15^2 -10^2+13^2+16^2 11^2-14^2+17^2 = 0 325 214 Now A^(')BA= 0 325 214 1 1 1 =325+214 =539

Q33 / 90 · JEE Main 2022 (26 Jul Shift 2) · Mathematics

Let $A = [\begin{array}{l} 1 \\ 1 \\ 1 \end{array}]$ and $B = [\begin{matrix} 9^{2} & - 10^{2} & 11^{2} \\ 12^{2} & 13^{2} & - 14^{2} \\ - 15^{2} & 16^{2} & 17^{2} \end{matrix}]$ , then the value of $A^{'} B A$ is;

A
$1224$
B
$1042$
C
$540$
Correct answer
D
$539$

Solution

Given $A = [\begin{array}{l} 1 \\ 1 \\ 1 \end{array}]$ and $B = [\begin{matrix} 9^{2} & - 10^{2} & 11^{2} \\ 12^{2} & 13^{2} & - 14^{2} \\ - 15^{2} & 16^{2} & 17^{2} \end{matrix}]$
then $A^{'} = [\begin{matrix} 1 & 1 & 1 \end{matrix}]$
So $A^{'} B = [\begin{array}{l} 1 & 1 & 1 \end{array}] [\begin{array}{lcc} 9^{2} & - 10^{2} & 11^{2} \\ 12^{2} & 13^{2} & - 14^{2} \\ - 15^{2} & 16^{2} & 17^{2} \end{array}]$
$= [\begin{matrix} 9^{2} + 12^{2} - 15^{2} & - 10^{2} + 13^{2} + 16^{2} & 11^{2} - 14^{2} + 17^{2} \end{matrix}]$
$= [\begin{matrix} 0 & 325 & 214 \end{matrix}]$
Now $A^{'} B A = [\begin{matrix} 0 & 325 & 214 \end{matrix}] [\begin{array}{l} 1 \\ 1 \\ 1 \end{array}]$
$= [325 + 214]$
$= [539]$

Hence, the correct answer is option: 'D'.

Let A=111 and B=92-102112122132-142-152162172, then the value of A'BA is;

Solution

Relevant Questions

Let $A = [\begin{array}{l} 1 \\ 1 \\ 1 \end{array}]$ and $B = [\begin{matrix} 9^{2} & - 10^{2} & 11^{2} \\ 12^{2} & 13^{2} & - 14^{2} \\ - 15^{2} & 16^{2} & 17^{2} \end{matrix}]$ , then the value of $A^{'} B A$ is;