JEE MAIN - Mathematics (2022 - 26th July Evening Shift - No. 3)
$$
\text { Let } A=\left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right] \text { and } B=\left[\begin{array}{ccc}
9^{2} & -10^{2} & 11^{2} \\
12^{2} & 13^{2} & -14^{2} \\
-15^{2} & 16^{2} & 17^{2}
\end{array}\right] \text {, then the value of } A^{\prime} B A \text { is: }
$$
1224
1042
540
539
Explanation
$$A'BA = \left[ {\matrix{ 1 & 1 & 1 \cr } } \right]\left[ {\matrix{ {{9^2}} & { - {{10}^2}} & {{{11}^2}} \cr {{{12}^2}} & {{{13}^2}} & { - {{14}^2}} \cr { - {{15}^2}} & {{{16}^2}} & {{{17}^2}} \cr } } \right]A$$
$$ = \left[ {\matrix{ {{9^2} + {{12}^2} - {{15}^2}} & { - {{10}^2} + {{13}^2} + {{16}^2}} & {{{11}^2} - {{14}^2} + {{17}^2}} \cr } } \right]\left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right]$$
$$ = \left[ {{9^2} + {{12}^2} - {{15}^2} - {{10}^2} + {{13}^2} + {{16}^2} + {{11}^2} - {{14}^2} + {{17}^2}} \right]$$
$$ = [({9^2} - {10^2}) + ({11^2} + {12^2}) + ({13^2} - {14^2}) + ({16^2} - {15^2}) + {17^2}]$$
$$ = [ - 19 + 265 + ( - 27) + 31 + 289]$$
$$ = [585 - 46] = [539]$$
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