STD 12 - 3 and 4 . determinant and metrices — JEE STD 11 Science — Question

$\mathrm{AA}^{\mathrm{T}}=\left[\begin{array}{cc}{25 \mathrm{a}^{2}+\mathrm{b}^{2}} & {15 \mathrm{a}-2 \mathrm{b}} \\ {15 \mathrm{a}-2 \mathrm{b}} & {13}\end{array}\right]$

Now, $A \,adj$ $\mathrm{A}=|\mathrm{A}| \mathrm{I}_{2}=\left[\begin{array}{cc}{10 \mathrm{a}+3 \mathrm{b}} & {0} \\ {0} & {10 \mathrm{a}+3 \mathrm{b}}\end{array}\right]$

Given $\mathrm{AA}^{\mathrm{T}}=\mathrm{A}$. adj $\mathrm{A}$

$15 a-2 b=0$         ........$(1)$

$10 a+3 b=13$         ...........$(2)$

Solving we get

$5 a=2$ and $b=3$

$\therefore 5 a+b=5$