MCQ
If $A=$ $\left[ {\begin{array}{*{20}{c}}{5a}&{ - b}\\3&2\end{array}} \right]$ and $A\;adj\;A = A\;{A^T},$ then $5a+b $ to :
- A$4$
- B$13$
- C$-1$
- ✓$5$
$\mathrm{AA}^{\mathrm{T}}=\left[\begin{array}{cc}{25 \mathrm{a}^{2}+\mathrm{b}^{2}} & {15 \mathrm{a}-2 \mathrm{b}} \\ {15 \mathrm{a}-2 \mathrm{b}} & {13}\end{array}\right]$
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 $b=3$
$\therefore 5 a+b=5$
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$3 x-y-z $$ =0 $, $-3 x+z $$ =0 $, $-3 x+2 y+z $$ =0 .$
Then the number of such points for which $x^2+y^2+z^2 \leq 100$ is