If $\mathrm{A}\left[\begin{array}{cc}5 \mathrm{a} & -\mathrm{b} \\ 3 & 2\end{array}\right]$ and $\mathrm{A}$…

If $\mathrm{A}\left[\begin{array}{cc}5 \mathrm{a} & -\mathrm{b} \\ 3 & 2\end{array}\right]$ and $\mathrm{A}$ adj $\mathrm{A}=\mathrm{AA}^{\mathrm{T}}$, then $5 \mathrm{a}+\mathrm{b}=$
  1. 13
  2. 4
  3. -1
  4. 5

Solution

$\begin{aligned} & \text { We have } \mathrm{A}\left[\begin{array}{cc} 5 \mathrm{a} & -\mathrm{b} \\ 3 & 2 \end{array}\right] \text { and } \mathrm{A} \operatorname{adj} \mathrm{A}=\mathrm{AA}^{\mathrm{T}} \\ & \therefore\left[\begin{array}{cc} 5 \mathrm{a} & -\mathrm{b} \\ 3 & 2 \end{array}\right]\left[\begin{array}{cc} 2 & \mathrm{~b} \\ -3 & 5 \mathrm{a} \end{array}\right]=\left[\begin{array}{cc} 5 \mathrm{a} & -\mathrm{b} \\ 3 & 2 \end{array}\right]\left[\begin{array}{cc} 5 \mathrm{a} & 3 \\ -\mathrm{b} & 2 \end{array}\right] \\ & {\left[\begin{array}{cc} 10 \mathrm{a}+3 \mathrm{~b} & 0 \\ 0 & 10 \mathrm{a}+3 \mathrm{~b} \end{array}\right]=\left[\begin{array}{cc} 25 \mathrm{a}^2+\mathrm{b}^2 & 15 \mathrm{a}-2 \mathrm{~b} \\ 15 \mathrm{a}-2 \mathrm{~b} & 9+4 \end{array}\right]} \\ & \therefore 10 \mathrm{a}+3 \mathrm{~b}=13 \text { and } 15 \mathrm{a}-2 \mathrm{~b}=0 \end{aligned}$ Solving these equations, we get $a=\frac{2}{5}$ and $b=3$ $\therefore 5 a+b=2+3=5$

Asked in: MHT CET 2021 (22 Sep Shift 2)

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