If $A=\left[\begin{array}{cc}5 a & -b \\ 3 & 2\end{array}\right]$ and $A \cdot \operatorname{adj} A=A^T$,…

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

Solution

$\begin{aligned} & \mathrm{A}_{11}=(-1)^{1+1}(2)=2, \mathrm{~A}_{12}=(-1)^{1+2}(3)=-3 \\ & \mathrm{~A}_{21}=(-1)^{2+1}(-\mathrm{b})=\mathrm{b}, \mathrm{A}_{22}=(-1)^{2+2}(5 \mathrm{a})=5 \mathrm{a}\end{aligned}$ $\therefore \quad \operatorname{adj} A=\left[\begin{array}{cc} 2 & -3 \\ b & 5 a \end{array}\right]^T=\left[\begin{array}{cc} 2 & b \\ -3 & 5 a \end{array}\right]$
Given, A adj $\mathrm{A}=\mathrm{AA}^{\mathrm{T}}$ $\begin{aligned} & \Rightarrow\left[\begin{array}{cc}5 \mathrm{a} & -\mathrm{b} \\ 3 & 2\end{array}\right]\left[\begin{array}{cc}2 & \mathrm{~b} \\ -3 & 5 \mathrm{a}\end{array}\right]=\left[\begin{array}{cc}5 \mathrm{a} & -\mathrm{b} \\ 3 & 2\end{array}\right]\left[\begin{array}{cc}5 \mathrm{a} & 3 \\ -\mathrm{b} & 2\end{array}\right] \\ & \Rightarrow\left[\begin{array}{cc}10 \mathrm{a}+3 \mathrm{~b} & 0 \\ 0 & 3 \mathrm{~b}+10 \mathrm{a}\end{array}\right]=\left[\begin{array}{cc}25 \mathrm{a}^2+\mathrm{b}^2 & 15 \mathrm{a}-2 \mathrm{~b} \\ 15 \mathrm{a}-2 \mathrm{~b} & 13\end{array}\right]\end{aligned}$ $\therefore \quad$ by the equality of matrices, $\begin{array}{ll} & 15 a-2 b=0 \text { and } 3 b+10 a=13 \\ & \Rightarrow a=\frac{2}{5} \text { and } b=3 \\ \therefore \quad & 5 a+b=5\left(\frac{2}{5}\right)+3=2+3=5 \end{array}$

Asked in: MHT CET 2024 (15 May Shift 2)

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