If \(\mathrm{A}=\left[\begin{array}{cc}2 \mathrm{a} & -3 \mathrm{~b} \\ 3 & 2\end{array}\right]\) and \(\mathrm{A} \cdot \operatorname{adj} \mathrm{A}=\mathrm{AA}^{\mathrm{T}}\), then \(2 a+3 b\) is

\(\begin{aligned} & A=\left[\begin{array}{cc}2 a & -3 b \\ 3 & 2\end{array}\right] \\ & A \cdot \operatorname{adj} A=\left[\begin{array}{cc}2 a & -3 b \\ 3 & 2\end{array}\right]\left[\begin{array}{cc}2 & 3 b \\ -3 & 2 a\end{array}\right] \\ & =\left[\begin{array}{cc}4 a+9 b & 0 \\ 0 & 9 b+4 a\end{array}\right] \\ & A \cdot A^T=\left[\begin{array}{cc}2 a & -3 b \\ 3 & 2\end{array}\right]\left[\begin{array}{cc}2 a & 3 \\ -3 b & 2\end{array}\right] \\ & =\left[\begin{array}{cc}4 a^2+9 b^2 & 6 a-6 b \\ 6 a-6 b & 13\end{array}\right] \\ & \therefore \quad \mathrm{A} \cdot \operatorname{adj} \mathrm{A}=\mathrm{A} \cdot \mathrm{A}^{\mathrm{T}} \\ & \Rightarrow\left[\begin{array}{cc}4 a+9 b & 0 \\ 0 & 4 a+9 b\end{array}\right]=\left[\begin{array}{cc}4 a^2+9 b^2 & 6 a-6 b \\ 6 a-6 b & 13\end{array}\right] \\ & \Rightarrow \mathrm{a}=\mathrm{b} \text { and } 4 \mathrm{a}+9 \mathrm{~b}=13 \\ & \Rightarrow \mathrm{a}=\mathrm{b}=1 \\ & \Rightarrow 2 a+3 b=5 \\ & \end{aligned}\)