If A = begin{bmatrix} 5a & -b 3 & 2 end{bmat | vedclass

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(D) We know that $A \cdot 	ext{adj}(A) = |A|I$,where $I$ is the identity matrix of order $2 	imes 2$. Given $A = \begin{bmatrix} 5a & -b \ 3 & 2 \e

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(D) We know that $A \cdot 	ext{adj}(A) = |A|I$,where $I$ is the identity matrix of order $2 	imes 2$. Given $A = \begin{bmatrix} 5a & -b \ 3 & 2 \end{bmatrix}$,the determinant $|A| = (5a)(2) - (-b)(3) = 10a + 3b$. So,$A \cdot 	ext{adj}(A) = (10a + 3b) \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 10a + 3b & 0 \ 0 & 10a + 3b \end{bmatrix}$. Also,$A^T = \begin{bmatrix} 5a & 3 \ -b & 2 \end{bmatrix}$. Then $AA^T = \begin{bmatrix} 5a & -b \ 3 & 2 \end{bmatrix} \begin{bmatrix} 5a & 3 \ -b & 2 \end{bmatrix} = \begin{bmatrix} 25a^2 + b^2 & 15a - 2b \ 15a - 2b & 9 + 4 \end{bmatrix} = \begin{bmatrix} 25a^2 + b^2 & 15a - 2b \ 15a - 2b & 13 \end{bmatrix}$. Equating $A \cdot 	ext{adj}(A) = AA^T$,we get: $10a + 3b = 13$ (from the bottom-right element). Also,$15a - 2b = 0$,which implies $b = \frac{15a}{2}$. Substituting $b$ into the first equation: $10a + 3(\frac{15a}{2}) = 13 \implies 20a + 45a = 26 \implies 65a = 26 \implies a = \frac{26}{65} = \frac{2}{5}$. Then $b = \frac{15}{2} 	imes \frac{2}{5} = 3$. Finally,$5a + b = 5(\frac{2}{5}) + 3 = 2 + 3 = 5$.

If $A = \begin{bmatrix} 5a & -b \\ 3 & 2 \end{bmatrix}$ and $A \cdot \text{adj}(A) = AA^T$,then $5a + b =$

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