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

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(C) Given $A = \begin{bmatrix} 5a & -b \ 3 & 2 \end{bmatrix}$.We know that $A(adj A) = |A|I$, where $I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}

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$5$

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(C) Given $A = \begin{bmatrix} 5a & -b \ 3 & 2 \end{bmatrix}$.We know that $A(adj A) = |A|I$, where $I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.$|A| = (5a)(2) - (-b)(3) = 10a + 3b$.So, $A(adj A) = \begin{bmatrix} 10a + 3b & 0 \ 0 & 10a + 3b \end{bmatrix}$.Also, $A^T = \begin{bmatrix} 5a & 3 \ -b & 2 \end{bmatrix}$.$AA^T = \begin{bmatrix} 5a & -b \ 3 & 2 \end{bmatrix} \begin{bmatrix} 5a & 3 \ -b & 2 \end{bmatrix} = \begin{bmatrix} 25a^2 + b^2 & 15a - 2b \ 15a - 2b & 9 + 4 \end{bmatrix} = \begin{bmatrix} 25a^2 + b^2 & 15a - 2b \ 15a - 2b & 13 \end{bmatrix}$.Equating $A(adj A) = AA^T$:$10a + 3b = 13$ (from the $(2,2)$ element).For the off-diagonal elements, $15a - 2b = 0$, so $b = \frac{15a}{2}$.Substitute $b$ into $10a + 3b = 13$:$10a + 3(\frac{15a}{2}) = 13 \implies 10a + \frac{45a}{2} = 13 \implies \frac{20a + 45a}{2} = 13 \implies 65a = 26 \implies a = \frac{26}{65} = \frac{2}{5}$.Then $b = \frac{15}{2} 	imes \frac{2}{5} = 3$.Thus, $5a + b = 5(\frac{2}{5}) + 3 = 2 + 3 = 5$.

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

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