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

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(D) Given $A = \begin{bmatrix} 5a & -b \ 3 & 2 \end{bmatrix}$.The transpose of $A$ is $A^T = \begin{bmatrix} 5a & 3 \ -b & 2 \end{bmatrix}$.Calculat

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$5$

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(D) Given $A = \begin{bmatrix} 5a & -b \ 3 & 2 \end{bmatrix}$.The transpose of $A$ is $A^T = \begin{bmatrix} 5a & 3 \ -b & 2 \end{bmatrix}$.Calculating $A \cdot A^T$:$A \cdot A^T = \begin{bmatrix} 5a & -b \ 3 & 2 \end{bmatrix} \begin{bmatrix} 5a & 3 \ -b & 2 \end{bmatrix} = \begin{bmatrix} 25a^2 + b^2 & 15a - 2b \ 15a - 2b & 13 \end{bmatrix}$.We know that $A \cdot 	ext{adj}(A) = |A| I$,where $|A| = (5a)(2) - (-b)(3) = 10a + 3b$.So,$A \cdot 	ext{adj}(A) = \begin{bmatrix} 10a + 3b & 0 \ 0 & 10a + 3b \end{bmatrix}$.Given $A \cdot 	ext{adj}(A) = A \cdot A^T$,we equate the corresponding elements:$1$) $15a - 2b = 0 \implies 15a = 2b \implies b = \frac{15a}{2}$.$2$) $10a + 3b = 13$.Substitute $b = \frac{15a}{2}$ into the second equation:$10a + 3(\frac{15a}{2}) = 13$$10a + \frac{45a}{2} = 13$$\frac{20a + 45a}{2} = 13$$65a = 26 \implies a = \frac{26}{65} = \frac{2}{5}$.Now find $b$:$b = \frac{15}{2} 	imes \frac{2}{5} = 3$.Therefore,$5a + b = 5(\frac{2}{5}) + 3 = 2 + 3 = 5$.

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

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