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

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

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$5$

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(C) Given $A = \begin{bmatrix} 2a & -3b \ 3 & 2 \end{bmatrix}$.We know that $A \cdot \operatorname{adj} A = |A| I = \begin{bmatrix} |A| & 0 \ 0 & |A| \end{bmatrix}$.Calculating the determinant $|A| = (2a)(2) - (-3b)(3) = 4a + 9b$.So,$A \cdot \operatorname{adj} A = \begin{bmatrix} 4a + 9b & 0 \ 0 & 4a + 9b \end{bmatrix}$.Now,calculate $A A^{T}$:$A A^{T} = \begin{bmatrix} 2a & -3b \ 3 & 2 \end{bmatrix} \begin{bmatrix} 2a & 3 \ -3b & 2 \end{bmatrix} = \begin{bmatrix} 4a^2 + 9b^2 & 6a - 6b \ 6a - 6b & 9 + 4 \end{bmatrix} = \begin{bmatrix} 4a^2 + 9b^2 & 6a - 6b \ 6a - 6b & 13 \end{bmatrix}$.Given $A \cdot \operatorname{adj} A = A A^{T}$,we equate the matrices:$\begin{bmatrix} 4a + 9b & 0 \ 0 & 4a + 9b \end{bmatrix} = \begin{bmatrix} 4a^2 + 9b^2 & 6a - 6b \ 6a - 6b & 13 \end{bmatrix}$.Comparing the elements:$6a - 6b = 0 \implies a = b$.$4a + 9b = 13$.Substituting $a = b$ into the second equation: $4a + 9a = 13 \implies 13a = 13 \implies a = 1$.Since $a = b$,we have $a = 1$ and $b = 1$.Therefore,$2a + 3b = 2(1) + 3(1) = 5$.

If $A=\begin{bmatrix} 2a & -3b \\ 3 & 2 \end{bmatrix}$ and $A \cdot \operatorname{adj} A = A A^{T}$,then $2a + 3b$ is

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