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(A) The given system of equations can be written in the form $AX = B$,where$A = \begin{bmatrix} 5 & 2 \ 3 & 2 \end{bmatrix}$,$X = \begin{bmatrix} x \

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$x = -1, y = 4$

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(A) The given system of equations can be written in the form $AX = B$,where$A = \begin{bmatrix} 5 & 2 \ 3 & 2 \end{bmatrix}$,$X = \begin{bmatrix} x \ y \end{bmatrix}$,and $B = \begin{bmatrix} 3 \ 5 \end{bmatrix}$.First,we calculate the determinant of $A$:$|A| = (5 	imes 2) - (2 	imes 3) = 10 - 6 = 4$.Since $|A| 
eq 0$,the inverse $A^{-1}$ exists.$A^{-1} = \frac{1}{|A|} 	ext{adj}(A) = \frac{1}{4} \begin{bmatrix} 2 & -2 \ -3 & 5 \end{bmatrix}$.Now,$X = A^{-1}B$:$X = \frac{1}{4} \begin{bmatrix} 2 & -2 \ -3 & 5 \end{bmatrix} \begin{bmatrix} 3 \ 5 \end{bmatrix}$$X = \frac{1}{4} \begin{bmatrix} (2 	imes 3) + (-2 	imes 5) \ (-3 	imes 3) + (5 	imes 5) \end{bmatrix}$$X = \frac{1}{4} \begin{bmatrix} 6 - 10 \ -9 + 25 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} -4 \ 16 \end{bmatrix} = \begin{bmatrix} -1 \ 4 \end{bmatrix}$.Thus,$x = -1$ and $y = 4$.

Solve the system of linear equations using the matrix method:
$5x + 2y = 3$
$3x + 2y = 5$

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Solve the system of linear equations using the matrix method:$5x + 2y = 3$$3x + 2y = 5$