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(B) The given system of equations can be written in the form $AX = B$,where$A = \begin{bmatrix} 5 & 2 \ 7 & 3 \end{bmatrix}$,$X = \begin{bmatrix} x \

Vedclass · Accepted Answer

$x = 2, y = -3$

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(B) The given system of equations can be written in the form $AX = B$,where$A = \begin{bmatrix} 5 & 2 \ 7 & 3 \end{bmatrix}$,$X = \begin{bmatrix} x \ y \end{bmatrix}$,and $B = \begin{bmatrix} 4 \ 5 \end{bmatrix}$.First,we calculate the determinant of $A$:$|A| = (5 	imes 3) - (2 	imes 7) = 15 - 14 = 1$.Since $|A| 
eq 0$,the matrix $A$ is non-singular and invertible.The inverse $A^{-1}$ is given by $\frac{1}{|A|} 	ext{adj}(A)$:$A^{-1} = \frac{1}{1} \begin{bmatrix} 3 & -2 \ -7 & 5 \end{bmatrix} = \begin{bmatrix} 3 & -2 \ -7 & 5 \end{bmatrix}$.Now,we find $X$ using $X = A^{-1}B$:$X = \begin{bmatrix} 3 & -2 \ -7 & 5 \end{bmatrix} \begin{bmatrix} 4 \ 5 \end{bmatrix} = \begin{bmatrix} (3 	imes 4) + (-2 	imes 5) \ (-7 	imes 4) + (5 	imes 5) \end{bmatrix} = \begin{bmatrix} 12 - 10 \ -28 + 25 \end{bmatrix} = \begin{bmatrix} 2 \ -3 \end{bmatrix}$.Thus,$x = 2$ and $y = -3$.

Solve the system of linear equations using the matrix method: $5x + 2y = 4$ and $7x + 3y = 5$.

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