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(D) The given system of equations can be written in the form $AX = B$,where $A = \begin{bmatrix} 4 & -3 \ 3 & -5 \end{bmatrix}$,$X = \begin{bmatrix}

Vedclass · Accepted Answer

$x = \frac{-6}{11}, y = \frac{-19}{11}$

Vedclass · Answer

(D) The given system of equations can be written in the form $AX = B$,where $A = \begin{bmatrix} 4 & -3 \ 3 & -5 \end{bmatrix}$,$X = \begin{bmatrix} x \ y \end{bmatrix}$,and $B = \begin{bmatrix} 3 \ 7 \end{bmatrix}$.First,calculate the determinant of $A$:$|A| = (4)(-5) - (-3)(3) = -20 + 9 = -11$.Since $|A| 
eq 0$,the matrix $A$ is non-singular and invertible.The inverse $A^{-1}$ is given by $\frac{1}{|A|} 	ext{adj}(A)$:$A^{-1} = \frac{1}{-11} \begin{bmatrix} -5 & 3 \ -3 & 4 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} 5 & -3 \ 3 & -4 \end{bmatrix}$.Now,solve for $X$ using $X = A^{-1}B$:$X = \frac{1}{11} \begin{bmatrix} 5 & -3 \ 3 & -4 \end{bmatrix} \begin{bmatrix} 3 \ 7 \end{bmatrix}$$X = \frac{1}{11} \begin{bmatrix} (5)(3) + (-3)(7) \ (3)(3) + (-4)(7) \end{bmatrix}$$X = \frac{1}{11} \begin{bmatrix} 15 - 21 \ 9 - 28 \end{bmatrix}$$X = \frac{1}{11} \begin{bmatrix} -6 \ -19 \end{bmatrix} = \begin{bmatrix} -\frac{6}{11} \ -\frac{19}{11} \end{bmatrix}$.Thus,$x = -\frac{6}{11}$ and $y = -\frac{19}{11}$.

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

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