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Question

(C) The given system of equations can be written in the matrix form $AX = B$,where:$A = \begin{bmatrix} 2 & -1 \ 3 & 4 \end{bmatrix}$,$X = \begin{bma

Vedclass · Accepted Answer

$x = \frac{-5}{11}, y = \frac{12}{11}$

Vedclass · Answer

(C) The given system of equations can be written in the matrix form $AX = B$,where:$A = \begin{bmatrix} 2 & -1 \ 3 & 4 \end{bmatrix}$,$X = \begin{bmatrix} x \ y \end{bmatrix}$,and $B = \begin{bmatrix} -2 \ 3 \end{bmatrix}$.First,calculate the determinant of $A$:$|A| = (2)(4) - (-1)(3) = 8 + 3 = 11$.Since $|A| 
eq 0$,the matrix $A$ is non-singular and invertible.The inverse of $A$ is given by $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$:$A^{-1} = \frac{1}{11} \begin{bmatrix} 4 & 1 \ -3 & 2 \end{bmatrix}$.Now,solve for $X$ using $X = A^{-1}B$:$X = \frac{1}{11} \begin{bmatrix} 4 & 1 \ -3 & 2 \end{bmatrix} \begin{bmatrix} -2 \ 3 \end{bmatrix}$$X = \frac{1}{11} \begin{bmatrix} (4)(-2) + (1)(3) \ (-3)(-2) + (2)(3) \end{bmatrix}$$X = \frac{1}{11} \begin{bmatrix} -8 + 3 \ 6 + 6 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} -5 \ 12 \end{bmatrix} = \begin{bmatrix} -\frac{5}{11} \ \frac{12}{11} \end{bmatrix}$.Thus,$x = -\frac{5}{11}$ and $y = \frac{12}{11}$.

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

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