If left[begin{array}{cc}1 & 1 -1 & 1end{arra | vedclass

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(D) Given the matrix equation: $\left[\begin{array}{cc}1 & 1 \ -1 & 1\end{array}ight]\left[\begin{array}{l}x \ y\end{array}ight]=\left[\begin{ar

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$-1, 3$

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(D) Given the matrix equation: $\left[\begin{array}{cc}1 & 1 \ -1 & 1\end{array}ight]\left[\begin{array}{l}x \ y\end{array}ight]=\left[\begin{array}{c}2 \ 4\end{array}ight]$.Performing matrix multiplication on the left side,we get:$\left[\begin{array}{c}1(x) + 1(y) \ -1(x) + 1(y)\end{array}ight] = \left[\begin{array}{c}2 \ 4\end{array}ight]$$\left[\begin{array}{c}x+y \ -x+y\end{array}ight] = \left[\begin{array}{c}2 \ 4\end{array}ight]$By equating the corresponding elements,we obtain the system of linear equations:$x + y = 2$ (Equation $1$)$-x + y = 4$ (Equation $2$)Adding Equation $1$ and Equation $2$:$(x + y) + (-x + y) = 2 + 4$$2y = 6 \implies y = 3$Substituting $y = 3$ into Equation $1$:$x + 3 = 2 \implies x = -1$Thus,the values are $x = -1$ and $y = 3$.

If $\left[\begin{array}{cc}1 & 1 \\ -1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}2 \\ 4\end{array}\right]$,then the values of $x$ and $y$ respectively are

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