ft[\begin{array}{l} 2 \\ 4 \end{array}\right] $

We can solve for the vector $\left[\begin{array}{l} x \\ y \end{array}\right]$ using the inverse of the coefficient matrix:

$ \left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{cc} 1 & 1 \\ -1 & 1 \end{array}\right]^{-1}\left[\begin{array}{l} 2 \\ 4 \end{array}\right] $

The inverse of the matrix $\left[\begin{array}{cc} a & b \\ c & d \end{array}\right]$ is given by:

$ \left[\begin{array}{cc} a & b \\ c & d \end{array}\right]^{-1} = \frac{1}{ad-bc}\left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right] $

For our matrix $\left[\begin{array}{cc} 1 & 1 \\ -1 & 1 \end{array}\right]$, the determinant $ ad-bc = (1)(1) - (1)(-1) = 2 $.

Thus, the inverse is:

$ \frac{1}{2}\left[\begin{array}{ll} 1 & -1 \\ 1 & 1 \end{array}\right] $

Substitute the inverse into the expression for $\left[\begin{array}{l} x \\ y \end{array}\right]$:

$ \left[\begin{array}{l} x \\ y \end{array}\right] = \frac{1}{2}\left[\begin{array}{ll} 1 & -1 \\ 1 & 1 \end{array}\right]\left[\begin{array}{l} 2 \\ 4 \end{array}\right] $

Calculate the matrix product:

$ \frac{1}{2}\left[\begin{array}{c} (1)(2) + (-1)(4) \\ (1)(2) + (1)(4) \end{array}\right] = \frac{1}{2}\left[\begin{array}{c} 2 - 4 \\ 2 + 4 \end{array}\right] = \frac{1}{2}\left[\begin{array}{c} -2 \\ 6 \end{array}\right] $

This simplifies to:

$ \left[\begin{array}{l} x \\ y \end{array}\right] = \left[\begin{array}{c} -1 \\ 3 \end{array}\right] $

Therefore, the values of $ x $ and $ y $ are $ x = -1 $ and $ y = 3 $.

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