1. **State the problem:** We need to find the values of $x$ and $y$ given the matrix equation: $$\left(4\begin{bmatrix}1 & 2 & 0 \\ 2 & -1 & 3\end{bmatrix} - 2 \begin{bmatrix}1 & 3 & -1 \\ 2 & -3 & 4\end{bmatrix}\right) \begin{bmatrix}2 \\ 0 \\ -1\end{bmatrix} = \begin{bmatrix}x \\ y\end{bmatrix}$$ 2. **Explain the formula and rules:** - Matrix multiplication is done by taking the dot product of rows of the first matrix with columns of the second. - Scalar multiplication means multiplying every element of the matrix by the scalar. - Subtraction of matrices is element-wise. 3. **Calculate each matrix multiplied by its scalar:** $$4 \begin{bmatrix}1 & 2 & 0 \\ 2 & -1 & 3\end{bmatrix} = \begin{bmatrix}4 & 8 & 0 \\ 8 & -4 & 12\end{bmatrix}$$ $$2 \begin{bmatrix}1 & 3 & -1 \\ 2 & -3 & 4\end{bmatrix} = \begin{bmatrix}2 & 6 & -2 \\ 4 & -6 & 8\end{bmatrix}$$ 4. **Subtract the two matrices:** $$\begin{bmatrix}4 & 8 & 0 \\ 8 & -4 & 12\end{bmatrix} - \begin{bmatrix}2 & 6 & -2 \\ 4 & -6 & 8\end{bmatrix} = \begin{bmatrix}4-2 & 8-6 & 0-(-2) \\ 8-4 & -4-(-6) & 12-8\end{bmatrix} = \begin{bmatrix}2 & 2 & 2 \\ 4 & 2 & 4\end{bmatrix}$$ 5. **Multiply the resulting matrix by the vector:** $$\begin{bmatrix}2 & 2 & 2 \\ 4 & 2 & 4\end{bmatrix} \begin{bmatrix}2 \\ 0 \\ -1\end{bmatrix} = \begin{bmatrix}2\times 2 + 2\times 0 + 2\times (-1) \\ 4\times 2 + 2\times 0 + 4\times (-1)\end{bmatrix} = \begin{bmatrix}4 + 0 - 2 \\ 8 + 0 - 4\end{bmatrix} = \begin{bmatrix}2 \\ 4\end{bmatrix}$$ 6. **Conclusion:** The vector on the right side is $\begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}2 \\ 4\end{bmatrix}$, so $$x = 2, \quad y = 4.$$