If $$\begin{bmatrix*}[r] 2 & 0 \\ 0 & 4 \end{bmatrix*}\begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 2 \\ -8 \end{bmatrix*}$$, the values of x and y respectively are :
- (a)1, -2
- (b)-2, 1
- (c)1, 2
- (d)-2, -1
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Solving,
$$\Rightarrow \begin{bmatrix*}[r] 2 & 0 \\ 0 & 4 \end{bmatrix*}\begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 2 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2.x + 0.y \\ 0.x + 4.y \end{bmatrix*} = \begin{bmatrix*}[r] 2 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x + 0 \\ 0 + 4y \end{bmatrix*} = \begin{bmatrix*}[r] 2 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x \\ 4y \end{bmatrix*} = \begin{bmatrix*}[r] 2 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow 2x = 2 \text{ and } 4y = -8 \\[1em] \Rightarrow x = \frac{2}{2} \text{ and } y = -\frac{8}{4} \\[1em] \Rightarrow x = 1 \text{ and } y = -2.$$
Hence, Option 1 is the correct option.