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# Find $X$ , if $Y = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}$ and $2X + Y = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix}$

$\begin{array}{1 1} X=\begin{bmatrix}1 & -1\\2 & -1\end{bmatrix} \\ X=\begin{bmatrix}-1 & -1\\2 & 0\end{bmatrix} \\ X=\begin{bmatrix}-1 & 1\\2 & -1\end{bmatrix} \\X=\begin{bmatrix}-1 & -1\\-2 & -1\end{bmatrix}\end{array}$

Toolbox:
• The scalar multiplication $cA$ of a matrix $A$ and a number $c$ (also called a scalar in the parlance of abstract algebra) is given by multiplying every entry of $A$ by $c$.
• The difference $A-B$ of two $m$-by-$n$ matrices $A$ and $B$ is calculated entrywise: $(A - B)_{i,j} = A_{i,j} - B_{i,j}$ where 1 ≤ i ≤ m and 1 ≤ j ≤ n.
Given $Y = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}$ and $2X+Y = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix}$:
$2X = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix} - Y$
Substituting for $Y$, we get $2X = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}$
$2X=\begin{bmatrix}1-3 & 0-2\\-3-1 & 2-4\end{bmatrix}.$
$2X=\begin{bmatrix}-2 & -2\\-4 & -2\end{bmatrix}.$
Therefore, $X=\frac{1}{2}\begin{bmatrix}-2 & -2\\-4 & -2\end{bmatrix}.$
$X=\begin{bmatrix}-1 & -1\\-2 & -1\end{bmatrix}.$