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# If $x\begin{bmatrix} 2 \\ 3 \end{bmatrix} + y\begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}$, find the values of $x$ and $y$.

$\begin{array}{1 1} x= - 4 y=3 \\ x= -3 y=4 \\ x=3 y= - 4 \\ x=3 y=4 \end{array}$

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 sum / 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.
• If the order of 2 matrices are equal, their corresponding elements are equal, i.e, if $A_{ij}=B_{ij}$, then any element $a_{ij}$ in matrix A is equal to corresponding element $b_{ij}$ in matrix B.
• We can then match the corresponding elements and solve the resulting equations to find the values of x and y.
Given $x\begin{bmatrix} 2 \\ 3 \end{bmatrix} + y\begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}$:
$\Rightarrow \begin{bmatrix} 2x \\ 3x \end{bmatrix} + \begin{bmatrix} -y \\ y \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}$
$2x-y=10 (i)$
$3x+y=5 (ii)$
Adding the equations (i) & (ii) and solving for $x$ we get:
$2x-y+3x+y = 10+5 \rightarrow 5x = 15 \rightarrow x = \frac{15}{5}$
$x = 3$
Substituting the value of $x$ in $(i)$:
$2 \times 3 - y = 10 \rightarrow 6 - y = 10$
$y = -4$