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# Given $3\begin{bmatrix} x & y \\ z & w \end{bmatrix}$ = $\begin{bmatrix} x & 6 \\ -1 & 2w \end{bmatrix}$ + $\begin{bmatrix} 4 & x + y \\ z + w & 3 \end{bmatrix}$, find the values of $x, y, z$ and $w$.

$\begin{array}{1 1} x=2 y=4 z=1 w=1 \\ x=2 y=4 z=1 w=3 \\ x=2 y=4 z=3 w=1 \\x=2 y=4 z=1 w=5 \end{array}$

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 the unknown variables.
Given $3\begin{bmatrix} x & y \\ z & w \end{bmatrix} = \begin{bmatrix} x & 6 \\ -1 & 2w \end{bmatrix} + \begin{bmatrix} 4 & x + y \\ z + w & 3 \end{bmatrix}$:
$\Rightarrow \begin{bmatrix} 3x & 3y \\ 3z &3 w \end{bmatrix} = \begin{bmatrix} x & 6 \\ -1 & 2w \end{bmatrix} + \begin{bmatrix} 4 & x + y \\ z + w & 3 \end{bmatrix}$
$\Rightarrow \begin{bmatrix} 3x & 3y \\ 3z &3 w \end{bmatrix} = \begin{bmatrix}x+4& 6+x+y\\-1+z+w & 2w+3\end{bmatrix}.$
Since the above matrices are equal there corresponding elements should be equal:
$3x=x+4----(1)$
$3y=6+x+y---(2)$
$3z=-1+z+w-----(3)$
$3w=2w+3----(4)$
Solving for $x$ in (1),
$3x=x+4 \rightarrow 2x = 4$
$x=2$.
Subsitituting $x=2$ and solving for $y$ in (2),
$3y=6+x+y \rightarrow 3y = 6 + 2 + y \rightarrow 2y = 8$
$y=4$
Solving for $w$ in (4),
$3w=2w+3 \rightarrow w = 3$
Substituting for $w$ in (3) and solving for $z$,
$3z=-1+z+w \rightarrow 3z = -1 + z + 3 \rightarrow 3z - z = 2z = -1 + 3 = 2$
$z = 1$
edited Feb 27, 2013