# How do you solve ((4, 8), (2, 5))((x), (y))=((0),(6))?

Feb 12, 2016

$\left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}- 12 \\ 6\end{matrix}\right)$

#### Explanation:

To solve $\left(\begin{matrix}4 & 8 \\ 2 & 5\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}0 \\ 6\end{matrix}\right)$, we need to first find inverse of the matrix $\left(\begin{matrix}4 & 8 \\ 2 & 5\end{matrix}\right)$.

The general form for inverse of matris $\left(\begin{matrix}a & b \\ c & d\end{matrix}\right)$ is $\frac{1}{a d - b c} \left(\begin{matrix}d & - b \\ - c & a\end{matrix}\right)$.

Hence inverse of $\left(\begin{matrix}4 & 8 \\ 2 & 5\end{matrix}\right)$ is $\frac{1}{4 \cdot 5 - 8 \cdot 2} \left(\begin{matrix}5 & - 8 \\ - 2 & 4\end{matrix}\right)$ or $\frac{1}{4} \left(\begin{matrix}5 & - 8 \\ - 2 & 4\end{matrix}\right)$. Multiplying both sides of $\left(\begin{matrix}4 & 8 \\ 2 & 5\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}0 \\ 6\end{matrix}\right)$ by this gives

$\left(\begin{matrix}x \\ y\end{matrix}\right) = \frac{1}{4} \left(\begin{matrix}5 & - 8 \\ - 2 & 4\end{matrix}\right) \left(\begin{matrix}0 \\ 6\end{matrix}\right)$ .............(A)

as $\frac{1}{4} \left(\begin{matrix}5 & - 8 \\ - 2 & 4\end{matrix}\right) \left(\begin{matrix}4 & 8 \\ 2 & 5\end{matrix}\right) = \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right)$

Solving (A) by simple matrix multiplication gives us

$\left(\begin{matrix}x \\ y\end{matrix}\right) = \frac{1}{4} \left(\begin{matrix}- 48 \\ 24\end{matrix}\right) = \left(\begin{matrix}- 12 \\ 6\end{matrix}\right)$

Feb 27, 2016

$\left(x , y\right) = \left(- 12 , 6\right)$

#### Explanation:

(this is just an alternative method to that provided by Shwetank Mauria that doesn't require the use of an inverse matrix)

$\left(\begin{matrix}4 & 8 \\ 2 & 5\end{matrix}\right) \times \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}4 x + 8 y \\ 2 x + 5 y\end{matrix}\right)$ (by standard matrix multiplication)

Therefore
$\textcolor{w h i t e}{\text{XXX}} \left(\begin{matrix}4 & 8 \\ 2 & 5\end{matrix}\right) \times \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}0 \\ 6\end{matrix}\right)$
is equivalent to
[1]$\textcolor{w h i t e}{\text{XXX}} 4 x + 8 y = 0$
[2]$\textcolor{w h i t e}{\text{XXX}} 2 x + 5 y = 6$

If we multiply equation [2] by $2$ and subtract the result form equation [1], we get
[3]$\textcolor{w h i t e}{\text{XXX}} - 2 y = - 12$
and from this
[4]$\textcolor{w h i t e}{\text{XXX}} y = 6$

Substituting $6$ for $y$ back in equation [1] gives
[5]$\textcolor{w h i t e}{\text{XXX}} \cancel{4} x + {\cancel{8}}^{2} \times \left(6\right) = 0$
and
[6]$\textcolor{w h i t e}{\text{XXX}} x = - 12$