# How do you solve x + y = 5  and 3x – y = –1 using matrices?

Jul 16, 2018

$x = 1 \mathmr{and} y = 4$

#### Explanation:

Here,

$x + y = 5 \to \left(1\right)$ , $\mathmr{and}$ $3 x - y = - 1 \to \left(2\right)$

Let us write in the matrix equation form :

$\left(\begin{matrix}1 & 1 \\ 3 & - 1\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}5 \\ - 1\end{matrix}\right)$

We take ,

$A = \left(\begin{matrix}1 & 1 \\ 3 & - 1\end{matrix}\right)$ , $X = \left(\begin{matrix}x \\ y\end{matrix}\right)$ ,$\mathmr{and} B = \left(\begin{matrix}5 \\ - 1\end{matrix}\right)$

$\therefore A X = B$

Now, $\det A = | \left(1 , 1\right) , \left(3 , - 1\right) | = - 1 - 3 = - 4 \ne 0$

$\therefore \text{We can say that , " A^-1 " exists}$

Now ,$a \mathrm{dj} A = \left(\begin{matrix}- 1 & - 1 \\ - 3 & 1\end{matrix}\right)$

$\therefore {A}^{-} 1 = \frac{1}{\det A} \cdot a \mathrm{dj} A$

$\therefore {A}^{-} 1 = - \frac{1}{4} \left(\begin{matrix}- 1 & - 1 \\ - 3 & 1\end{matrix}\right)$

We have,

$A X = B \implies X = {A}^{-} 1 B$

$\implies X = - \frac{1}{4} \left(\begin{matrix}- 1 & - 1 \\ - 3 & 1\end{matrix}\right) \left(\begin{matrix}5 \\ - 1\end{matrix}\right)$

Using Product of two matrices :

$X = - \frac{1}{4} \left(\begin{matrix}- 5 + 1 \\ - 15 - 1\end{matrix}\right)$

$\implies X = - \frac{1}{4} \left(\begin{matrix}- 4 \\ - 16\end{matrix}\right)$

$\implies \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}1 \\ 4\end{matrix}\right)$

$\implies x = 1 \mathmr{and} y = 4$