# How do you solve 3x-y=3 and -2x+y=2 using matrices?

Aug 4, 2016

The Soln. is $x = 5 , y = 12$.

#### Explanation:

We can write the given system of eqns., using the Matrix Form, as :

$\left[\begin{matrix}3 & - 1 \\ - 2 & 1\end{matrix}\right] \left[\begin{matrix}x \\ y\end{matrix}\right] = \left[\begin{matrix}3 \\ 2\end{matrix}\right] \ldots \ldots \ldots \ldots . \left(1\right)$.

Let $A = \left[\begin{matrix}3 & - 1 \\ - 2 & 1\end{matrix}\right] , X = \left[\begin{matrix}x \\ y\end{matrix}\right] , \mathmr{and} , B = \left[\begin{matrix}3 \\ 2\end{matrix}\right]$.

Then, $\left(1\right)$ becomes, $A X = B$.

Now, we know from Algebra that the soln. of $\left(1\right)$ exists iff ${A}^{-} 1$ exists.

Also, ${A}^{-} 1$ exists iff $\det A \ne 0$, where,

$\det A = \det | \left(3 , - 1\right) , \left(- 2 , 1\right) | = 3 - 2 = 1 \ne 0$. Hence, ${A}^{-} 1$ exists, and, so does the unique soln. of the eqns.

Now, ${A}^{-} 1 = \frac{1}{\det} A \cdot a \mathrm{dj} A = \frac{1}{1} \left[\begin{matrix}1 & 1 \\ 2 & 3\end{matrix}\right] = \left[\begin{matrix}1 & 1 \\ 2 & 3\end{matrix}\right]$

Therefore, the soln. is $X = {A}^{-} 1 B = \left[\begin{matrix}1 & 1 \\ 2 & 3\end{matrix}\right] \left[\begin{matrix}3 \\ 2\end{matrix}\right]$

$\therefore \left[\begin{matrix}x \\ y\end{matrix}\right] = \left[\begin{matrix}1 & 1 \\ 2 & 3\end{matrix}\right] \left[\begin{matrix}3 \\ 2\end{matrix}\right] = \left[\begin{matrix}5 \\ 12\end{matrix}\right]$

So, the Soln. is $x = 5 , y = 12$.