Answer
$\left\{\left(\dfrac{1}{2},3\right)\right\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
6x+3y=12\\
2x-y=-2
\end{cases}$
The matrices associated with the system are:
$A=\begin{bmatrix}6&3\\2&-1\end{bmatrix}$
$X=\begin{bmatrix}x\\y\end{bmatrix}$
$B=\begin{bmatrix}12\\-2\end{bmatrix}$
We have:
$AX=B$
$X=A^{-1}B$
Determine $A^{-1}$ if it exists.
$detA=6(-1)-2(3)=-12$
As $det A\not=0$, $A^{-1}$ exists. Determine $A^{-1}$:
$A^{-1}=\dfrac{1}{det A}\begin{bmatrix}-1&-3\\-2&6\end{bmatrix}=\dfrac{1}{-12}\begin{bmatrix}-1&-3\\-2&6\end{bmatrix}=\begin{bmatrix}\dfrac{1}{12}&\dfrac{1}{4}\\\dfrac{1}{6}&-\dfrac{1}{2}\end{bmatrix}$
Determine $X$:
$X=A^{-1}B=\begin{bmatrix}\dfrac{1}{12}&\dfrac{1}{4}\\\dfrac{1}{6}&-\dfrac{1}{2}\end{bmatrix}\begin{bmatrix}12\\-2\end{bmatrix}$
$=\begin{bmatrix}\dfrac{1}{2}\\3\end{bmatrix}$
$x=\dfrac{1}{2}$
$y=3$
The solution set is:
$\left\{\left(\dfrac{1}{2},3\right)\right\}$
