Answer:
x = 2, y = -2
Step-by-step explanation:
This is an alternate way. You can also solve simultaneous equations using matrices. First, rewrite the simultaneous equations in matrix form.
[tex]\displaystyle \large{\left[\begin{array}{ccc}2&3\\1&1\end{array}\right] \left[\begin{array}{ccc}x\\y\end{array}\right] =\left[\begin{array}{ccc}2\\0\end{array}\right]}[/tex]
Let:
[tex]\displaystyle \large{A= \left[\begin{array}{ccc}2&3\\1&1\end{array}\right]}\\\displaystyle \large{X = \left[\begin{array}{ccc}x\\y\end{array}\right]}\\\displaystyle \large{B =\left[\begin{array}{ccc}2\\0\end{array}\right]}[/tex]
So now we have [tex]\displaystyle \large{AX = B}[/tex] and we have to solve for X which:
[tex]\displaystyle \large{A^{-1}AX = A^{-1}B}\\\displaystyle \large{X = A^{-1} B}[/tex]
Where:
[tex]\displaystyle \large{A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{ccc}d&-b\\-c&a\end{array}\right]}[/tex]
**ad-bc is determinant of 2*2 matrices and shortened as det(var) or |var|**
Therefore, from A:
Find detA:
[tex]\displaystyle \large{\det A = |A| = ad-bc = 2(1)-3(1) = 2-3 = -1}[/tex]
Therefore, detA = -1. Hence:
[tex]\displaystyle \large{A^{-1}=-1\left[\begin{array}{ccc}1&-3\\-1&2\end{array}\right]}[/tex]
And [tex]\displaystyle \large{X = A^{-1}B}[/tex]:
[tex]\displaystyle \large{X = -1\left[\begin{array}{ccc}1&-3\\-1&2\end{array}\right] \left[\begin{array}{ccc}2\\0\end{array}\right] }[/tex]
Evaluate the matrices:
[tex]\displaystyle \large{X = -1\left[\begin{array}{ccc}1 \cdot 2-3 \cdot 0\\-1 \cdot 2+2\cdot 0\end{array}\right] }\\\displaystyle \large{X = -1\left[\begin{array}{ccc}2-0\\-2+0\end{array}\right] }\\\displaystyle \large{X = -1\left[\begin{array}{ccc}2\\-2\end{array}\right] }\\\displaystyle \large{X = \left[\begin{array}{ccc}-2\\2\end{array}\right] }\\\displaystyle \large{\left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}-2\\2\end{array}\right] }[/tex]
Therefore, x = -2 and y = 2.
