Answer:
[tex]x_1=-4[/tex] and [tex]x_2=5[/tex]
Step-by-step explanation:
Consider given system ,
[tex]5x_1-2x_2=-30\\\\\\2x_1-x_2=-13[/tex]
We have to solve for [tex]x_1[/tex] and [tex]x_2[/tex]
In matrix form,
[tex]\left[\begin{array}{cc}5&-2\\2&-1\end{array}\right]\left[\begin{array}{c}x_1\\x_2\end{array}\right] =\left[\begin{array}{c}-30\\-13\end{array}\right][/tex]
then this is in form of AX = b,
A=[tex]\left[\begin{array}{cc}5&-2\\2&-1\end{array}\right][/tex]
X=[tex]\left[\begin{array}{c}x_1\\x_2\end{array}\right][/tex]
b=[tex]\left[\begin{array}{c}-30\\-13\end{array}\right][/tex]
Pre-multiply by [tex]A^{-1}[/tex] both side, we get,
[tex]X=A^{-1}B[/tex] ..............(1)
First finding inverse of A ,
[tex]\mathrm{Find\:2x2\:matrix\:inverse\:according\:to\:the\:formula}:\quad \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}^{-1}=\frac{1}{\det \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}}\begin{pmatrix}d\:&\:-b\:\\ -c\:&\:a\:\end{pmatrix}[/tex]
[tex]=\frac{1}{\det \begin{pmatrix}5&-2\\ 2&-1\end{pmatrix}}\begin{pmatrix}-1&-\left(-2\right)\\ -2&5\end{pmatrix}[/tex]
[tex]\det \begin{pmatrix}5&-2\\ 2&-1\end{pmatrix}=-1[/tex]
Thus, inverse of A is ,
[tex]\frac{1}{-1}\begin{pmatrix}-1&-\left(-2\right)\\ -2&5\end{pmatrix}=\begin{pmatrix}1&-2\\ 2&-5\end{pmatrix}[/tex]
Now substitute [tex]A^{-1}[/tex] in (1) , we get,
[tex]\left[\begin{array}{c}x_1\\x_2\end{array}=\left[\begin{array}{cc}1&-2\\2&-5\end{array}\right]\left[\begin{array}{c}-30\\-13\end{array}\right][/tex]
Multiply , we get,
[tex]=\begin{pmatrix}1\cdot \left(-30\right)+\left(-2\right)\left(-13\right)\\ 2\left(-30\right)+\left(-5\right)\left(-13\right)\end{pmatrix}[/tex]
[tex]\left[\begin{array}{c}x_1\\x_2\end{array}\right]=\left[\begin{array}{c}-4\\5\end{array}\right][/tex]
Thus, [tex]x_1=-4[/tex] and [tex]x_2=5[/tex]
