Question

Use matrices to solve this linear system:

Answer 1

Answer:

$x_1=-4$ and $x_2=5$

Step-by-step explanation:

Consider  given system ,  

$5x_1-2x_2=-30\\\\\\2x_1-x_2=-13$

We have to solve for $x_1$ and $x_2$

In matrix form,

$\left[\begin{array}{cc}5&-2\\2&-1\end{array}\right]\left[\begin{array}{c}x_1\\x_2\end{array}\right] =\left[\begin{array}{c}-30\\-13\end{array}\right]$

then this is in form of AX = b,

A=$\left[\begin{array}{cc}5&-2\\2&-1\end{array}\right]$

X=$\left[\begin{array}{c}x_1\\x_2\end{array}\right]$

b=$\left[\begin{array}{c}-30\\-13\end{array}\right]$

Pre-multiply by $A^{-1}$ both side, we get,

$X=A^{-1}B$   ..............(1)

First finding inverse of A ,

$\mathrm{Find\:2x2\:matrix\:inverse\:according\:to\:the\:formula}:\quad \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}^{-1}=\frac{1}{\det \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}}\begin{pmatrix}d\:&\:-b\:\\ -c\:&\:a\:\end{pmatrix}$

$=\frac{1}{\det \begin{pmatrix}5&-2\\ 2&-1\end{pmatrix}}\begin{pmatrix}-1&-\left(-2\right)\\ -2&5\end{pmatrix}$

$\det \begin{pmatrix}5&-2\\ 2&-1\end{pmatrix}=-1$

Thus, inverse of A is ,

$\frac{1}{-1}\begin{pmatrix}-1&-\left(-2\right)\\ -2&5\end{pmatrix}=\begin{pmatrix}1&-2\\ 2&-5\end{pmatrix}$

Now substitute $A^{-1}$ in (1) , we get,

$\left[\begin{array}{c}x_1\\x_2\end{array}=\left[\begin{array}{cc}1&-2\\2&-5\end{array}\right]\left[\begin{array}{c}-30\\-13\end{array}\right]$

Multiply , we get,

$=\begin{pmatrix}1\cdot \left(-30\right)+\left(-2\right)\left(-13\right)\\ 2\left(-30\right)+\left(-5\right)\left(-13\right)\end{pmatrix}$

$\left[\begin{array}{c}x_1\\x_2\end{array}\right]=\left[\begin{array}{c}-4\\5\end{array}\right]$

Thus,  $x_1=-4$ and $x_2=5$