Question

Use matrix method to solve the following system of equations:
$x+y+z=0$
$2x-y+z=-3$
$x+2y-z=8$

Answer 1

$\left(\begin{array}{cccc}1& 1& 1& 0\\ 2& -1& 1& -3\\ 1& 2& -1& 8\end{array}\right)$

$R_3\to R_3-R_1$

$R_2\to R_2-R_1$

$\left(\begin{array}{cccc}1& 1& 1& 0\\ 0& -3& -1& -3\\ 0& 1& -2& 8\end{array}\right)$

$R_2\to R_2+2R_3$

$\left(\begin{array}{cccc}1& 1& 1& 0\\ 0& -1& -5& 13\\ 0& 1& -2& 8\end{array}\right)$

$R_3\to R_3+R_2$

$\left(\begin{array}{cccc}1& 1& 1& 0\\ 0& -1& -5& 13\\ 0& 0& -7& 21\end{array}\right)$ i.e $EcholonFormofamatrix$

From the matrix, we get the equations

$x+y+z=\mathrm{0............}\left(1\right)$

$-y-5z=\mathrm{13............}\left(2\right)$

$-7z=21\Rightarrow z=-\mathrm{3................}\left(3\right)$

Put z=-3 in equation 2

$-y-5(-3)=13\Rightarrow y=2$

Put $z=-3,y=2$ in equation 1

$x+2-3=0\Rightarrow x=1$

Therefore $(x,y,z)are(1,2,-3)$