Kaycee Roche

2021-02-08

Use matrices to solve the given system of equations. Use back-substitution Gaussian elimination or Gauss-Jordan elimination.

cheekabooy

So,

$\left[A|B\right]=\left[\begin{array}{cccc}1& 3& 0& 0\\ 1& 1& 1& 1\\ 3& -1& -1& 11\end{array}\right]$

$⇒\left[A|B\right]=\left[\begin{array}{cccc}1& 3& 0& 0\\ 0& -2& 1& 1\\ 0& -10& -1& 11\end{array}\right]$
${R}_{2}\to -\frac{1}{2}{R}_{2}$
$⇒\left[A|B\right]=\left[\begin{array}{cccc}1& 3& 0& 0\\ 0& 1& -\frac{1}{2}& -\frac{1}{2}\\ 0& -10& -1& 11\end{array}\right]$
${R}_{3}\to {R}_{3}+10{R}_{2}$
$⇒\left[A|B\right]=\left[\begin{array}{cccc}1& 3& 0& 0\\ 0& 1& -\frac{1}{2}& -\frac{1}{2}\\ 0& 0& -6& 6\end{array}\right]$
${R}_{3}\to -\frac{1}{6}{R}_{3}$
$⇒\left[A|B\right]=\left[\begin{array}{cccc}1& 3& 0& 0\\ 0& 1& -\frac{1}{2}& -\frac{1}{2}\\ 0& 0& 1& -1\end{array}\right]$
$x+3y=0$
$y-\frac{1}{2}z=-\frac{1}{2}$
$z=-1$
So,
$y-\frac{1}{2}\left(-1\right)=-\frac{1}{2}$
$⇒y+\frac{1}{2}=-\frac{1}{2}$
$⇒y=-\frac{1}{2}-\frac{1}{2}$
$⇒y=-1$
$x+3\left(-1\right)=0$
$⇒x-3=0$
$⇒x=3$

Jeffrey Jordon