Quintin Stafford

2022-06-24

Finding the amount of solutions in a 3 equation solution
$x+2y+3z=7$
$3x+4y-z=4$
$3x+2y-11z=-13$

Elianna Douglas

Expert

We have the system
$x+2y+3z=7$
$3x+4y-z=4$
$3x+2y-11z=-13$
The associated augmented coefficient matrix is given by:
$\left(\begin{array}{ccccc}1& 2& 3& |& 7\\ 3& 4& -1& |& 4\\ 3& 2& -11& |& -13\end{array}\right)$
Use Gaussian Elimination (row reduction using elementary row operations). If you encounter a row with all zeros, then you'll know there are infinitely many solutions. You'll have an inconsistent system, for which there are NO solutions, if you obtain a row such as
$\left(0\phantom{\rule{thickmathspace}{0ex}}0\phantom{\rule{thickmathspace}{0ex}}0\phantom{\rule{thickmathspace}{0ex}}|\phantom{\rule{thickmathspace}{0ex}}a\right)$
where $a\ne 0$. Otherwise, the system will have one unique solution.
If you row reduce correctly, you should obtain a reduced row form as follows:
$\left(\begin{array}{ccccc}1& 0& -7& |& -10\\ 0& 1& 5& |& \frac{17}{2}\\ 0& 0& 0& |& 0\end{array}\right)$

flightwingsd2

Expert

$\left[\begin{array}{rrrr}1& 2& 3& 7\\ 3& 4& -1& 4\\ 3& 2& -11& -13\end{array}\right]$
$\left[\begin{array}{rrrr}1& 2& 3& 7\\ 0& -2& -10& -17\\ 0& -4& -20& -34\end{array}\right]$
$\left[\begin{array}{rrrr}1& 2& 3& 7\\ 0& -2& -10& -17\\ 0& 0& 0& 0\end{array}\right]$
$0=0$
Infinite amount of solutions

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