Jamiya Weber

2022-06-21

How to solve this particular system of linear equations?
$\left[\begin{array}{ccccc}1& 0& 0& -1& 2\\ 0& 1& 0& 2& -3\\ 0& 0& 1& -1& 2\end{array}\right]$

Carmelo Payne

From
$\left[\begin{array}{ccccc}1& 0& 0& -1& 2\\ 0& 1& 0& 2& -3\\ 0& 0& 1& -1& 2\end{array}\right]$
you only have to paint, no to calculate, to get the solution: expand the matrix in this
$\left[\begin{array}{ccccc}1& 0& 0& -1& 2\\ 0& 1& 0& 2& -3\\ 0& 0& 1& -1& 2\\ 0& 0& 0& -1& 0\\ 0& 0& 0& 0& -1\end{array}\right]$
way and convince yourself that the solution set is
$\left\{\left[\begin{array}{c}2\\ -3\\ 2\\ 0\\ -1\end{array}\right]+t\left[\begin{array}{c}-1\\ 2\\ -1\\ -1\\ 0\end{array}\right]\mid t\in R\right\}.$

Manteo2h

You found
$\left\{\begin{array}{ll}{x}_{1}-{x}_{4}& =2\\ {x}_{2}+2{x}_{4}& =-3\\ {x}_{3}-{x}_{4}& =2\end{array}$
which is equivalent to
$\left\{\begin{array}{ll}{x}_{1}& ={x}_{3}\\ {x}_{1}& =2+{x}_{4}\\ {x}_{2}& =-3-2{x}_{4}\end{array}$
So the set of solutions is
$\left\{\left(2+{x}_{4},-3-2{x}_{4},2+{x}_{4},{x}_{4}\right):{x}_{4}\in \mathbb{F}\right\}.$

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