The BackwardSubstitute(A, b) command, where A is an upper row echelon Matrix and b is a Vector, returns a solution to the equation $A·x=b$.

If b is not included in the calling sequence, A is assumed to be an augmented Matrix.

$\mathrm{with}\left(\mathrm{Student}\left[\mathrm{LinearAlgebra}\right]\right)\:$

$M≔⟨⟨2\,4\,-1⟩\left|⟨4\,-2\,-2⟩\right|⟨0\,2\,5⟩⟩$

$M≔\left[\begin{array}{ccc}2& 4& 0\\ 4& \mathrm{-2}& 2\\ \mathrm{-1}& \mathrm{-2}& 5\end{array}\right]$

$A≔\mathrm{GaussianElimination}\left(M\right)$

$A≔\left[\begin{array}{ccc}2& 4& 0\\ 0& \mathrm{-10}& 2\\ 0& 0& 5\end{array}\right]$

$\mathrm{BackwardSubstitute}\left(A\,⟨4\,-5\,-5⟩\right)$

$\left[\begin{array}{c}\frac{7}{5}\\ \frac{3}{10}\\ \mathrm{-1}\end{array}\right]$

$M≔⟨⟨2\,0⟩\left|⟨3\,0⟩\right|⟨-1\,3⟩⟩$

$M≔\left[\begin{array}{ccc}2& 3& \mathrm{-1}\\ 0& 0& 3\end{array}\right]$

$A≔⟨M|⟨4\,1⟩⟩$

$A≔\left[\begin{array}{cccc}2& 3& \mathrm{-1}& 4\\ 0& 0& 3& 1\end{array}\right]$

$\mathrm{BackwardSubstitute}\left(A\right)$

$\left[\begin{array}{c}\frac{13}{6}-\frac{3{\mathrm{\_t0}}_1}{2}\\ {\mathrm{\_t0}}_1\\ \frac{1}{3}\end{array}\right]$