The BackSubstitution command returns a solution to the equation A.x=b, where A is an upper-triangular matrix, using the back substitution algorithm.

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

$A≔\mathrm{Matrix}\left(\left[\left[1.1\,4.9\,7.7\right]\,\left[4.4\,7.5\,8.1\right]\,\left[2.7\,8\,10\right]\right]\right)$

$A≔\left[\begin{array}{ccc}1.1& 4.9& 7.7\\ 4.4& 7.5& 8.1\\ 2.7& 8& 10\end{array}\right]$

$b≔\mathrm{Vector}\left(\left[3.4\,5.4\,3.3\right]\right)$

$b≔\left[\begin{array}{c}3.4\\ 5.4\\ 3.3\end{array}\right]$

$\mathrm{Lower},\mathrm{Upper}≔\mathrm{MatrixDecomposition}\left(A\,\mathrm{method}=\mathrm{LU}\,\mathrm{output}=\left['L'\,'U'\right]\right)$

$\mathrm{Lower},\mathrm{Upper}≔\left[\begin{array}{ccc}1& 0& 0\\ 4.000000000& 1& 0\\ 2.454545455& 0.3328324570& 1\end{array}\right],\left[\begin{array}{ccc}1.1& 4.9& 7.7\\ 0& \mathrm{-12.10000000}& \mathrm{-22.70000000}\\ 0& 0& \mathrm{-1.344703226}\end{array}\right]$

$c≔\mathrm{ForwardSubstitution}\left(\mathrm{Lower}\,b\right)$

$c≔\left[\begin{array}{c}3.4\\ \mathrm{-8.20000000}\\ \mathrm{-2.316228400}\end{array}\right]$

$\mathrm{BackSubstitution}\left(\mathrm{Upper}\,c\right)=\mathrm{LinearSolve}\left(A\,b\,\mathrm{method}=\mathrm{LU}\right)$

$\left[\begin{array}{c}2.409319491\\ \mathrm{-2.553749034}\\ 1.722482965\end{array}\right]=\left[\begin{array}{c}2.409319491\\ \mathrm{-2.553749034}\\ 1.722482965\end{array}\right]$