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Student[LinearAlgebra]

 BackwardSubstitute
 solve A . x = b where A is in upper row echelon form

 Calling Sequence BackwardSubstitute(A, b, options)

Parameters

 A - upper row echelon form Matrix b - (optional) Vector options - (optional) parameters; for a complete list, see LinearAlgebra[BackwardSubstitute]

Description

 • The BackwardSubstitute(A, b) command, where A is an upper row echelon Matrix and b is a Vector, returns a solution to the equation $A\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}.\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}x=b$.
 If b is not included in the calling sequence, A is assumed to be an augmented Matrix.

Examples

 > $\mathrm{with}\left(\mathrm{Student}[\mathrm{LinearAlgebra}]\right):$
 > $M≔⟨⟨2,4,-1⟩|⟨4,-2,-2⟩|⟨0,2,5⟩⟩$
 ${M}{≔}\left[\begin{array}{rrr}{2}& {4}& {0}\\ {4}& {-}{2}& {2}\\ {-}{1}& {-}{2}& {5}\end{array}\right]$ (1)
 > $A≔\mathrm{GaussianElimination}\left(M\right)$
 ${A}{≔}\left[\begin{array}{rrr}{2}& {4}& {0}\\ {0}& {-}{10}& {2}\\ {0}& {0}& {5}\end{array}\right]$ (2)
 > $\mathrm{BackwardSubstitute}\left(A,⟨4,-5,-5⟩\right)$
 $\left[\begin{array}{c}\frac{{7}}{{5}}\\ \frac{{3}}{{10}}\\ {-}{1}\end{array}\right]$ (3)
 > $M≔⟨⟨2,0⟩|⟨3,0⟩|⟨-1,3⟩⟩$
 ${M}{≔}\left[\begin{array}{rrr}{2}& {3}& {-}{1}\\ {0}& {0}& {3}\end{array}\right]$ (4)
 > $A≔⟨M|⟨4,1⟩⟩$
 ${A}{≔}\left[\begin{array}{rrrr}{2}& {3}& {-}{1}& {4}\\ {0}& {0}& {3}& {1}\end{array}\right]$ (5)
 > $\mathrm{BackwardSubstitute}\left(A\right)$
 $\left[\begin{array}{c}\frac{{13}}{{6}}{-}\frac{{3}}{{2}}{}{{\mathrm{_t0}}}_{{1}}\\ {{\mathrm{_t0}}}_{{1}}\\ \frac{{1}}{{3}}\end{array}\right]$ (6)