The GaussianEliminationTutor command by default opens a Maplet window which allows you to work interactively through Gaussian elimination. Options provide other ways to show the step-by-step solutions, as described below.

The GaussianEliminationTutor(M) command calls the Matrix Property Analyzer form of the tutor. The tutor allows you to interactively reduce the Matrix M to row echelon form using Gaussian elimination.  You can then query for the rank, nullity, and bases for the row, column, and null spaces. It returns the reduced Matrix.

The GaussianEliminationTutor(M, v) command calls the System Solver form of the tutor. The tutor allows you to interactively solve the system $M·x=v$ by reducing the augmented Matrix <M | v> to row echelon form using Gaussian elimination followed by backwards substitution. It returns the solution as a column Vector.

The above two forms of the tutor are mutually exclusive, and certain commands from one form are not available in the other. To access the other form, close the tutor, and call it again using the appropriate arguments. If no arguments are given, the tutor will start in the System Solver form by default.

Floating-point numbers in M or v are converted to rationals before computation begins.

The dimensions of the Matrix must be no greater than 5x5.

Note: There are two definitions of row echelon form. This tutor uses this definition, which requires only that pivot elements be nonzero.  (The other definition of row echelon form requires the pivot elements to be equal to 1.)

When the tutor is closed, the solution steps displayed in the tutor are re-displayed in the worksheet, with annotations showing the operations applied. Note that this display only and is not a value which can be further manipulated from within the worksheet. The value returned from the tutor, which can be used (for example, by referencing its equation label), is the state of the problem at the time the tutor was closed.

The following options can be used to control how the problem is displayed and what output is returned, giving the ability to generate step-by-step solutions directly without going through the Maplet tutor interface:

output = steps,canvas,script,record,list,print,printf,typeset,link (default: maplet)

displaystyle= columns,compact,linear,brief (default: linear)

$\mathrm{with}\left(\mathrm{Student}\left[\mathrm{LinearAlgebra}\right]\right)\&colon;$

$M≔⟨⟨1\&comma;2\&comma;0⟩\left|⟨2\&comma;3\&comma;2⟩\right|⟨0\&comma;2\&comma;1⟩|⟨3\&comma;5\&comma;5⟩⟩$

$M≔\left[\begin{array}{cccc}1& 2& 0& 3\\ 2& 3& 2& 5\\ 0& 2& 1& 5\end{array}\right]$

$v≔⟨5\&comma;4\&comma;2⟩$

$v≔\left[\begin{array}{c}5\\ 4\\ 2\end{array}\right]$

$\mathrm{GaussianEliminationTutor}\left(M\right)$

$\mathrm{GaussianEliminationTutor}\left(M\&comma;v\right)$

$\mathrm{GaussianEliminationTutor}\left(M\&comma;\mathrm{output}=\mathrm{steps}\right)$

$\begin{array}{lll}& & \text{Gaussian Reduce}\\ & & \left[\begin{array}{cccc}1& 2& 0& 3\\ 2& 3& 2& 5\\ 0& 2& 1& 5\end{array}\right]\\ \text{•}& & \text{Subtract 2 times row 1 from row 2; (R2 = R2-2*R1)}\\ & & \left[\begin{array}{cccc}1& 2& 0& 3\\ 0& \mathrm{-1}& 2& \mathrm{-1}\\ 0& 2& 1& 5\end{array}\right]\\ \text{•}& & \text{Add 2 times row 2 to row 3; (R3 = 2*R2+R3)}\\ & & \left[\begin{array}{cccc}1& 2& 0& 3\\ 0& \mathrm{-1}& 2& \mathrm{-1}\\ 0& 0& 5& 3\end{array}\right]\end{array}$

$\mathrm{GaussianEliminationTutor}\left(M\&comma;v\&comma;\mathrm{output}=\mathrm{steps}\right)$

$\begin{array}{lll}& & \text{Gaussian Solve}\\ & & \left[\begin{array}{ccccc}1& 2& 0& 3& 5\\ 2& 3& 2& 5& 4\\ 0& 2& 1& 5& 2\end{array}\right]\\ \text{•}& & \text{Subtract 2 times row 1 from row 2; (R2 = R2-2*R1)}\\ & & \left[\begin{array}{ccccc}1& 2& 0& 3& 5\\ 0& \mathrm{-1}& 2& \mathrm{-1}& \mathrm{-6}\\ 0& 2& 1& 5& 2\end{array}\right]\\ \text{•}& & \text{Add 2 times row 2 to row 3; (R3 = 2*R2+R3)}\\ & & \left[\begin{array}{ccccc}1& 2& 0& 3& 5\\ 0& \mathrm{-1}& 2& \mathrm{-1}& \mathrm{-6}\\ 0& 0& 5& 3& \mathrm{-10}\end{array}\right]\\ \text{•}& & \text{Extract equations from the matrix}\\ & & \left[\begin{array}{c}{x}_1+2x_2+3x_4=5\\ -x_2+2x_3-x_4=\mathrm{-6}\\ 5x_3+3x_4=\mathrm{-10}\end{array}\right]\\ \text{•}& & \text{Introduce free variables}\left[x_4=t\right]\\ & & \left[\begin{array}{c}{x}_1+2x_2+3t=5\\ -x_2+2x_3-t=\mathrm{-6}\\ 5x_3+3t=\mathrm{-10}\end{array}\right]\\ \text{•}& & \text{Solve}{x}_3\\ & & x_3=-2-\frac{3t}{5}\\ \text{•}& & \text{Substitute solved value back into equations}\\ & & \left[\begin{array}{c}{x}_1+2x_2+3t=5\\ -x_2-4-\frac{11t}{5}=\mathrm{-6}\\ x_3=-2-\frac{3t}{5}\end{array}\right]\\ \text{•}& & \text{Solve}{x}_2\\ & & x_2=2-\frac{11t}{5}\\ \text{•}& & \text{Substitute solved value back into equations}\\ & & \left[\begin{array}{c}{x}_1+4-\frac{7t}{5}=5\\ x_2=2-\frac{11t}{5}\\ x_3=-2-\frac{3t}{5}\end{array}\right]\\ \text{•}& & \text{Solve}{x}_1\\ & & x_1=1+\frac{7t}{5}\\ \text{•}& & \text{Substitute solved value back into equations}\\ & & \left[\begin{array}{c}{x}_1=1+\frac{7t}{5}\\ x_2=2-\frac{11t}{5}\\ x_3=-2-\frac{3t}{5}\end{array}\right]\end{array}$