The InverseTutor command by default opens a Maplet window which allows you to interactively find the inverse of a matrix. Options provide other ways to show the step-by-step solutions, as described below.

The InverseTutor(M) command allows you to interactively find the inverse, if it exists, of the square Matrix M by using row operations to reduce the augmented Matrix $⟨M|\mathrm{Id}⟩$ to $⟨\mathrm{Id}|\frac{1}{M}⟩$ where Id is the identity Matrix of the same dimension as M. It returns the inverse Matrix.

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

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

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)

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

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

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

$M^{-1}$

$\left[\begin{array}{ccc}\frac{1}{5}& \frac{2}{5}& -\frac{4}{5}\\ \frac{2}{5}& -\frac{1}{5}& \frac{2}{5}\\ -\frac{4}{5}& \frac{2}{5}& \frac{1}{5}\end{array}\right]$

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

$\mathrm{InverseTutor}\left(M\,\mathrm{output}=\mathrm{steps}\right)$

$\begin{array}{lll}& & \text{Compute the inverse of this matrix}\\ & & \left[\begin{array}{ccc}1& 2& 0\\ 2& 3& 2\\ 0& 2& 1\end{array}\right]\\ \text{•}& & \text{Augment matrix with the identity matrix}\\ & & \left[\begin{array}{cccccc}1& 2& 0& 1& 0& 0\\ 2& 3& 2& 0& 1& 0\\ 0& 2& 1& 0& 0& 1\end{array}\right]\\ \text{•}& & \text{Subtract 2 times row 1 from row 2; (R2 = R2-2*R1)}\\ & & \left[\begin{array}{cccccc}1& 2& 0& 1& 0& 0\\ 0& \mathrm{-1}& 2& \mathrm{-2}& 1& 0\\ 0& 2& 1& 0& 0& 1\end{array}\right]\\ \text{•}& & \text{Multiply row 2 by -1; (R2 = -1*R2)}\\ & & \left[\begin{array}{cccccc}1& 2& 0& 1& 0& 0\\ 0& 1& \mathrm{-2}& 2& \mathrm{-1}& 0\\ 0& 2& 1& 0& 0& 1\end{array}\right]\\ \text{•}& & \text{Subtract 2 times row 2 from row 1; (R1 = R1-2*R2)}\\ & & \left[\begin{array}{cccccc}1& 0& 4& \mathrm{-3}& 2& 0\\ 0& 1& \mathrm{-2}& 2& \mathrm{-1}& 0\\ 0& 2& 1& 0& 0& 1\end{array}\right]\\ \text{•}& & \text{Subtract 2 times row 2 from row 3; (R3 = R3-2*R2)}\\ & & \left[\begin{array}{cccccc}1& 0& 4& \mathrm{-3}& 2& 0\\ 0& 1& \mathrm{-2}& 2& \mathrm{-1}& 0\\ 0& 0& 5& \mathrm{-4}& 2& 1\end{array}\right]\\ \text{•}& & \text{Multiply row 3 by 1/5; (R3 = 1/5*R3)}\\ & & \left[\begin{array}{cccccc}1& 0& 4& \mathrm{-3}& 2& 0\\ 0& 1& \mathrm{-2}& 2& \mathrm{-1}& 0\\ 0& 0& 1& -\frac{4}{5}& \frac{2}{5}& \frac{1}{5}\end{array}\right]\\ \text{•}& & \text{Subtract 4 times row 3 from row 1; (R1 = R1-4*R3)}\\ & & \left[\begin{array}{cccccc}1& 0& 0& \frac{1}{5}& \frac{2}{5}& -\frac{4}{5}\\ 0& 1& \mathrm{-2}& 2& \mathrm{-1}& 0\\ 0& 0& 1& -\frac{4}{5}& \frac{2}{5}& \frac{1}{5}\end{array}\right]\\ \text{•}& & \text{Add 2 times row 3 to row 2; (R2 = 2*R3+R2)}\\ & & \left[\begin{array}{cccccc}1& 0& 0& \frac{1}{5}& \frac{2}{5}& -\frac{4}{5}\\ 0& 1& 0& \frac{2}{5}& -\frac{1}{5}& \frac{2}{5}\\ 0& 0& 1& -\frac{4}{5}& \frac{2}{5}& \frac{1}{5}\end{array}\right]\\ \text{•}& & \text{We have found the inverse}\\ & & \left[\begin{array}{ccc}\frac{1}{5}& \frac{2}{5}& -\frac{4}{5}\\ \frac{2}{5}& -\frac{1}{5}& \frac{2}{5}\\ -\frac{4}{5}& \frac{2}{5}& \frac{1}{5}\end{array}\right]\end{array}$