In-class Demonstration: Inverses

By Russell Blyth

restart: with(LinearAlgebra):

We start with a random 3 x 3 matrix.

A:= RandomMatrix(3,3,generator=rand(-3..3));

Our task is to find the inverse of the matrix A.

The equation Y = AX computes Y given X. We need to compute X given Y. Define a general Y, and solve for X:

Y := Vector(3,<y1,y2,y3>); AugA := <A|Y>; RedA := ReducedRowEchelonForm(AugA);

Next we extract the expression for X in terms of Y

Xvec := DeleteColumn(RedA,1..3);

y1vec := eval(Xvec,{y1=1,y2=0,y3=0}); y2vec := eval(Xvec,{y1=0,y2=1,y3=0}); y3vec := eval(Xvec,{y1=0,y2=0,y3=1});

Thus Xvec = y1*y1vec + y2*y2vec + y3*y3vec. If we make a matrix with y1vec, y2vec, y3vec as the columns, then y1, y2 and y3 are the coefficients of the linear combination giving X. We write this as BY, where

B := <y1vec | y2vec | y3vec>;

Check:

B.Y;

This is just the transpose of Xvec. Thus X = BY. B is the inverse of A!

Check the products AB and BA:

A.B; B.A;

Note the first column of B arises from solving AX = NiMpNyUiIiIiIiFGJiUidEc=; the second column from solving AX =NiMpNyUiIiEiIiJGJSUidEc= and the third column from solving AX = NiMpNyUiIiFGJSIiIiUidEc=. These three systems may be solved simultaneously, by finding the reduced row echelon form of the matrix [A,I]

I3 := IdentityMatrix(3); BigA := <A | I3>; RedBigA := ReducedRowEchelonForm(BigA);

Note the final three columns are precisely the inverse matrix B.

Maple computes matrix inverses directly:

A^(-1);