Richard Skinner		rls2723
Week 8		5.4.13
http://208.180.24.221/5_4_13.htm

Prove that if A and B are square matrices satisfying AB = I, then A and B are invertible (and therefore BA = I also).

Solution:
Although this is a proof by example, one can see that it is extendable to an infinite degree within square matrices. All the same, I wouldn't suggest it as the math is horrendous. For this reason I decided to use Maple to help me with the algebra on a 3x3 matrix.

> A := matrix(3,3,[a11,a12,a13,a21,a22,a23,a31,a32,a33]);

> B := matrix(3,3,[b11,b12,b13,b21,b22,b23,b31,b32,b33]);

> multiply(B,A);

> eq1 := a11*b11+a21*b12+a31*b13=1;

> eq2 := b11*a12+b12*a22+b13*a32=0;

> eq3 := b11*a13+b12*a23+b13*a33=0;

> eq4 := b21*a11+b22*a21+b23*a31=0;

> eq5 := a12*b21+a22*b22+a32*b23=1;

> eq6 := b21*a13+b22*a23+b23*a33=0;

> eq7 := b31*a11+b32*a21+b33*a31=0;

> eq8 := b31*a12+b32*a22+b33*a32=0;

> eq9 := a13*b31+a23*b32+a33*b33=1;

> solve({eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9},{a11,a12,a13,a21,a22,a23,a31,a32,a33});

> multiply(A,B);

> eq1 := a11*b11+a12*b21+a13*b31=1;

> eq2 := a11*b12+a12*b22+a13*b32=0;

> eq3 := a11*b13+a12*b23+a13*b33=0;

> eq4 := a21*b11+a22*b21+a23*b31=0;

> eq5 := a21*b12+a22*b22+a23*b32=1;

> eq6 := a21*b13+a22*b23+a23*b33=0;

> eq7 := a31*b11+a32*b21+a33*b31=0;

> eq8 := a31*b12+a32*b22+a33*b32=0;

> eq9 := a31*b13+a32*b23+a33*b33=1;

> solve({eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9},{a11,a12,a13,a21,a22,a23,a31,a32,a33});

You can see by comparing the specific solutions of A within AB and BA that they are equal when both multiplications are set equal to the indentity matrix I. Thus it is proved for a 3x3 square matrix. This, however, is extendible to an infinitely large square matrix.