Properties of the inverse

So, at least the algorithm is clear. If we can keep on finding pivots successfully times, then we will get the desired result. Thus the only reason to fail is if one of the rows of becomes zero and yields no pivot. This is the working test and many times the only useful one.

A matrix having an inverse is said to be invertible or nonsingular . It is said to be singular otherwise . In a sense all nonsquare matrices should be called singular, however, the term is reserved for square matrices only.

Equations with singular left hand sides pose special problems in solving equations. We discuss a sample for now, but the full treatment is done in the third chapter.

Example: Solve where:

> A:=matrix(3,3,[1,1,2,2,1,3,3,4,7]);B:=matrix(3,1,[p,q,r]);

> M:=augment(A,B);

> M1:=addrow(M,1,2,-2);

> M2:=addrow(M1,1,3,-3);

> M3:=addrow(M2,2,3,1);

Thus there shall be no solution, unless the RHS of the last equation vanishes, i.e. . We now assume this condition and continue . But we have only two pivots and we need to clean out only the first two columns.

> M4:=addrow(M3,2,1,1);

>

Assuming the variables to be , we can write the solution as , and is arbitrary, provided, of course the consistency condition holds.

The variable is said to be free and the solutions are either an infinite set or an empty set in case the consistency condition fails.

If the right hand side has only independent unknown quantities, and if some row becomes zero on the left hand side, then we will always get consistency conditions . If the conditions are satisfied, then the whole rows of the augmented matrix become zero.

Some simple observations follow.

If the number of pivots is less than the total number of rows (number of equations), then we get consistency conditions and some right hand sides will lead to a no solution situation.

If the number of pivots is less than the number of columns of left hand side (number of variables), then there is at least one free variable and the solution set is either infinite or empty - so the solution is never unique!

It is not hard to convince yourself that the system has a unique solution for all possible (of correct height, of course) iff
the number of equations=number of variables=number of pivots .
In other words:
the number of rows=number of columns=number of pivots
Equivalently, . is invertible .
In particular, a nonsquare matrix cannot have an inverse!

Think about this. We will give more proof later.

It is possible to make a formula for the inverse of a matrix in terms of its coefficients or as a polynomial expression in the matrix. We only give a calculation to try for an invertible .

> A:=matrix(3,3,[1,2,1,-2,3,1,3,1,1]);charpoly(A,t);

> AA:=evalm(A^2-5*A+7*eye(3));

> evalm(AA&*A);

The polynomial charpoly(A,t) is a polynomial satisfied by the matrix and this is a very useful fundamental theorem, the so-called Cayley-Hamilton Theorem, to be learnt later. But using the fact that , we can guess the inverse of . We simply rewrite the equation as: or . Thus is the desired inverse!

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