Suppose A is the 3x3 matrix whose rows are [2 4 1], [3 -2 2], and [-1 3 4], and suppose X is the 3x1 matrix whose entries are 2, 1, 5. Then AX is (enter A and X and the command Y = A*X) the 3x1 matrix whose entries are 13, 14, 21. The goal is to use A and Y to reconstruct X.
The command [D, M, N] = quick(A) gives the diagonal matrix D with diagonal entries 1, 1, -77 equivalent to A and gives unimodular matrices M and N for which MAN = D. (This is fairly typical. If you just pick a 3x3 matrix at random, you will very likely find that `quick' gives a diagonal matrix whose first two diagonal entries are 1 and whose last diagonal entry is a rather large number, possibly negative. The reason for this phenomenon will become clear in Chapter 5.) That's all there is to step (1).
Step (2) is to find all solutions Z of DZ = MY. The command M*Y immediately shows that MY is the 3x1 matrix whose entries are 13, 1, 0. Since the diagonal entries of Z are nonzero, this unambiguously determines Z. Since Z is the same as DZ (zero divided by -77 is zero), enter the command Z = M*Y. That is all there is to step (2).
Finally, since there is just one possible Z, there is only one possible X, namely X = NZ, found by entering the command N*Z. This matrix is, of course, the matrix X with which we started, and the problem is solved.
Note that there will be no solution Z of DZ = MY unless all entries in the last row of MY are divisible by 77. This means that if Y is an arbitrarily chosen matrix with 3 rows, AX = Y is unlikely to have a solution X. (If Y has one column, the chances are 1 in 77; if it has two columns, they are 1 in 5929 = 77*77.) However, if, as suggested, you start with A and X, you know there is at least one solution of AX = Y and therefore at least one solution of DZ = MY. In other words, the entries in the last row of MY must all be divisible by 77.
In the same way, for this A you can always recover X when AX is given. Try this out by picking a 3x2 matrix X, finding Y = AX, and using the solution procedure to recover X. In particular, if AX is 0, then X must be 0. That is (see Proposition 2), this matrix A is not a zero divisor. If you pick a 3x3 matrix at random, you can be almost sure it will not be a zero divisor.
Return to Chapter 3.