General Solution to a Singular System (or no solution)

[MurdocJensen]

When I realize that I am going to have a singular matrix (after exhausting row swap options and maybe even some elimination steps) what about the matrix tells me whether or not I can have a general solution?

[Mark44]

Your question is sparse on details, so I'll answer what I think you're asking. Let's say we have these matrix equations:
Ax = 0
Ax = b

In both equations I'm assuming that A is a square, noninvertible matrix (i.e., |A| = 0).

Because |A| = 0, row reducing A will yield at least one row whose entries are all zero. This means that we have a system of equations with fewer equation than variables, meaning that at least one variable is free, so there are an infinite number of solutions for x.

If we represent the second matrix equation by an augmented matrix, row reducing A will still yield at least one row of zeroes on the left side of the augmented matrix. If the element of b that corresponds to that zero row is not zero, that row of the augmented matrix represents the equation
0x1 + 0x2 + ... + 0xn = bk, which has no solution. In this case, the system of equations is inconsistent.

[HallsofIvy]

Nothing about the coefficient matrix tells you that! If you are talking about the "augmented" matrix, where you have added the vector "b" (assuming your equation is Ax= b) as an additional column to the coefficient matrix, then the fact that the coefficient matrix is singular tells you that a row reduction will reduce the last row of the coefficient matrix to all "0"s. If there exist a row of the row-reduced augmented matrix where all entries except the last are "0" and the last column is not, there is NO solution. If, whenever all entries in a row, up to the last column, are "0" the last column entry in that row is also, then there exist an infinite number of solutions (so you can find a "general solution").

[MurdocJensen]

quickreply: i want to post an example system. anyone know how to type and copy matrices?