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 + ... + 0x_n = b_k, 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?

 

[blue_raver22]

The fact that a matrix is singular, meaning it has no inverse, indicates that the system of equations it represents has either no solution or an infinite number of solutions. In order to determine which of these scenarios is the case, we can look at the reduced row echelon form (RREF) of the matrix.

If the RREF of the matrix has a row of all zeros, then the system has no solution. This is because in the RREF, this row represents an equation such as 0x + 0y + 0z = k, where k is a non-zero number. This is impossible to satisfy, as any value for x, y, or z will result in 0 on the left side of the equation, not k.

On the other hand, if the RREF of the matrix has no rows of all zeros, then the system has an infinite number of solutions. This is because in the RREF, each non-zero row represents an equation such as x + 0y + 0z = k, where k is a constant. This means that x can take on any value, and y and z are free variables. Therefore, there are infinitely many solutions to the system.

In summary, the RREF of a singular matrix can provide information about the existence and nature of solutions to the corresponding system of equations. It is important to carefully analyze the RREF in order to determine if the system has no solution or an infinite number of solutions.