# Row reduced matrix has coefficents

1. Jan 26, 2005

### 3.14159265358979

I just want to confirm these two questions. Thanks in advance.

(1) Describe all solutions of Ax = 0 in parametric vector form, where A is row equivalent to the given matrix.

$$\left(\begin{array}{uvwxyz}1 & 5 & 2 & -6 & 9 & 0 \\0 & 0 & 1 & -7 & 4 & -8\\0 & 0 & 0 & 0 & 0 & 1\\0 & 0 & 0 & 0 & 0 & 0\end{array}\right)$$

There are no solutions because row 3 and 4 contradict eachother. Row 3 implies no solution.

(2) Suppose A is a 3x3 matrix and y is a vector in R^{3} such that the equation Ax = y does not have a solution. Does there exist a vector z in R^{3} such that the equation Ax = z has a unique solution?

I said no because if the vector y does not have a solution in R^{3}, then this implies the last row of the row reduced matrix has coefficents that are all zero. Therefore, it either has no solution or an infinite number of solutions.

2. Jan 26, 2005

### matt grime

Rows three and four do not contradict themselves. All it says is that, with the obvious notation x_6=0 from row 3, and, from row 4, that 0=0. Row 4 contradicts nothing. Besides which, x=0 is always a solution, or otherwise you are saying that the kernel is empty, and since it is a nonempty subspace....

3. Jan 29, 2005

### HallsofIvy

Staff Emeritus
Problem 2 is related to the "Fredholm Alternative". The equation Ax= b has a unique solution if A is "non-singular". If A is singular then Ax= b has either no solution or an infinite number of solutions depending on b. In this case, since Ax= y has no solution, it might be (and in fact must be) the case that there exist z such that Ax= z has an infinite number of solutions but there cannot exist z such that Ax= z has a unique solution.