Row reduced matrix has coefficents

[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 each other. 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.

 

[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...

 

[HallsofIvy]

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.