Suppose A is a 5 x 5 matrix and suppose that det(A) = 0. I have three questions...
1.) What could be said of the dimensions of the row space, column space, and the null space?


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*I have that the dimeniosn of the row and column space would have to be less than 5 and the null space would be 5 minus the dimension of the row space. 


2.) For each b, can you tell if there is at least one solution to Ax = b?


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*I have that you can know if there is a solution by finding whether or not b is in the column space (how do you know if b is in the column space?).


3.) For each b, can you tell if there is at most one solution to Ax = b?


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*I don't have anything for this one. 

 A: We have $\det(A)=0$ iff $A$ is singular iff $\ker A\ne\{0\}$ so by the rank-nullity theorem $\dim\ker A\ge1$ and the dimension of rows space equal to dimension of columns space which is less than $4$. The equation $Ax=b$ has a solution iff $b\in \operatorname{Im}(A)$ and in this case there are infinity number of solutions:
if $x_0$ is a solution then $x_0+u, u\in\ker A$ is also a solution.
A: 1) What you have is great.
2) Yes - There is a solution to $Ax=b$ if and only if $b$ is in the column space. However, since the dimension of the latter is less than $5$, there are vectors which are not in the column space. For such values of $b$, there is no solution to the above equation.
3) As follows from 1), the null space is not trivial, and so there are many solutions to $Ax=0$.
A: One way to determine if $b$ is in the column space of $A$ is to go about solving the equation by Gauss-Jordan elimination.  At some point in triangularizing the matrix (also referred to as LU decomposition) you will reach a point where all the remaining rows of the matrix are $0$.  If the coefficients of the (transformed) $b$ corresponding to the $0$ rows in the matrix are all $0$ then $b$ is in the column space of $A$.
From that point on, if $b$ is in the column space of $A$, for all the coefficients of $x$ that correspond to the $0$ rows of the transformed $A$ you have a free choice.  Choose any convenient values for those coefficients of $x$ then proceed with back substitution.  This gives you a particular value of $x$ that is a valid solution for the equation $Ax=b$.  Then for all $y$ for which $Ay=0$, $x+ky$ is a valid solution to the equation also.
