<s>[INST] How to Solve a AU=B System when Determinant of A=0 Let
$$A=\begin{pmatrix} 1 & 5 & -1\\ -2 & -10 & 5\\ -2 & -10 & -1\end{pmatrix}$$
and 
$$B=\begin{pmatrix} 0 \\ 15 \\ -15\end{pmatrix}.$$
To find a vector $U$ such that $A$ maps $U$ to $B$ then you must solve the system $AU = B$.
Isolating for U yields $U = A^{-1}B$
However, when attempting to solve the inverse of $A$ produces all $0$'s in the bottom row during matrix inversion algorithm, otherwise known as a determinant of $0$.
 [/INST] After Gaussian elimination we get a system in echelon form,
$$\begin{pmatrix}1&5&-1\\0&0&3\\0&0&0\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\15\\0\end{pmatrix}.$$
So by the second equation, $z=5$ and by the first $x+5y=5$. This is all you can say.
</s>