Determine values of k for a matrix to have a unique solution I have the following system and need to find for what values of $k$ does the system have 
i) a unique solution
ii)  no solution
iii) an infinite number of solutions
$(k^3+3k)x + (k+5)y + (k+3)z = k^5+(k+3)^2
ky + z = 3
(k^3+k^2-6k)z = k(k^2-9)$
Putting this into a matrix I get   
$M =\left[
\begin{array}{cc}
k^3 + 3k & k+5 & k+3 & k^5+(k+3)^2 \\
0 & k & 1 & 3 \\
0 & 0 & k^3+k^2-6k & k(k^2-9)\\
\end{array}\right]$
I understand for ii) we need the bottom row to read $0 \ 0 \ 0  \ k(k^2-9)$ which it does for values $k=2$.
For iii) we need the bottom row to read $0  \  0  \ 0  \ 0$ which we do when $k=0$ and $k=-3$.
Because the first column only contains a value for the top row, I can't use elementary row operations to chance the value in $a_{11}$, so I don't know how I can make it into the identity matrix?
 A: Coefficient matrix is given by
$A = \begin{pmatrix}
k^3+3k & k+5 & k+3\\
0      & k   & 1   \\
0      & 0   & k^3+k^2-6k
\end{pmatrix}$
And, augmented matrix is given by 
$[A:b] = \begin{pmatrix}
k^3+3k & k+5 & k+3 & k^5+(k+3)^2\\
0      & k   & 1   & 3\\
0      & 0   & k^3+k^2-6k & k(k^2-9)
\end{pmatrix}$
(1) UNIQUE SOLUTION, if $rank(A) = rank([A:b]) =$ number of variables (here, number of variables is $3$).
Find out the values of $k$ for which $det(A) \neq 0$. 
Values of $k$ are = $\mathbb{C}-\{0, \pm\sqrt{3} i, 2, -3\}$, where $\mathbb{C}$ denotes the set of complex numbers.
(2) NO SOLUTION, if $rank(A) \neq rank([A:b])$.
Check that, for $k = \pm\sqrt{3} i, 2$, $rank(A) = 2$ but $rank([A:b]) = 3$.
(3) INFINITELY MANY SOLUTION, if $rank(A) = rank ([A:b]) <$ number of variables.
Check that, for $k= 0, -3$, $rank(A) = rank ([A:b]) <$ number of variables. 
A: 
$\textbf{HINT:}$ Check the determinant. if $\det=0$ you will have infinite solutions, also you can check the last file, if $a_{32}=0$, then check $a_{34}:$ 
  
  
*
  
*if $a_{34}\neq0$ then there are no solutions. 
  
*if $a_{34}=0$ there are infinite solutions.
  

