For which values of $\alpha \in \mathbb R$ is the following system of linear equations solvable? The problem I was given: 
Calculate the value of the following determinant:
$\left|
\begin{array}{ccc}
\alpha & 1 & \alpha^2 & -\alpha\\
1 & \alpha & 1 & 1\\
1 & \alpha^2 & 2\alpha & 2\alpha\\
1 & 1 & \alpha^2 & -\alpha
\end{array}
\right|$
For which values of $\alpha \in \mathbb R$ is the following system of linear equations solvable?
$\begin{array}{lcl} 
\alpha x_1 & + & x_2 & + & \alpha^2 x_3 & = & -\alpha\\ 
x_1 & + & \alpha x_2 & + & x_3 & = & 1\\
x_1 & + & \alpha^2 x_2 & + & 2\alpha x_3 & = & 2\alpha\\
x_1 & + & x_2 & + & \alpha^2 x_3 & = & -\alpha\\
\end{array}$
I got as far as finding the determinant, and then I got stuck.
So I solve the determinant like this: 
$\left|
\begin{array}{ccc}
\alpha & 1 & \alpha^2 & -\alpha\\
1 & \alpha & 1 & 1\\
1 & \alpha^2 & 2\alpha & 2\alpha\\
1 & 1 & \alpha^2 & -\alpha
\end{array}
\right|$ = 
$\left|
\begin{array}{ccc}
\alpha - 1 & 0 & 0 & 0\\
1 & \alpha & 1 & 1\\
1 & \alpha^2 & 2\alpha & 2\alpha\\
1 & 1 & \alpha^2 & -\alpha
\end{array}
\right|$ = 
$(\alpha - 1)\left|
\begin{array}{ccc}
\alpha & 1 & 1\\
\alpha^2 & 2\alpha & 2\alpha \\
1 & \alpha^2 & -\alpha
\end{array}
\right|$ =
$(\alpha - 1)\left|
\begin{array}{ccc}
\alpha & 1 & 0\\
\alpha^2 & 2\alpha & 0 \\
1 & \alpha^2 & -\alpha - \alpha^2
\end{array}
\right|$  = $-\alpha^3(\alpha - 1) (1 + \alpha)$
However, now I haven't got a clue on solving the system of linear equations... It's got to do with the fact that the equations look like the determinant I calculated before, but I don't know how to connect those two.
Thanks in advance for any help. (:
 A: Let me first illustrate an alternate approach. You're looking at $$\left[\begin{array}{ccc} 
\alpha & 1 & \alpha^2\\ 
1 & \alpha & 1\\
1 & \alpha^2 & 2\alpha\\
1 & 1 & \alpha^2
\end{array}\right]\left[\begin{array}{c} x_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c} -\alpha\\ 1\\ 2\alpha\\ -\alpha\end{array}\right].$$ We can use row reduction on the augmented matrix $$\left[\begin{array}{ccc|c} 
\alpha & 1 & \alpha^2 & -\alpha\\ 
1 & \alpha & 1 & 1\\
1 & \alpha^2 & 2\alpha & 2\alpha\\
1 & 1 & \alpha^2 & -\alpha
\end{array}\right].$$ In particular, for the system to be solvable, it is necessary and sufficient that none of the rows in the reduced matrix is all $0$'s except for in the last column. Subtract the bottom row from the other rows, yielding $$\left[\begin{array}{ccc|c} 
\alpha-1 & 0 & 0 & 0\\ 
0 & \alpha-1 & 1-\alpha^2 & 1+\alpha\\
0 & \alpha^2-1 & 2\alpha-\alpha^2 & 3\alpha\\
1 & 1 & \alpha^2 & -\alpha
\end{array}\right].$$
It's clear then that if $\alpha=1$, the second row has all $0$s except in the last column, so $\alpha=1$ doesn't give us a solvable system. Suppose that $\alpha\neq 1$, multiply the top row by $\frac1{\alpha-1}$, and subtract the new top row from the bottom row, giving us $$\left[\begin{array}{ccc|c} 
1 & 0 & 0 & 0\\ 
0 & \alpha-1 & 1-\alpha^2 & 1+\alpha\\
0 & \alpha^2-1 & 2\alpha-\alpha^2 & 3\alpha\\
0 & 1 & \alpha^2 & -\alpha
\end{array}\right].$$
Swap the second and fourth rows and add the new second row to the last two rows, giving us $$\left[\begin{array}{ccc|c} 
1 & 0 & 0 & 0\\ 
0 & 1 & \alpha^2 & -\alpha\\
0 & \alpha^2 & 2\alpha & 2\alpha\\
0 & \alpha & 1 & 1
\end{array}\right],$$ whence subtracting $\alpha$ times the fourth row from the third row gives us $$\left[\begin{array}{ccc|c} 
1 & 0 & 0 & 0\\ 
0 & 1 & \alpha^2 & -\alpha\\
0 & 0 & \alpha & \alpha\\
0 & \alpha & 1 & 1
\end{array}\right].$$
Note that $\alpha=0$ readily gives us the solution $x_1=x_2=0$, $x_3=1$. Assume that $\alpha\neq 0,$ multiply the third row by $\frac1\alpha$, subtract the new third row from the fourth row, and subtract $\alpha^2$ times the new third row from the second row, yielding $$\left[\begin{array}{ccc|c} 
1 & 0 & 0 & 0\\ 
0 & 1 & 0 & -\alpha^2-\alpha\\
0 & 0 & 1 & 1\\
0 & \alpha & 0 & 0
\end{array}\right],$$ whence subtracting $\alpha$ times the second row from the fourth row yields $$\left[\begin{array}{ccc|c} 
1 & 0 & 0 & 0\\ 
0 & 1 & 0 & -\alpha^2-\alpha\\
0 & 0 & 1 & 1\\
0 & 0 & 0 & \alpha^3+\alpha^2
\end{array}\right].$$ The bottom right entry has to be $0$, so since $\alpha\neq 0$ by assumption, we need $\alpha=-1$, giving us $$\left[\begin{array}{ccc|c} 
1 & 0 & 0 & 0\\ 
0 & 1 & 0 & 0\\
0 & 0 & 1 & 1\\
0 & 0 & 0 & 0
\end{array}\right].$$
Hence, the two values of $\alpha$ that give the system a solution are $\alpha=0$ and $\alpha=-1$, and in both cases, the system has solution $x_1=x_2=0$, $x_3=1$. (I think all my calculations are correct, but I'd recommend double-checking them.)

The major upside of the determinant approach is that it saves you time and effort, since you've already calculated it. If we assume that $\alpha$ is a constant that gives us a solution, then since we're dealing with $4$ equations in only $3$ variables, we have to have at least one of the rows in the reduced echelon form of the augmented matrix be all $0$s--we simply don't have enough degrees of freedom otherwise. The determinant of the reduced matrix will then be $0$, and since we obtain it by invertible row operations on the original matrix, then the determinant of the original matrix must also be $0$.
By your previous work, then, $-\alpha^3(\alpha-1)(1+\alpha)=0$, so the only possible values of $\alpha$ that can give us a solvable system are $\alpha=0$, $\alpha=-1$, and $\alpha=1$. We simply check the system in each case to see if it actually is solvable. If $\alpha=0$, we readily get $x_1=x_2=0$, $x_3=1$ as the unique solution; similarly for $\alpha=-1$. However, if we put $\alpha=1$, then the second equation becomes $$x_1+x_2+x_3=1,$$ but the fourth equation becomes $$x_1+x_2+x_3=-1,$$ so $\alpha=1$ does not give us a solvable system.
