Gaussian elimination involving parameters The problem is :Solve the given system of equations involving the parameter a :
$$x+y+az=1\\
x+ay+z=a\\
ax+y+z=a^2\\
ax+ay+az=a^3 .$$
I tried to solve this using the Gaussian method but I'm stuck because this is $4\times3$ matrix, and the Gaussian process is used for square matrix ? Please help..... 
 A: recommend rewriting the first three in order
$$
ax+y+z=a^2\\
x+ay+z=a\\
x+y+az=1
$$
since the eigenvalues, determinant, and inverse (when there is one) of
$$
\left(
\begin{array}{ccc}
a & 1 & 1\\
1 & a & 1 \\
1 & 1 & a
\end{array}
\right)
  $$
are pretty easy to find. It is of the form $ T +(a-1)I,$ where $T$ is the 3 by 3 matrix consisting of all entries equal to $1.$
Now that i think of it, the matrix has no inverse when $a=1$ or $a=-2,$ you might as well do those two cases using the actual numbers involved, no symbols.
A: You can solve the system formed by the first three equations, which will let you express $x,y,z$ as a function of $a$.
Then plugging these expressions in the fourth equation, you will get a condition on $a$ for the system to be resolvable.
A: Subsitute $y=1-x-az$ from the first equation and then $z=  - (a - x + 1)$ from the
second equation, assuming $a\neq 1$. Then the third equation gives $x=(a^2+2a+1)/(a+2)$, the case $a=-2$ being impossible. Then the fourth equation gives
$$
a(a+1)(a-1)=0.
$$
For $a=0$ we have a $3\times 3$ system with unique solution $(x,y,z)=(\frac{1}{2},-\frac{1}{2},\frac{1}{2})$. For $a=1$ we obtain $(x,y,z)=(1-y-z,y,z)$ with two free parameters $x,y$. Otherwise it follows that $a=-1$ and $(x,y,z)=(0,1,0)$.
