<s>[INST] How to solve a matrix system using Gauss elimination $$\left(\begin{array}{ccc|c}
-1 & 2 & 1 & 3\\
3 & \alpha & -2 & \beta\\
-1 & 5 & 2 & 9
\end{array}\right)$$
I am struggling to solve this system $Ax=b$. I understand the basics of Gauss elimination but am not sure how to handle it with the alpha and beta. It needs to be solved giving conditions for alpha and beta for no solutions, one solution and infinitely many solutions. 
I also need to compute the determinant of A and give a condition on $\alpha$ such that $A$ is invertible.
 [/INST] $$\det\left( 
\begin{array}{rrr}
 -1 & 2 & 1 \\
 3 & a & -2 \\
 -1 & 5 & 2 \\
\end{array}
\right)=-3-\alpha$$
if $-3-\alpha\ne 0$ that is $\alpha\ne -3$ there exists one and only one solution
$$\left\{\frac{3 a-b+6}{a+3},\frac{b+3}{a+3},\frac{3 (2 a-b+3)}{a+3}\right\}$$
if $\alpha=-3$ the determinant of the matrix $A$ of coefficients is zero $\text{rank}(A)=2$, so we must consider the augmented matrix $A|B$
$$\left(
\begin{array}{rrr|r}
 -1 & 2 & 1 & 3 \\
 3 & -3 & -2 & b \\
 -1 & 5 & 2 & 9 \\
\end{array}
\right)$$
if $\text{rank}(A|B)\ne \text{rank}(A)$ the system has no solutions, therefore to have infinite solutions we must have $\text{rank}(A|B)=2$ so all $3-$rd order determinants extracted from $A|B$ must be zero
$$
\det \left(
\begin{array}{rrr}
 2 & 1 & 3 \\
 -3 & -2 & \beta \\
 5 & 2 & 9 \\
\end{array}
\right) =\beta+3
$$
if $\beta\ne -3$ the system is impossible
if $\beta+3=0\to \beta=-3$ the determinant is zero the rank of the augmented matrix is equal to the rank of $A$ and we have infinite solutions
$\{t,\;t,\;3-t\}$
</s>