Finding the rank of the Matrix $M$ for the different values of $\alpha$ Let's have $\
  M=
  \left[ {\begin{array}{ccccc}
   1 & \alpha & -1 & 2 \\
   2 & -1 & \alpha & 5 \\
   1 & 10 & -6 & 1 \\
  \end{array} } \right]$.
Using Gaussian elimination I've come to have the next matrix
 $\
  M'=
  \left[ {\begin{array}{ccccc}
   1 & \alpha & -1 & 2 \\
   0 & 1 & \frac{-5}{10- \alpha} & \frac{-1}{10 - \alpha} \\
   0 &  0 & \frac{-5(1+2\alpha)}{10- \alpha}+\alpha +2 & 1 \\
  \end{array} } \right]$
So:


*

*$\alpha \neq 10$ 

*$\frac{-5(1+2\alpha)}{10- \alpha}+\alpha +2 = 1$ and then i calculate the values of $\alpha$


Is it ok what I did? Is the rank always going to be 3 no matter the value of $\alpha \neq 10$ ? What if $\alpha = 10$ ? 
Since it is not a square matrix this is the only way to find out the rank, or is there another way?
 A: A good strategy for these problems is to do as many operations as you can without dividing by an expression involving a variable.
In particular, I find:
$$
\pmatrix{
 1 & \alpha & -1 & 2 \\
   2 & -1 & \alpha & 5 \\
   1 & 10 & -6 & 1 \\
} \to
\pmatrix{
 1 & \alpha & -1 & 2 \\
   0 & -1 - 2\alpha & 2+\alpha & 1 \\
   0 & 10-\alpha & -5 & -1 \\
}
$$
The rank of this matrix is always at least $2$, since the first and last rows are linearly independent.  We can only have rank exactly two if the matrix
$$
\pmatrix{-1-2\alpha & 2+\alpha & 1\\
10-\alpha & -5 & -1}
$$
has rank $1$.  That is, we can only achieve rank $2$ if the first and second columns are multiples of the third.  In other words, we must have
$$
-(-1-2\alpha) = 10 - \alpha \implies \alpha = 3\\
(-(-5)) = 2 + \alpha \implies \alpha = 3
$$
So indeed, when $\alpha = 3$, the matrix has rank $2$.  In all other cases, the matrix has rank $3$.

A more computational approach is to note that permuting the columns (or more generally "applying column operations") doesn't change the rank.  So, we can have
$$
\pmatrix{
 1 & \alpha & -1 & 2 \\
   2 & -1 & \alpha & 5 \\
   1 & 10 & -6 & 1 \\
} \to
\pmatrix{
 1 & \alpha & -1 & 2 \\
   0 & -1 - 2\alpha & 2+\alpha & 1 \\
   0 & 10-\alpha & -5 & -1 \\
}
\\ \to 
\pmatrix{
 1 &2 & \alpha & -1  \\
   0 &1 & -1 - 2\alpha & 2+\alpha \\
   0 & -1 & 10-\alpha & -5 \\
}
\\ \to 
\pmatrix{
 1 &2 & \alpha & -1  \\
   0 &1 & -1 - 2\alpha & 2+\alpha \\
   0 & 0 & 9-3\alpha & -3 + \alpha \\
}
$$
It is clear that this matrix will have full rank whenever $\alpha \neq 3$, and rank $2$ when $\alpha = 3$.
