Different way to determine rank of $3\times3$-matrix with parameter by using row transformations

Question

For $a \in \mathbb{R}$, the matrix $A$ is given with $$A = \begin{pmatrix} 2-a & 1-a & a \\ 1-a & 1-a & 0 \\ 2-2a & 1-a & -(1-a) \end{pmatrix}.$$

For all possible $a$, determine the rank of $A$ by only using row transformations.

One thing that always works would be the Gauss algorithm, but that would be relatively cumbersome as it requires a lot of case differentiations when dividing by something that includes an $a$.

So my question is: Is there a way to solve this problem with row transformations without using the effective but laborious Gauss algorithm?

Thanks in advance for any help!

Answer 1

You can simplify Gauß' algorithm, borrowing techniques used for the computation of determinants:

\begin{align} \begin{bmatrix} 2-a & 1-a & a \\ 1-a & 1-a & 0 \\ 2-2a & 1-a & -(1-a) \end{bmatrix}&\rightsquigarrow \begin{bmatrix} 1-a & 1-a & 0 \\ 2-2a & 1-a & -(1-a)\\2-a & 1-a & a \end{bmatrix}\rightsquigarrow \begin{bmatrix} 1-a & 1-a & 0 \\ 0 & -(1-a) & -(1-a)\\2-a & 1-a & a \end{bmatrix} \end{align} Now, if $a\ne 1$, this matrix has the same rank as \begin{align} \begin{bmatrix} 1 & 1 & 0 \\ 0 & -1 & -1\\2-a & 1-a & a \end{bmatrix}\rightsquigarrow \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & -1 & a \end{bmatrix} \rightsquigarrow \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & a+1 \end{bmatrix}. \end{align} You can now conclude (not forgetting to examine the case $a=1$).

Answer 2

Gauss is very comfortable here; just switch column two and one. Consider then the case $a=1$; here the rank is $1$. Otherwise you will be done in two very easy steps without considering further cases.