xonycutieoxl1

2022-06-19

I have a system of linear equations:
$x-y+2z-t=1$
$2x-3y-z+t=-1$
$x+\left(\alpha -4\right)z=\alpha -3$
I have already found that this system has a solution for any value of $\alpha$. Now I need to find the $\alpha$ for which the matrix of the system has a rank=2. The row echelon form of the matrix looks like this:
$\left[\begin{array}{ccccccccc}1& & -1& & 2& & -1& & 1\\ 0& & -1& & -5& & 3& & -3\\ 0& & 0& & \alpha -11& & 4& & \alpha -7\end{array}\right]$
I don't think that the rank of this matrix could be 2 for any value of $\alpha$ but the problem specifically asks for me to prove that it can. Maybe I'm missing something. Any help is appreciated.

timmeraared

It looks like your transformation to row echolon form is correct.
You are correct, there is no choice for $\alpha$ such that the matrix has rank 2 (both the matrix of the coefficients as well as the augmented including the r.h.s.) since there is no choice for $\alpha$ which will make the last row linear dependent of the first two rows, as can be seen in the echelon form.

Do you have a similar question?