xonycutieoxl1

2022-06-19

I have a system of linear equations:

$x-y+2z-t=1$

$2x-3y-z+t=-1$

$x+(\alpha -4)z=\alpha -3$

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.

$x-y+2z-t=1$

$2x-3y-z+t=-1$

$x+(\alpha -4)z=\alpha -3$

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

Beginner2022-06-20Added 22 answers

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.

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.