How to find parameters such that rank of linear equation system is 1?

Question

I'm given a system of linear equations with parameters a and b, and I'm supposed to find values for the parameters such that two variables can be chosen freely. So, n = 3 (x1, x2, x3) and m=3 (number of equations).

For two variables to be free to chose, $n-r=2$, so I must find values for $a$ and $b$ such that the rank $r$ equals 1.

The system is:

$$ \begin{pmatrix} & ax_2 &+ x_3 &= -b \\ ax_1 & &+ bx_3 &= -1 \\ ax_1 & + ax_2 &+ 2x_3 &= -2 \end{pmatrix}$$

I applied Gauss elimination procedure, but am a bit confused about what it gives me to the end (just showing coefficients):

$$ \begin{pmatrix} a & a & 2 & -2 \\ 0 & -a & b-2 & 1 \\ 0 & 0 & b+1 & b-1 \end{pmatrix} $$

So from the last equation, $(b+1)x_3 = b-1$, and so $x_3 = (b-1)/(b+1)$.

But I'm not sure what to do now, setting $b=1$ means $x_3 = 0$. And then what do I do next?

Answer 1

First, I think you forgot some signs in the Gauss elimination, you should get: $$ \begin{pmatrix} a & a & 2 & -2 \\ 0 & -a & b-2 & 1 \\ 0 & 0 & b+1 & -b-1 \\ \end{pmatrix} $$

Now, we are looking for the parameters $a$ an $b$ such that the rank of the system is one. The rank of a system represents the number of independent equations it has. In this case it is one, so all equations are linearly dependent on one equation, and that means they are proportional.

We can use this fact to solve for the parameters. For example the first equation being proportional to the second one implies $$ \frac{b-2}{2}=\frac{1}{-2} \implies b=1 $$ Now plugging in $b=1$ in the system we get $$ \begin{pmatrix} a & a & 2 & -2 \\ 0 & -a & -1 & 1 \\ 0 & 0 & 2 & -2 \end{pmatrix} $$

So you can easily see that the only option for the rows to be proportional is $a=0$, and it is done.

In general this problem is that of determining the rank of a linear system depending on some parameters. This is solved by using the fact that the determinant of a matrix is zero when it contains linearly dependent vectors (see this fact discussed in this post) and in our case we only needed to compute the determinant of a $2\times 2$ matrix. You can also see a similar example in this post.