Problem about a linear system

Question

Given the linear system $$ \begin{cases} x+y-az=b\\ ax+y-z=1\\ 2x+2y-(a+1)z=2b \end{cases} $$ I have to determine which of the following equations

1. $a^2-2ab+b=1$
2. $a^2+b^2=0$
3. $3a-b^2+2=0$
4. $b^2+ab-a=1$
5. $2b^2-a^2+3ab=0$

is satisfied by all the pairs $(a, b)$ such that the linear system has infinite solutions.

My attempt. Written the augmented matrix $$ \begin{pmatrix} 1 & 1 & -a & b\\ a & 1 & -1 & 1\\ 2 & 2 & -(a+1) & 2b \end{pmatrix}, $$ by reduction by rows I found (subtracting $2$ times the first row to the third row) $$ \begin{pmatrix} 1 & 1 & -a & b\\ a & 1 & -1 & 1\\ 0 & 0 & a-1 & 0 \end{pmatrix}. $$ Can I conclude that the system has infinite solutions if $a=1$ (and what about $b$?) ? But why does the exercise require an equation to be satisfied by all the pairs $(a, b)$?

Thank You

Answer 1

From your last matrix, if you subtract from the second row $a$ times the first row you'll get$$\begin{pmatrix}1&1&-a&b\\0&-a+1&a^2-1&-ab+1\\0&0&a-1&0\end{pmatrix}.$$Now, it is clear that the system has one and only one solution if $a\neq1$. If $a=1$, you get the matrix$$\begin{pmatrix}1&1&-1&b\\0&0&0&-b+1\\0&0&0&0\end{pmatrix},$$and therefore the system has infinitely many solutions if and only if $b=1$. Therefore, the correct option is the fourth one and only that one.

Answer 2

If $a=1$ then

$\begin{pmatrix} 1 & 1 & -a & b\\ a & 1 & -1 & 1\\ 0 & 0 & a-1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 1 & -1 & b\\ 1 & 1 & -1 & 1\\ 1 & 0 & 0 & 0 \end{pmatrix}$ and you see that $b=1$.

The pair $(a,b)=(1,1)$ satiesfies $b^2+ab-a=1$.

Answer 3

By careful observation, if $a=1$, the first and the third equations will be dependent. So the third equation can be ignored, leaving: $$\begin{cases} x+y-z=b \\ x+y-z=1\end{cases}$$ If $b\ne 1$, there is no solution. If $b=1$, there is one equation with three unknowns. Thus it has infinitely many solutions. The result follows.