The system:
$x+ay=1$
$bx+y=2$
Note: I have to find the unique and infinitely many solutions
I set-up the augmented matrix but my answers didn't come from any operations:
What I got: -The system has a unique solution (x=1, y=2) if a=b=0.
-The system has no solution if a=b=1, then we will have a contradiction.
-The system has infinitely many solutions when $a ≠ 0, a ≠ 1, b ≠ 0, b ≠ 1,$ then solutions will be $ay+x=1$ and $bx+y=2$.
What am I doing wrong and is there a way I could get to this other than simply observing since the matrix already seems reduced?
Best Answer
You can write your system as $Au=v$, where $A = \begin{bmatrix}1 & a \\ b & 1\end{bmatrix}$, $u=\begin{bmatrix}x \\ y\end{bmatrix}$ and $v=\begin{bmatrix}1 \\ 2\end{bmatrix}$.
Case 1: Unique solution
It is known that if $A$ is invertible, then the system has unique solution. This is because we can write $$u=A^{-1}v.$$ Therefore, we look for a condition that $A$ is invertible. Such condition is that $\det{A} \neq 0$. In your case: $$1\cdot1-a\cdot b \neq 0.$$
Case 2: Multiple solutions
This happens when the two equations describe the same situation. Each of $x+ay=1$ and $bx+y=2$ describe two lines in the $\mathbb{R}^2$ plane. The solution is given by the points $(x,y)$ which describe the intersection of the two lines. If we seek the case where we have infinite solution points $(x,y)$, we mean that the two lines are identical. This happens if the equations are the same. As $x+ay=1$ is equivalent to $2x+2ay=2$ (observe how the right-hand side is the same for both equations now), we can deduce that $b=2$ and $a=\dfrac{1}{2}$ yield the same equation. In conseqence, we have infinitely-many solutions.
Case 3: No solutions
This part does not need to be treated as you already determined the values for $a,b$ which yield solutions (unique or infinitely many). So, to have no solutions, you require $$ab = 1\text{ and } b\neq2.$$
Observe how the condition above excludes first unique solution and then infinite case.
Hope it helps and clarifies the process of finding $a$ and $b$.