I'm doing the following question here:
Let M=$\left( \begin{array}{ccc|c} 1 & 1 & 3a+1 & 2-3a \\ 2 & a & 2 & -8\\ -1 & -1 & a^2-2a-3 & a^2 \end{array} \right)$.
Find the values of $a$, $a \in \mathbb{R},$ such that:
i) The system has infinitely many solutions.
ii) The system has no solution.
iii) The system has an unique solution.
So I row reduced the matrix until I got this:
$\left( \begin{array}{ccc|c} 1 & 1 & 3a+1 & 2-3a \\ 0 & a-2 & -6a & 6a-12\\ 0 & 0 & a^2+a-2 & a^2-3a+2 \end{array} \right)$.
i) There will be infinitely many solutions if we have a non pivot column (A free variable).
This means $a^2+a-2 = a^2-3a+2$. Solving, this gives $a=1$.
ii) There is no solution if $a^2+a-2=0$ but $a^2-3a+2 \neq 0.$ Solving, we get $(a-1)(a+2)=0$ or $a=1,-2$ for the first equation and $(a-2)(a-1) \neq 0$ or $a\neq 2,1$ for the second equation. Since $a$ cannot be be equal to $1$ and equal to $1$ at the same time, this means that $a=-2$ and $a\neq 1$ will ensure that we have no solution.
iii) Since this imples we will eventually get to RREF, this means that $a-2 \neq 0$, $a^2+a-2 \neq 0$ or $a\neq 2,1,-2$ for unique solutions.
Can someone tell me if my process is correct? Thanks!
Your process is almost correct. (i) You must show that $a=2$ does not meet the requirement.
(ii) $a=2$ is in this case.
(iii) You are right.