I was working on this problem, to which I have the answer, but it is just that, an answer with no explanation, and I am stuck on how the answer was arrived at, for a few parts of this question.
In this question, we have the following matrix, with constants $a$ and $b$.
$$\left[ \begin{array}{cccc|c} 1 & 0 & -3 & 0 &3a\\ 0 & 1 & b & 0 &b-4\\ 0 & 0 & 0 & (a-2)(b+1) &a-2\end{array} \right]$$
We are trying to find, if any, what values of $a$ and $b$ give the following:
- No solution,
- A unique solution,
- A one parameter solution,
- A two parameter solution.
I have worked out the following, which matches up with the answer given:
A1. No solution when $a \neq2$ and $b=-1$
A3. One parameter solution when $a \neq2$ and $b \neq-1$
But I am stuck on Q2 and Q4 and would be very grateful if someone could give me an insight into how they would tackle these parts of this question.
Thank you in advance!
PS: If you would like, I can add the answers to the remaining parts of the question, which I am stuck on. Edit: Answers, as requested:
A2: No such $a$ and $b$.
A4: $a=2$ and b free. (I assume it means b is a parameter?)
$$\left[ \begin{array}{cccc|c} 1 & 0 & -3 & 0 &3a\\ 1 & 1 & b & 0 &b-4\\ 0 & 0 & 0 & (a-2)(b+1) &a-2\end{array} \right] \to \left[ \begin{array}{cccc|c} 1 & 0 & -3 & 0 &3a\\ 0 & 1 & b+3 & 0 &-3a+b-4\\ 0 & 0 & 0 & (a-2)(b+1) &a-2\end{array} \right] $$
first let us look at $(a-2)(b+1)$ is zero if $a = 2$ or $b = -1.$
in the case $a = 2,$ there are infinitely many solutions. the reason is there are two free variables, third and fourth column variables.
the case $a \ne2, b = -1,$ the last equation reduces to $0 = 1$ so there are no solutions.
the remaining case $a \neq 2 \text{ and } b \neq - 1$ there is free variable, so the system has infiniteley many solutions.