Find parameters $a,b,c \in\mathbb{R}$, such that following system is compatible Given $$\begin{cases}2x-3y+4z-5t = 1 \\ x+9y+az+t=-3 \\ 5x-6y+10z-bt=c\end{cases}$$
how to find $a,b,c\in\mathbb{R}$ so that the system is compatible? How should one approach such a problem? Where to start from?
I managed to find the cases themselves which are $$a\neq 2, b,c\in\mathbb{R}, \text{ or } b\neq 12, a,c\in\mathbb{R} 
\text{ or } a=2, b=12,c=2$$ but my solution is anything but proper. I started by stating $\text{rank} A \geq 2$, which is obvious, then calculated $$\det\left(\begin{matrix}2 & -3 & 
4\\ 1 &9&a\\1&0&2 \end{matrix}\right) = -3a+6=0 \iff a = 2$$ split the problem in two cases, did this for $b$ as well and somehow got to the right answer but I'm lost as to why what I did works. Rather, I'd want to know the common way/approach and reasoning behind solving this.
I am also confused as to why couldn't I compute, according to Rouche' theorem, $c$ from $$\det\left(\begin{matrix}2 &-3&1\\1&9&-3\\5&-6&c\end{matrix}\right) = 0$$
and let $a,b$ be any real number?
 A: We always can write a linear system of equations using matrix notation
as $Ax=B$ or $[A|B]$ and then to study the compatibility of the system using elimination by row (see more details compatibility using Elimination by Row). In this case we have
$$\left(\begin{array}{cccc|c}2&-3&4&-5&1\\1&9&a&1&-3\\5&-6&10&-b&c\end{array}\right)\sim \left(\begin{array}{cccc|c}1&*&*&*&*\\0&1&*&*&*\\0&0&a-2&7(b-12)&7(2-c)\end{array}\right)$$
Thus we we look in the last row,
$$(a=2)\wedge (b=12)\wedge (c\not=2)\iff \text{incompatibility},$$since we would have in that case that $0x+0y+0z+0t=k$ for $k\not=0$ which has no solution. Hence the contrapositive give the compatibility.
A: If we exchange the first two equations, the system becomes$$\left\{\begin{array}{l}x+9y+az+t=-3\\2x-3y+4z-5t=1\\5x-6y+10z-bt=c.\end{array}\right.$$Now, if we add to the second equation the first one times $-2$ and if we add to the third equation the first equation times $-5$, we get the equivalent system:$$\left\{\begin{array}{l}x+9y+az+t=-3\\-21y+(4-2a)z-7t=7\\-51y+(10-5a)z-(b+5)t=c+15.\end{array}\right.$$Now, if we divide the second equation by $-21$, we get the equivalent system$$\left\{\begin{array}{l}x+9y+az+t=-3\\y-\frac{4-2a}{21}z+\frac t3=-\frac13\\-51y+(10-5a)z-(b+5)t=c+15.\end{array}\right.$$The next step is to add to the first equation the second one multiplied by $-9$ and to add to the third equation the second one multiplied by $51$. We get:$$\left\{\begin{array}{l}x+\frac{a+12}7z-2t=0\\y-\frac{4-2a}{21}z+\frac t3=-\frac13\\\frac{2-a}7z+(12-b)t=c-2.\end{array}\right.\tag1$$Now, if $a\ne2$, it is clear that, for each $t$, the system $(1)$ has one and only one solution. On the other hand, if $a=2$, the system $(1)$ becomes$$\left\{\begin{array}{l}x+2z-2t=0\\y+\frac t3=-\frac13\\(12-b)t=c-2.\end{array}\right.\tag2$$It is clear now that if $b\ne12$, then the system $(2)$ is clearly compatible. And, on the other hand, if $b=12$; then the system $(2)$ is compatible if and only if $c=2$.
