Find the value of parameter $a$, so that the system has no solution.
\begin{bmatrix} a & 3 & 2 & & 5 \\ 1 & 7 & 3 & = & 13 \\ 3 & 1 & a & & 3 \\ \end{bmatrix}
How to do that? I know I am supposed to get 0 0 0 = non-zero value, but how to actually get to the solution? The answer is 13/7.
from equation (2) we get $$x_1=13-7x_2-3x_3$$ plugging this in the first and third equation and we have $$x_2(3-7a)+x_3(2-3a)=5-13a$$ $$-20x_2+x_3(a-9)=-36$$ from our last equation $$x_2=\frac{9}{5}+\left(\frac{a-9}{20}\right)x_3$$ and now we can eliminate $$x_2$$ $$x_3((a+1)(7a-13)=8(1+a)$$ can you finish? if $$7a-13=0$$ so $$a=\frac{13}{7}$$ and we get no solutions.