# Matrix inconsistent & dependent

I've understood a system to be

Dependent if at least one row is $\begin{array}{ccc|c} 0 & 0 & 0 & 0 \\ \end{array}$

Inconsistent if atleast one row is $\begin{array}{ccc|c} 0 & 0 & 0 & b \\ \end{array}, b\in\mathbb{Z}\neq0$

In the matrix below, I've used Gauss elimination and found out that if a=2 it's inconsistent and if a=1 it's dependent. But I believe that there are more solutions to a that makes it either dependent or inconsistent. How do I find out all solutions if I have an unknown variable in the matrix? Is it just by trial and error through Gauss elim?

$\begin{array}{ccc|c} 2 & 3 & a & 5 \\ 1 & 2 & a & 3 \\ 3 & a & -2 & 4 \\ \end{array}$

• The determinat is $$-(a-2) (a-1)$$ and the RREF is $$\begin{bmatrix} 1 & 0 & 0 & \dfrac{2 (a-1)}{a-2} \\ 0 & 1 & 0 & -\dfrac{2}{a-2} \\ 0 & 0 & 1 & \dfrac{1}{a-2} \\ \end{bmatrix}$$ Your conclusions are correct, but the above should resolve your questions.
– Moo
Nov 22, 2017 at 20:23

Gauß' elimination leads to (without any hypothesis on the values of $a$): $$\begin{bmatrix} 1&2&a&|&3\\ 0&1&a&|&1\\ 0&0&(1-a)(a-2)&|&1-a \end{bmatrix}$$ This is enough to conclude, without any trial and errors.
• There's a moment when the last two rows are $$\begin{bmatrix}0&1&a&1\\0&a-6&-2-3a&-5\end{bmatrix}$$ if I'm not mistaken. Initially, I swapped rows 1and 2. Nov 22, 2017 at 20:44
• I simply replaced $R_3$ with $r_3-(a-6)r_2$ and factored out the third coefficient of the new last row. Nov 22, 2017 at 20:52