# How to find the value of an arbitrary constant for which a system of linear equations has no solution?

I'm trying to find the value of the constant a for which the following system of linear equations has no solution:

$\left\{ \begin{array}{c} x_1+0x_2+x_3=1 \\ 2x_1+x_2+3x_3=1 \\ 0x_1+x_2+ax_3=3 \end{array} \right.$

I made this into an augmented matrix:

$\left[\begin{array}{rrr|r}1 & 0 & 1 & 1\\2 & 1 & 3 & 1\\0 & 1 & a & 3 \end{array} \right]$

...and tried to bring it into rref:

$\left[\begin{array}{rrr|r}1 & 0 & 1 & 1\\0 & 1 & 1 & -1\\0 & 0 & a-1 & 4 \end{array} \right]$

But I'm at a loss for how to proceed. I searched this problem and found this and this but didn't understand the answers provided. Could anyone tell me how to proceed and find the value of a for which the system has no solution?

• The system is inconsistent if there’s no pivot in the third row, but the entry on the right side is non-zero. – amd Apr 20 '17 at 19:15

multiplying the first equation with $-2$ and adding to the secvond we have $$x_2+x_3=-1$$ $$x_2+ax_3=1$$ and from here we obtain $$x_3(a-1)=2$$ can you finish?
• Realised I incorrectly copied the linear equations. It's actually $\left\{ \begin{array}{c} x_1+0x_2+x_3=1 \\ 2x_1+x_2+3x_3=1 \\ 0x_1+x_2+ax_3=3 \end{array} \right.$ What operation did you perform to get x_3(a−1)=2? – Eamon Cullen Apr 20 '17 at 10:37