Why is this happening? I am not sure what do they mean by why is this happening?

The system is $x_1+2x_2+3x_3=b_1$ $2x_1+5x_2+3x_3=b_2$ $x_2-3x_3=b_3$

I did Gauss method $-2p_1+p_2$ to obtain $x_2-3x_3=-2b_1+b_2$ . $p_1+p_3$ to obtain $x_1+3x_2=b_1+b_3$. $-p_2+p_1$ to obtain $-x_1-3x_2=b_1-b_2$ $p_1+p_3$ to obtain $0=2b_1-b_2+b_3$

I need to give specific values for $b_1$,$b_2$ and $b_3$ that makes the system have no solution.

I will give you 4 examples or cases: Case 1: $b_1=-1$ $b_2=0$ $b_3=2$

Case 2: $b_1=0$ $b_2=2$ $b_3=2$

Case 3: $b_1=1$ $b_2=2$ $b_3=0$

Case 4: $b_1=1$ $b_2=0$ $b_3=-2$

The problem is $b_1$,$b_2$ and $b_3$ can be any value and not a specific one. So it has infinitely many solutions instead right?

  • $\begingroup$ If $b_1,b_2,b_3$, doesn't satisfy the condition $b_2 = 2b_1 + b_3$, then the system of equation doens't have a solution. $\endgroup$
    – Stefan4024
    Commented Dec 17, 2013 at 23:40
  • $\begingroup$ Have you considered using matrices for the questions? They are a lot more pleasant to the eye/brain.(imo) $\endgroup$
    – Asinomás
    Commented Dec 17, 2013 at 23:41
  • $\begingroup$ You can actually use my answer to your question and check under what condition the system doesn't have solution by setting one of the determinants not equal to 0. $\endgroup$
    – Stefan4024
    Commented Dec 17, 2013 at 23:47

1 Answer 1


First note that $$2\cdot(x_1 + 2x_2 + 3x_3) + 1\cdot(x_2-3x_3) = 2x_1 + 5x_2 + 3x_3$$ Hence, we have $$2b_1 +b_3 = b_2 \tag{$\star$}$$ Any choice of $(b_1,b_2,b_3)$ violating $(\star)$ will have no solution.

  • $\begingroup$ what do you mean by star? $\endgroup$
    – user983246
    Commented Dec 17, 2013 at 23:50
  • $\begingroup$ See the star off to the right of the equation? $\endgroup$
    – Amzoti
    Commented Dec 17, 2013 at 23:51
  • $\begingroup$ I never listed a star. I see a star underneath the arrow toggles but I'm pretty sure that is not what you are asking about because that is suppose to designate a favorite question...Other than that star, I'm don't know of any other star. I thought since b_1, b_2 and b_3 can have any value would make it have infinitely many solutions but apparently this is not the case. how come? $\endgroup$
    – user983246
    Commented Dec 17, 2013 at 23:53
  • $\begingroup$ @user983246 Do you see two ($\star$)'s in my answer? One to the extreme right of $2b_1 + b_3 = b_2$ and other in the last statement. $\endgroup$
    – user17762
    Commented Dec 18, 2013 at 0:07
  • $\begingroup$ Oops...that star was so far away I did not see it until you pointed it out. Awesome heads-up Amzoti and ShikariShambu! $\endgroup$
    – user983246
    Commented Dec 18, 2013 at 0:11

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