# Give an example of b_1,b_2 and b_3 so that no solutions exists

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?

• 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. Commented Dec 17, 2013 at 23:40
• Have you considered using matrices for the questions? They are a lot more pleasant to the eye/brain.(imo) Commented Dec 17, 2013 at 23:41
• 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. Commented Dec 17, 2013 at 23:47

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.
• @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.