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$$x_1 + 2x_2 + 3x_3 = b_1 \\ 2x_1 + 5x_2 + 3x_3 = b_2 \\ x_2 - 3x_3 = b_3$$

Use Gauss method: $-2p_1+p_2$ to produce $x_2-x_3=-2b_1+b_2$. $p_1+p_3$ to produce $x_1+3x_2=b_1+b_3$ $-p_2+p_1$ to produce $-x_1-3x_2= b_1-b_2$

The above produces the system: the first row $-x_1-3x_2=b_1-b_2$ second row $x_2-x_3=-2b_1+b_2$ third row $x_1+3x_2=b_1+b_3$ Do Gauss method again. $p_1+p_3$ to produce $0=2b_1-b_2+b_3$

Since I got a homogeneous equation from the above. I can just stop and rewrite the system. first row $-x_1-3x_2=b_1-b_2$ second row $x_2-x_3=-2b_1+b_2$ third row $0 = 2b_1-b_2+b_3.$ Write the homogeneous equation in terms of $b_3. b_3 = -2b_1+b_2$

So I can say the system is consistent if and only if $b_3=-2b_1+b_2.$

May I get a verification of my answer?

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2 Answers 2

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In fact, you can directly check it yourself.

The condition means that LHS of $L_3$ is LHS of $-2 L_1 + L_2$, where $L_i$ is the i-th line of the system, and this is true. Moreover, since LHS of $L_1$ and $L_2$ are independant, there is exactly one linear condition on the $b_i$ in order to have a solution.

So your answer is correct.

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  • $\begingroup$ if I need to give a specific value of b_1,b_2 and b_3 and show a solution does not exist are they asking for an actual value for b_1,b_2 and b_3? I forgot to put down the question also said "why does this happen if I have to show a solution does not exist?" $\endgroup$
    – user983246
    Dec 17, 2013 at 23:02
  • $\begingroup$ If you have actual values for the $b_i$, you can show easily that the system has no solution if the $b_i$ don't respect your condition. Indeed, you just have to say that any relation for the LHS of the system has to be respected by the RHS of the system. $\endgroup$ Dec 18, 2013 at 8:21
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You can also apply Cramer's rule. For this system we have:

$$\Delta = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 5 & 3 \\ 0 & 1 & -3 \end{vmatrix}$$

If the determinant is not equal to zero, then the system is consistent and has one solution. Unfortunally if you calculate it you'll see that the determinant is equal to zero. So it'll have either infinite amount of solution or it won't have a solution.

In order to have an infinite amount of solution we need to have:

$$\Delta_x = \Delta_y = \Delta_z = 0$$

Choose one, it won't change anything so you have:

$$\Delta_x= \begin{vmatrix} b_1 & 2 & 3 \\ b_2 & 5 & 3 \\ b_3 & 1 & -3 \end{vmatrix} = 0$$

Calculate the value of the determinant and you'll end up with:

$$-15b_1 + 6b_3 + 3b_2 - 15b_3 + 6b_2 - 3 b_1 = 0$$ $$-18b_1 - 9b_3 + 9b_2 = 0$$ $$b_2 = 2b_1 + b_3$$

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