# Will any value of a free variable satisfy a system of equation?

Say I have a reduced row echelon form matrix like this: $$A=\begin{bmatrix} 1 & 0 & \frac{1}{2} & 0\\ 0 & 1 & -\frac{1}{3} & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$$ The number of unknowns is more than the number of known equations. So I can expect an infinite number of solutions. And $Ax=b$ is like this: $$\begin{bmatrix} 1 & 0 & \frac{1}{2} & 0\\ 0 & 1 & -\frac{1}{3} & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4} \end{bmatrix} = \begin{bmatrix} b_{1}\\ b_{2}\\ b_{3} \end{bmatrix}$$

Then I can say that my $x$ is like this with $x_{3}$ being a free variable in the equation: $$\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4} \end{bmatrix} = x_{3}\begin{bmatrix} \frac{1}{2}\\ \frac{1}{3}\\ 1\\ 0 \end{bmatrix}+ \begin{bmatrix} b_{1}\\ b_{2}\\ 0\\ b_{3} \end{bmatrix}$$ Now, if I let $\; \begin{bmatrix} b_{1}\\ b_{2}\\ b_{3} \end{bmatrix}=\begin{bmatrix} 5\\ 2\\ 7 \end{bmatrix}$, then... $$\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4} \end{bmatrix} = x_{3}\begin{bmatrix} \frac{1}{2}\\ \frac{1}{3}\\ 1\\ 0 \end{bmatrix}+ \begin{bmatrix} 5\\ 2\\ 0\\ 7 \end{bmatrix}$$

At this stage, I can say that for any value that I put into the variable $x_{3}$, I would get an answer that is equals to $b$, right?

So assume I just randomly throw a value into $x_{3}=2$, then... $$\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4} \end{bmatrix} = \begin{bmatrix} 1\\ \frac{2}{3}\\ 2\\ 0 \end{bmatrix}+ \begin{bmatrix} 5\\ 2\\ 0\\ 7 \end{bmatrix}= \begin{bmatrix} 6\\ \frac{8}{3}\\ 2\\ 7 \end{bmatrix}$$ From here, I need to tally if the equation really gets back my intended values of $b$, which is $\begin{bmatrix} b_{1}\\ b_{2}\\ b_{3} \end{bmatrix}=\begin{bmatrix} 5\\ 2\\ 7 \end{bmatrix}$. So,

$x_{4} = b_{3} =7$

This is correct.

$b_{2}=x_{2}-\frac{1}{3}x_{3}$

$b_{2}=\frac{8}{3}-\frac{2}{3}=2$

This is also correct.

BUT NOW,

$b_{1}=x_{1}+\frac{1}{2}x_{3}$

$b_{1}=6+1=7$

$b_{1}$ suppose to be 6 but somehow, why am I getting 7 instead? This is weird. Where did I go wrong?

Thanks for any help!

You simply made a sign error in your third matrix equation, you should have $x_1 = \frac{-1}{2} x_3 + b_1$ instead of $x_1 = \frac{1}{2} x_3 + b_1$. That will give you the right answer which is $b_1=5$ by the way.

• OMG! you are right! this is madness! I didn't realise this. thanks for highlighting this to me! Is this the only issue to this whole problem? Or it is as what aengle's answer says that there is some mapping error too? Commented Jun 28, 2011 at 20:17

I think you've confused yourself somewhere.

You have a mapping from $\mathbb{R}^4 \to \mathbb{R}^3$. $x_3$ is free $x_4 = b_3$, the only stipulation is that:

$x_1/3 + x_2/2 = 8/3$.

The vector you've given: $[6, 8/3, 2, 7]^T$ does not satisfy the last condition.

Edit: Solve for $[x_1,x_2,x_3,x_4]^T$ given a vector $[b_1,b_2,b_3]^T$:

$x_4=b_3$

$x_1 + \dfrac{x_3}{2} = b_1$

$x_2 - \dfrac{x_3}{3} = b_2$

These all imply that

$\dfrac{x_1}{3} + \dfrac{x_2}{2} = \dfrac{b_1}{3} + \dfrac{b_2}{2}$

Now you want the solution to the vector $[5,2,7]^T$, so

$\dfrac{x_1}{3} + \dfrac{x_2}{2} = \dfrac{5}{3} + 1 = \dfrac{8}{3}$.

The vector you said mapped to $\vec{b}$ was $[6, 8/3, 2, 7]$.

Your proposed $x_1 = 6$ and $x_2 = \dfrac{8}{3}$ do not satisfy this.

• How did you get $x_1/3 + x_2/2 = 8/3$? I am also not too sure why I didn't satisfy the last condition. Commented Jun 28, 2011 at 20:15
• I edited the question. Let me know if you have any more questions... Commented Jun 28, 2011 at 20:53
• You got a point there! It indeed doesn't map correctly. But how come when I do it the way I had in my question post, it seems to work after I correct the sign error? That's weird. So is this system of equation even considered valid? Commented Jun 29, 2011 at 9:57