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!