Given the system of linear equations:
$\hspace{20pt}6x_2 + 2x_3 + 10x_4 = b_1$
$x_1 +x_2+4x_3- 2x_4 \hspace{5pt}=b_2$
$x_1 - 2x_2 + 3x_3 - 7x_4 = b_3$
(a) Find all possible Values of $b_1$,$b_2$,and $b_3$ for which this system has solutions;
(b)Find all possible solutions of this system if $b_1=6$,$b_2=7$, and $b_3 = 4$.
I was able to solve (a), but not sure what the solution should look like for (b)?
I set up an augmented matrix obtain rref of the matrix:(edit after comment)
$\begin{bmatrix} 0&6&2&10&b_1\\ 1&1&4&-2&b_2\\ 1&-2&3&-7&b_3\\ \end{bmatrix} \Rightarrow \begin{bmatrix} 1&0&\frac{11}{3}&-\frac{11}{3}&b_2-\frac{1}{6}b_1\\ 0&1&\frac{1}{3}&\frac{5}{3}&\frac{1}{6}b_1\\ 0&0&0&0&b_3-b_2+\frac{1}{2}b_1\\ \end{bmatrix}$
So for (a): $b_2 = \frac{1}{6}b_1$; $b_3=b_2-\frac{1}{2}b_1=-\frac{1}{3}b_1$.
We get, $\begin{bmatrix} \frac{1}{6}b_1\\ \frac{1}{6}b_1\\ -\frac{1}{3}b_1 \end{bmatrix}$ the column space of matrix is the line containing vector $\begin{bmatrix} \frac{1}{6}\\ \frac{1}{6}\\ -\frac{1}{3} \end{bmatrix}$.
For part (b), the rank of the matrix is $2 +$ nullspace $= 4$ (number of columns). Thus $x_3$, and $x_4$ will be scalars so let $x_3=s$,$x_4=t$.
For given values of $b_1=6$,$b_2=7$,$b_3=4$ we have the augmented matrix: \begin{bmatrix} 1&0&\frac{11}{3}&-\frac{11}{3}&6\\ 0&1&\frac{1}{3}&\frac{5}{3}&1\\ 0&0&0&0&0\\ \end{bmatrix}
$\Rightarrow \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}x_1+\begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}x_2+\begin{bmatrix} \frac{1}{3}\\ \frac{1}{3}\\ 0 \end{bmatrix}s+\begin{bmatrix} \frac{-11}{3}\\ \frac{5}{3}\\ 0 \end{bmatrix}t=\begin{bmatrix} 6\\ 1\\ 0 \end{bmatrix}$
$x_1+0x_2+\frac{11}{3}s - \frac{11}{3}t = 6$; $x_1 = 6-\frac{11}{3}s+\frac{11}{3}t$
$0x_1+x_2+\frac{1}{3}s+\frac{5}{3}t = 1$; $x_2= 1- \frac{1}{3}s-\frac{5}{3}t$
$x_3=s$
$x_4=t$
Not sure, does this answer (b)?