Please comment on my answer and your thoughts on improving my understanding of this linear algebra concept.
Show the complete solution for this system of equations (Anton 8th Ed Ex Set 1.6 Q17):
\begin{align*} x_1 - x_2 +3x_3 +2x_4 = b_1 \\ -2x_1 + x_2 +5x_3 + x_4 = b_2 \\ -3x_1 + 2x_2 + 2x_3 - x_4 = b_3 \\ 4x_1 -3x_2 + x_3 + 3x_4 = b_4 \end{align*}
A consistent linear system is one which has a unique or infinite set of solutions. Therefore answer should show either reduced row echelon matrix form for each of $b_1,b_2,b_3$ and $b_4$ or all zero in matrix rows to indicate an infinite range of parametrized values for $b_1,b_2,b_3$ and $b_4$, suitably expressed in an equation.
For the system of equations above, the augmented matrix is as follows. $r_1$ and $r_2$ indicate the rows for the rowops.
Matrix 1:
\begin{array}{rrrrr|r} r_1 & 1 & -1 & 3 & 2 & b_1 \\ r_2 & -2 & 1 & 5 & 1 & b_2 \\ r_3 & -3 & 2 & 2 & -1 & b_3 \\ r_4 & 4 & -3 & 1 & 3 & b_4 \end{array}
rowops $ r_2 = 2r_1 + r_2 $
rowops $ r_3 = 3r_1 + r_3 $
rowops $ r_4 = -4r_1 + r_4 $
Yields Matrix 2:
\begin{array}{rrrrr|r} r_1 & 1 & -1 & 3 & 2 & b_1 \\ r_2 & 0 & -1 & 11 & 5 & 2b_1 + b_2 \\ r_3 & 0 & -1 & 11 & 5 & 3b_1 + b_3 \\ r_4 & 0 & 1 & -11 & -5 & -4b_1 + b_4 \end{array}
rowops $ r_4 = r_3 + r_4 $
Yields Matrix 3:
\begin{array}{rrrrr|r} r_1 & 1 & -1 & 3 & 2 & b_1 \\ r_2 & 0 & -1 & 11 & 5 & 2b_1 + b_2 \\ r_3 & 0 & -1 & 11 & 5 & 3b_1 + b_3 \\ r_4 & 0 & 0 & 0 & 0 & -b_1 + b_3 + b_4 \end{array}
Since $0 = -b_1 + b_3 + b_4 $
Therefore $$ b_1 = b_3 + b_4 $$
Now to find $b_2$:
rowops $ r_2 = r_2 - r_3 $
Yields Matrix 4:
\begin{array}{rrrrr|r} r_1 & 1 & -1 & 3 & 2 & b_1 \\ r_2 & 0 & 0 & 0 & 0 & -b_1 + b_2 - b_3 \\ r_3 & 0 & -1 & 11 & 5 & 3b_1 + b_3 \\ r_4 & 0 & 0 & 0 & 0 & -b_1 + b_3 + b_4 \end{array}
Since $ b_1 = b_3 + b_4 $
And Since $0 = -b_1 + b_2 - b_3 $
Therefore $0 = -b_3 - b_4 + b_2 -b_3$
Which is $$b_2 = 2b_3 + b_4$$
Yes, it looks right in this scenario.
Just a note to mention: You should reduce the matrix to row reduced echelon form just to be sure! Then you can read the solutions to this equation off easily, and make sure you don't have to deal with a third condition or even a fourth condition on b1,b2,b3,b4. As of right now, you have no justification as to why you stopped doing row ops at that point!
If this system of equations has a solution, then the vector $b = [b_1, b_2, b_3, b_4]^T $ must belong to the columns space of $A$ (the coefficient matrix).
By row reduce into reduced row echelon form (RREF), we get the RREF matrix
$ \begin{bmatrix} 1 && 0 && -8 && -3 \\ 0 && 1 && -11 && -5 \\ 0 && 0 && 0 && 0 \\ 0 && 0 && 0 && 0 \end{bmatrix} $
The two leading $1's$ are in the first and second position, therefore, to have a solution, we must have
$ b = \lambda_1 \ C_1 + \lambda_2 C_2 = \lambda_1 \begin{bmatrix} 1 \\ -2 \\ -3 \\ 4 \end{bmatrix} + \lambda_2 \begin{bmatrix} -1 \\ 1 \\ 2 \\ -3 \end{bmatrix} $
The orthogonal space to the space spanned by these two vectors is the span of the following two vectors,
$ V_1 = \begin{bmatrix} 1 \\ -1 \\ 1 \\ 0 \end{bmatrix} $ and $ V_2 = \begin{bmatrix} -2 \\ 1 \\ 0 \\ 1 \end{bmatrix} $
Therefore, the entries of vector $b$ must satisfy:
$ b_1 - b_2 + b_3 = 0 $
and
$ -2 b_1 + b_2 + b_4 = 0 $
