Rijad Hadzic said:
A must follow the 4 conventions:
- leading ones in every row
- Above and below those leading ones, in the same column, are all 0's
- Bottom row's are 0 rows
- The second leading one is to the right of the first, the 3d is to the right of the second.. and so on.
Since A is a 3x3 matrix, we're not going to be left with any free variables, thus having multiple solutions would be impossible. Not sure if this is right but it's how I'm viewing this question currently..
An augmented matrix does not need to follow those rules you wrote. Before you perform row-operations you can have a general 3x3 matrix A.
To answer the question, look at how you would deal with the 3x3 linear system ##Ax = b##. When you write it all out you get three linear equations in the three unknowns. If we write the system as
$$ \begin{bmatrix} a_{11} & a_{12} & a_{13}\\
a_{12} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{bmatrix}
\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}
$$
then in order to have a unique solution, at least one element ##a_{i1}## in the first column must be non-zero (because if not, ##x_1## is not actually part of the system of equations and so can be anything---making the solution non-unique). So, without loss of generality we can assume that ##a_{11} \neq 0##; if not, interchange the rows to make that happen. The first equation reads as ##a_{11} x_1 + a_{12} x_2 + a_{13} x_3 = b_1##. The coefficient of ##x_1## is nonzero, so we can solve for ##x_1## in terms of ##x_2## and ##x_3##:
$$x_1 = \frac{b_1}{a_{11}} - \frac{a_{12}}{a_{11}} x_2 - \frac{a_{13}}{a_{11}} x_3$$
Substituting that expression for ##x_1## into the other two equations will give us a 2x2 system involving ##x_2, x_3## alone. This new system will have an altered 2x2 coefficient matrix on the left and a new 2x1 right-hand-side, say
$$\begin{bmatrix} a'_{22} & a'_{23} \\ a'_{32} & a'_{33} \end{bmatrix} \begin{bmatrix} x_2 \\ x_3 \end{bmatrix}
= \begin{bmatrix} b'_2 \\ b'_3 \end{bmatrix}
$$
Again, if the solution is unique, one of the elements ##a'_{22}, a'_{32}## must be nonzero, and so we can solve for ##x_2## in terms of ##x_3##. When we substitute that expression for ##x_2## into the other equation we obtain a single equation involving ##x_3## alone. If that has a unique solution the coefficient of ##x_3## must be nonzero.
All that is what happens if we have a right-hand-side ##b_1, b_2, b_3## in the original system. However, if we change to a new right-hand-side ##d_1, d_2, d_3##, nothing at all changes on the left when we repeat the steps outlined above. What does that tell you about solutions to the new system?
What happens if the original system does not have a unique solution, or does not have any solution at all? Well, in the course of our solution procedure we must hit a matrix or submatrix in which one of the columns is all zero, so that a particular ##x## does not appear at all in the system. Again, that affects only the left-hand-sides of the equations and so does not change when we go to a new right-hand-side. Think carefully about what that would mean about properties of solutions for the original linear system.
