Yes, in what's called an augmented matrix (Mao, you'll come to it). You could get a system of equations:
1 2 3 | 0
2 2 2 | 2
4 4 4 | 4
The first 3/4 of the matrix represents x, y and z. And the number proceeding it represents the number it equals. I.E. x + 2y + 3z = 0 is represented by the first line 1 2 3 | 0
Now, in that 2nd row it's 2x + 2y + 2z = 2
In the 3rd row, 4x + 4y + 4z = 4
This matrix has a rank of TWO because when you row-reduce it (that is, perform elementary row operations in order to obtain leading ones), you may only obtain 2 leading ones. I will perform the row reductions below:
1 2 3 | 0
2 2 2 | 2
4 4 4 | 4
Row 1 is represented by 1 2 3 | 0, Row 2 is 2 2 2 | 1 and row 3 is 0 0 0 | 0
If you minus 2xR2 from R3 you get:
1 2 3 | 0
2 2 2 | 2
0 0 0 | 0
If you minus 2 x R1 from R2, you obtain the matrix:
1 2 3 | 0
0 -2 -4 | 2
0 0 0 | 0
Now divide R2 by -2 to obtain:
1 2 3 | 0
0 1 2 | -1
0 0 0 | 0
If you notice, we have two leading ones. The best explanation of this that I can give is a '1' in the left-hand most COLUMN, when ONLY preceded by a 0 or is itself in the first column. (A shitty definition, I know, maybe Ahmad can clear it up better).
Anyway from this you see that in a 3x3 matrix the rank is only TWO. The rank of a matrix is how many leading 1s it has. For there to be a distinct solution, we need Rank(M) >= Nj, where Nj = the number of columns in the matrix. However, for this matrix, Nj = 3 and Rank(M) = 2. Rank(M) < Nj, therefore we have more unknowns than we have equations to solve them! Thus, we have infinitely many solutions.
ALSO, this leads to a bit of a lemma. For any system of equations for which the number of equations is equal to the number of unknowns, if one equation is a scalar multiple of another, then the system of equations has INFINITELY many solutions.