The reduced row echelon form of the matrix
-1/2 1/2 3/2
-3/2 2 3/2
1/2 -3/2 9/2
is?
I don't like to work with matrices containing fractions, so I would factor out the common factor 1/2:
-1/2 1/2 3/2.... (1)
-3/2 2 3/2.......(2)
1/2 -3/2 9/2 ....(3)
becomes
(1/2)*
-1 1 3 ....(1A)
-3 4 3 ....(2A)
1 -3 9 ....(3A)
Eliminate first column (except first row):
(1/2)*
1 -1 -3 ....(-1)*(1A)=(1B)
0 1 -6 ....(2A)-3*(1A)=(2B)
0 -2 12 ....(3A)+(1A)=(3B)
Eliminate second column (3rd row only)
(1/2)*
1 -1 -3 ....(1C)=(1B)
0 1 -6 ....(2C)=(2B)
0 0 0 ....(3C)=(3B+2B)
The third row is (0,0,0), which means that the solution of the homogeneous equation has infinite number of roots.
Continuing, after eliminating the third row, we obtain the row echelon form:
1 -1 -3 ....(1C)=(1B)
0 1 -6 ....(2C)=(2B)
To obtain the reduced row echelon form, we now need to work with columns containing leading coefficients so that each column is zero except the leading coefficient (the first "1" of each row, counting from the left).
This means we have to eliminate -1 from the first row (equation (1C):
1 0 -9 ....(1D)=(1C)+(2C)
0 1 -6 ....(2D)=(2C)
The above set of equations is now in the row echelon form, namely: all elements in columns containing the leading coefficient are zero except the leading coefficient itself.