Not getting the right answer for a matrix in reduced column echelon form. Introduction to Linear Algebra by Donald J. Wright has this question in section 1.7: 

Find a matrix in column reduced echelon form that is column equivalent to 
  $
        \begin{bmatrix}
        2 & 1 & 1   \\
        -1 & 1 & -2 \\
        2 & 3 & -1  \\
        \end{bmatrix}
$.

The back of the book says the answer is 
$
        \begin{bmatrix}
        1 & 0 & 0   \\
        0 & 1 & 0 \\
        \frac{5}{3} & {4\over3} & 0  \\
        \end{bmatrix}
$
, but the answer I keep getting is $I_3$. Here's my reasoning: Set A to the original matrix. $$ A*F_{12} * F_{12}(-2) * F_{13}(-1) * F_{32}(-1) * F_3({-1\over2}) * F_{32}(2) * F_{31}(-3) * F_2({1\over3}) * F_{21}(-1) = I_3$$
Where $I_n$ is the identity matrix, $F_{ab}(k)$ is the matrix that results from adding column a of $I_n$ multiplied by k to column b, $F_{ab}$ is the result of swapping columns a and b of $I_n$, and $F_{a}(k)$ is the result of multiplying column a of I_n by k. 
Please show me what am I doing wrong.
P.S.: This is my first post in math.stackexchange.com so please let me know if my question can be improved somehow. 
 A: We are doing Column Reduced Echelon Form, which is a basic analog of row reduced echelon form (in fact, if you take the transpose of $(RREF)^T = CREF$).
There is also a definition for CREF to note.
Try these steps ($C_x =$ Column $x$):
$
        \begin{bmatrix}
        2 & 1 & 1   \\
        -1 & 1 & -2 \\
        2 & 3 & -1  \\
        \end{bmatrix}
$


*

*$C_1: C_1 - C_2$ (this reads $C_1 = C_1 - 2 C_2$)


$
        \begin{bmatrix}
        1 & 1 & 1   \\
        -2 & 1 & -2 \\
        -1 & 3 & -1  \\
        \end{bmatrix}
$


*

*$C_2: C_2-C_3$


$
        \begin{bmatrix}
        1 & 0 & 1   \\
        -2 & 3 & -2 \\
        -1 & 4 & -1  \\
        \end{bmatrix}
$


*

*$C_3: C_3 - C_1$


$
        \begin{bmatrix}
        1 & 0 & 0   \\
        -2 & 3 & 0 \\
        -1 & 4 & 0  \\
        \end{bmatrix}
$


*

*$C_1: C_1+C_2$


$
        \begin{bmatrix}
        1 & 0 & 0   \\
        1 & 3 & 0 \\
        3 & 4 & 0  \\
        \end{bmatrix}
$


*

*$C_2: \dfrac{C_2}{3}$


$
        \begin{bmatrix}
        1 & 0 & 0   \\
        1 & 1 & 0 \\
        3 & \dfrac{4}{3} & 0  \\
        \end{bmatrix}
$


*

*$C_1: C_1 - C_2$


$
        \begin{bmatrix}
        1 & 0 & 0   \\
        0 & 1 & 0 \\
        \dfrac{5}{3} & \dfrac{4}{3} & 0  \\
        \end{bmatrix}
$
This gives the book answer of:
$$ CREF = 
        \begin{bmatrix}
        1 & 0 & 0   \\
        0 & 1 & 0 \\
        \frac{5}{3} & {4\over3} & 0  \\
        \end{bmatrix}
$$ 
