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.
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} $
$ \begin{bmatrix} 1 & 1 & 1 \\ -2 & 1 & -2 \\ -1 & 3 & -1 \\ \end{bmatrix} $
$ \begin{bmatrix} 1 & 0 & 1 \\ -2 & 3 & -2 \\ -1 & 4 & -1 \\ \end{bmatrix} $
$ \begin{bmatrix} 1 & 0 & 0 \\ -2 & 3 & 0 \\ -1 & 4 & 0 \\ \end{bmatrix} $
$ \begin{bmatrix} 1 & 0 & 0 \\ 1 & 3 & 0 \\ 3 & 4 & 0 \\ \end{bmatrix} $
$ \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 3 & \dfrac{4}{3} & 0 \\ \end{bmatrix} $
$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ \dfrac{5}{3} & \dfrac{4}{3} & 0 \\ \end{bmatrix} $
This gives the book answer of:
$$ CREF = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ \frac{5}{3} & {4\over3} & 0 \\ \end{bmatrix} $$