Answer
See answers below
Work Step by Step
$\begin{bmatrix}
3 & -3 &6\\
2 &-2& 4 \\
6 & -6 &12
\end{bmatrix} \approx^1 \begin{bmatrix}
3 & -3 &6\\
2 &-2& 4 \\
1 & -1 &2
\end{bmatrix} \approx^2\begin{bmatrix}
1 & -1 &2\\
3 &-3& 6 \\
2 & -2 &4
\end{bmatrix} \approx^3 \begin{bmatrix}
1 & -1 &2\\
0 &0& 0 \\
0 & 0 &0
\end{bmatrix} $
Rank (A)= 1
The last matrix is in a row-echelon form.
$M_3(\frac{1}{6})$
$P_{13}$
$A_{12}(-3), A_{13}(-2)$
