TITLE: Express the following invertible matrix A as a product of elementary matrices
QUESTION [14 upvotes]: I've been at this for a while... I tried to the inverse method but it keeps on saying I'm getting it wrong... Can anyone show me a step-by-step solution? The matrix I have is a $3\times 3$ square one(sorry for formatting):
$$
\begin{pmatrix}
6  & 6 & -2 \\
-1 & 0 & 0 \\
-1 & 1 & 0
\end{pmatrix}
$$
I'm starting to go crazy, I honestly have a few pages of written work and the marker keeps on saying I got it wrong.. I guess it's too late to get my marks since I used all my attempts but I want to see how to do it for future reference.

REPLY [31 votes]: The idea is to row-reduce the matrix to its reduced row echelon form, keeping track of each individual row operation.
Call the original matrix $A$.
Step 1. Switch $\operatorname{Row}_1$ and $\operatorname{Row}_2$. This corresponds to multiplying $A$ on the left by the elementary matrix
$$
E_1=
\begin{pmatrix}
0 & 1 & 0 \\
1 & 0 & 0 \\ 
0 & 0 & 1
\end{pmatrix}
$$
and the result is
$$
E_1A=
\begin{pmatrix}
-1 & 0 & 0 \\
6 & 6 & -2 \\ 
-1 & 1 & 0
\end{pmatrix}
$$
Step 2. Multiply $\operatorname{Row}_1$ by $-1$. This corresponds to multiplying $E_1A$ on the left by the elementary matrix 
$$
E_2=
\begin{pmatrix}
-1 & 0 & 0 \\
0 & 1 & 0 \\ 
0 & 0 & 1
\end{pmatrix}
$$
and the result is
$$
E_2E_1A
=
\begin{pmatrix}
1 & 0 & 0 \\
6 & 6 & -2 \\ 
-1 & 1 & 0
\end{pmatrix}
$$
Step 3. Subtract $6\cdot\operatorname{Row}_1$ from $\operatorname{Row}_2$. This corresponds to multiplying $E_2E_1A$ on the left by the elementary matrix
$$
E_3
\begin{pmatrix}
1 & 0 & 0 \\
-6 & 1 & 0 \\ 
0 & 0 & 1
\end{pmatrix}
$$
and the result is
$$
E_3E_2E_1A
=
\begin{pmatrix}
1 & 0 & 0 \\
0 & 6 & -2 \\ 
-1 & 1 & 0
\end{pmatrix}
$$
Step 4. Multiply $\operatorname{Row}_2$ by $\displaystyle\frac{1}{6}$. This corresponds to multiplying $E_3E_2E_2A$ on the left by the elementary matrix
$$
E_4
\begin{pmatrix}
1 & 0 & 0 \\
0 & \frac{1}{6} & 0 \\ 
0 & 0 & 1
\end{pmatrix}
$$
and the result is
$$
E_4E_3E_2E_1A
=
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & -\frac{1}{3} \\ 
-1 & 1 & 0
\end{pmatrix}
$$
Step 5. Add $\operatorname{Row}_1$ to $\operatorname{Row}_3$. This corresponds to multiplying $E_4E_3E_2E_1A$ on the left by the elementary matrix
$$
E_5=
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\ 
1 & 0 & 1
\end{pmatrix}
$$
and the result is
$$
E_5E_4E_3E_2E_1A
=
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & -\frac{1}{3} \\ 
0 & 1 & 0
\end{pmatrix}
$$
Step 6.
Subtract $\operatorname{Row}_2$ from $\operatorname{Row}_3$. This corresponds to multiplying $E_5E_4E_3E_2E_1A$ on the left by the elementary matrix 
$$
E_6=
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\ 
0 & -1 & 1
\end{pmatrix}
$$
and the result is
$$
E_6E_5E_4E_3E_2E_1A
=
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & -\frac{1}{3} \\ 
0 & 0 & \frac{1}{3}
\end{pmatrix}
$$
Step 7. Multiply $\operatorname{Row}_3$ by $3$. This corresponds to multiplying $E_6E_5E_4E_3E_2E_1A$ on the left by the elementary matrix 
$$
E_7=
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\ 
0 & 0 & 3
\end{pmatrix}
$$
and the result is
$$
E_7E_6E_5E_4E_3E_2E_1A
=
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & -\frac{1}{3} \\ 
0 & 0 & 1
\end{pmatrix}
$$
Step 8. Add $\displaystyle\frac{1}{3}\cdot\operatorname{Row}_3$ to $\operatorname{Row}_2$. This corresponds to multiplying $E_7E_6E_5E_4E_3E_2E_1A$ on the left by the elementary matrix 
$$
E_8=
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & \frac{1}{3} \\ 
0 & 0 & 1
\end{pmatrix}
$$
and the result is
$$
E_8E_7E_6E_5E_4E_3E_2E_1A
=
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\ 
0 & 0 & 1
\end{pmatrix}
$$
Finally, we have the equation
$$
E_8E_7E_6E_5E_4E_3E_2E_1A=I
$$
where each $E_i$ is an elementary matrix. To finish the problem, we write
$$
A=E_1^{-1}E_2^{-1}E_3^{-1}E_4^{-1}E_5^{-1}E_6^{-1}E_7^{-1}E_8^{-1}
$$
Can you invert each $E_i$ and carry out the matrix multiplication?