# Express the following invertible matrix A as a product of elementary matrices

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.

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?

• Thia should work for any invertibale matrix right? – Number4 May 7 '19 at 14:59
• Yes this is the generalise-able process. – john Oct 22 '19 at 4:45