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I apologise in advance if this is simple, but I'm losing my brain over this question. I'm unsure how to make the matrix format work either.

I'm trying to find elementary matrices so that $E_n\times...\times E_1\times A=I$.

The given matrix $A$ is \begin{pmatrix} 0&1&1\\2&2&0\\2&1&1\end{pmatrix} and my first step was $R_1=R_1+\dfrac 12 R_2$. This gives me the result of \begin{pmatrix}1&2&1\\2&2&0\\2&1&1 \end{pmatrix}.

However using the instructions given on This Website, I get \begin{pmatrix}1&1/2&0\\0&1&0\\0&0&1\end{pmatrix} for the elementary matrix, which does not result in the same answer when multiplied with $A$.

Am I missing something painfully obvious? Further down the line I was able to reduce $A$ to $RREF$ without too much trouble, yet online calculators are telling me it isn't consistent, and none of my elementary matrices are correct.

Looking online I could only find the same information on how to get elementary matrices, and other examples where it works fine, which leads me to believe I have made an error somewhere.

Thanks in advance.

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    $\begingroup$ Little unsure what your asking. You have to find all of them and multiply them by $A$ to satisfy your equation, you only found one. Try finding all of them and trying it out. Your matrix does reduce to the identity. $\endgroup$ – HighSchool15 Aug 23 '15 at 8:36
  • $\begingroup$ Sorry for it being unclear. The problem is that the elementary matrix I found for the first step (third one pictured above), does not give the required result when multiplied with matrix A. Thus I am unsure if the website I linked is giving me the wrong information, or if I have made an error somewhere. $\endgroup$ – Dwayne H Aug 23 '15 at 9:41
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I hope that the following argumentation will solve the problem and, at the same time, it will show the method of solving similar problems:

So, we start with this matrix: $$A=\begin{pmatrix} 0&1&1\\2&2&0\\2&1&1\end{pmatrix}$$

First step: Let's subtract the first row from the third one. Multiplying with the matrix

$$E_1=\begin{pmatrix}1&0&0\\0&1&0\\-1&0&1 \end{pmatrix}$$ will do that. Indeed,

$$E_1A=\begin{pmatrix}1&0&0\\0&1&0\\-1&0&1 \end{pmatrix}\begin{pmatrix} 0&1&1\\2&2&0\\2&1&1\end{pmatrix}=\begin{pmatrix} 0&1&1\\2&2&0\\2&0&0\end{pmatrix}.$$ Second step: Exchange the first row with the third one. Multiplying with

$$E_2=\begin{pmatrix} 0&0&1\\0&1&0\\1&0&0\end{pmatrix}$$ will do that. Indeed,

$$E_2E_1A=\begin{pmatrix} 0&0&1\\0&1&0\\1&0&0\end{pmatrix}\begin{pmatrix} 0&1&1\\2&2&0\\2&0&0\end{pmatrix}=\begin{pmatrix} 2&0&0\\2&2&0\\0&1&1\end{pmatrix}.$$

Third step: Subtract the first row from he second one. Multiplying with the matrix

$$E_3=\begin{pmatrix} 1&0&0\\-1&1&0\\0&0&1\end{pmatrix}$$ will do that. Indeed,

$$E_3E_2E_1A=\begin{pmatrix} 1&0&0\\-1&1&0\\0&0&1\end{pmatrix}\begin{pmatrix} 2&0&0\\2&2&0\\0&1&1\end{pmatrix}=\begin{pmatrix} 2&0&0\\0&2&0\\0&1&1\end{pmatrix}.$$

Fourth step: Divide the first two rows by $\frac12$. The matrix

$$E_4=\begin{pmatrix} \frac12&0&0\\0&\frac12&0\\0&0&1\end{pmatrix}$$ will do that. Indeed,

$$E_4E_3E_2E_1A=\begin{pmatrix} \frac12&0&0\\0&\frac12&0\\0&0&1\end{pmatrix}\begin{pmatrix} 2&0&0\\0&2&0\\0&1&1\end{pmatrix}=\begin{pmatrix} 1&0&0\\0&1&0\\0&1&1\end{pmatrix}.$$

*Fifth step:** Subtract the second row from the third one. Multiplying by

$$E_5=\begin{pmatrix} 1&0&0\\0&1&0\\0&-1&1\end{pmatrix}$$ will do that. Indeed,

$$E_5E_4E_3E_2E_1A=\begin{pmatrix} 1&0&0\\0&1&0\\0&-1&1\end{pmatrix}\begin{pmatrix} 1&0&0\\0&1&0\\0&1&1\end{pmatrix}=\begin{pmatrix} 1&0&0\\0&1&0\\0&0&1\end{pmatrix}=I.$$

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  • $\begingroup$ Thanks for your answer! This does indeed solve the problem, and does explain the process well. I have just realised that I made the simple error of multiplying in the form of AE, rather than EA, which explains why my elementary matrices were giving wrong results. Gotta love making basic errors. $\endgroup$ – Dwayne H Aug 23 '15 at 13:04

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