Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have this series of operations. How can I obtain the matrix?

$$ K := E_{12}(27) \cdot E_{13}(2) \cdot E_{15}(7) \cdot E_{23}(23) \cdot E_{45}(3) \cdot E_{41}(12) \cdot E_{52}(2). $$

I tried to multiply from right to left (starting with the matrix I listed below). But I don't obtain the matrix I need.

I want to check with you if this is right:

$$ E_{52}(2)=\left[ \begin{array}{ c c } 1&0&0&0&0 \\ 0&1&0&0&0 \\ 0&0&1&0&0 \\ 0&0&0&1&0\\ 0&2&0&0&1 \end{array} \right] $$

Am I right?

share|cite|improve this question
up vote 1 down vote accepted

I presume that your $E$'s are elementary matrices. So, $E_{52}(2)$ is as you have (put "2" in the $(5,2)$ position of the identity matrix $I$).

But this is just a guess... You should refer to your text/notes for the definition of the $E$'s.

Note that if the $E$'s are as described above, then multiplication of $A$ by one of these on the left (i.e., $EA$) corresponds to a row operation on $A$ (the same row operation needed to obtain $E$ from $I$).

Using this, you can compute $K$ in a manner less prone to arithmetic errors.

share|cite|improve this answer
Yes, E's are elementary matrices. I don't have an A. I have only these ones. So I should start with the $E_{52}(2)$ right? Instead of A – Andrew Nov 29 '11 at 9:27
Yes. Then multiply on the left by $E_{41}(12)$. So just take $E_{52}(2)$ and add 12 times row 1 to row 4. – David Mitra Nov 29 '11 at 9:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.