\begin{pmatrix}3&-1&3\\ -2&3&2\\ 1&-3&1\end{pmatrix}

Any help would be greatly appreciated! I have done this before with other matrices, yet this one is bringing me trouble. I have spent too long on this as I believe it should be easy and yet my answers are always wrong. What Do you do after you get the identity matrix?

What I am doing: So I first row reduce it to the identity matrix:

  • I first switch $R_1$ and $R_3$

  • Then $R_3-3R_1$

  • then $R_2+2R_1$

  • then $-1/3R_2$

  • then $R_3-8R_2$

  • then $R_2-2R_1$

  • then $3/32R_3$

  • then $R_1+3R_3$

  • and finally $R_2+4/3R_3$

I know there is many possible ways to get this too the identity matrix but this worked for me. I then apply all of the above steps to identity matrices and multiply them together to check my work but it never works out to be the original matrix so I am forced to believe something is wrong.

  • $\begingroup$ "Some reason" is not helpful. What are you doing that's causing trouble? $\endgroup$ – fonini Nov 8 '17 at 4:16
  • $\begingroup$ Sorry, my apologies for being vague. Basically when I row reduce it like normal I get that the solution set is inconsistent in the last row. $\endgroup$ – G Muf Nov 8 '17 at 4:17
  • $\begingroup$ @GMuf Are you saying that this matrix is singular? Only non-singular matrices are products of elementary matrices. $\endgroup$ – Lord Shark the Unknown Nov 8 '17 at 4:30
  • $\begingroup$ Then it must be non singular as it does have an answer, I am just not doing it correctly. $\endgroup$ – G Muf Nov 8 '17 at 5:44
  • $\begingroup$ If you have all of the operations that reduce $A$ to $I$ Then performing the reverse operation in reverse order, will give you the elementary operations to get from I back to $A.$ $\endgroup$ – Doug M Nov 8 '17 at 20:26

Since you are not showing you matrices, I am guessing the most likely mistake is


Should be



Most likely another mistake is $R_3-8R_1$ should be $R_3-8R_\color{red}2$


After you perform row operations:

$$E_k\ldots E_1 A = I$$

$$A =E_1^{-1} \ldots E_k^{-1} $$

  • $\begingroup$ Sorry typo! I did add it. $\endgroup$ – G Muf Nov 8 '17 at 20:07
  • 1
    $\begingroup$ do you mind sharing the intermediate matrices. $\endgroup$ – Siong Thye Goh Nov 8 '17 at 20:07
  • $\begingroup$ Again, another typo. I know I did row reduce it correctly to the identity. I Just am not sure what to do from there. $\endgroup$ – G Muf Nov 8 '17 at 20:17
  • $\begingroup$ did you solve for the matrix $A$ as described in my answer to your duplicate post? what is the answer to that question? $\endgroup$ – Siong Thye Goh Nov 8 '17 at 20:19
  • $\begingroup$ I agree with Doug, but there could be a careless mistake lurking around in any of your steps... Including a careless mistake in multiplying matrices or taking inverse. I don't think we have sufficient info to pinpoint the critical mistake yet. I am not sure if the mistake is of a careless nature of a conceptual error for now. $\endgroup$ – Siong Thye Goh Nov 8 '17 at 20:30

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