-1
$\begingroup$

\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.

$\endgroup$
  • $\begingroup$ "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
0
$\begingroup$

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

$$R_2-2R_1$$

Should be

$$R_2\color{red}+2R_1$$

Edit:

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

Edit:

After you perform row operations:

$$E_k\ldots E_1 A = I$$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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