What are the necessary steps and reasoning for calculating the following matrix in GL(2,$\Bbb Z_{11}$):

$M = \begin{pmatrix} 2&6 \\3&5 \end{pmatrix}$.

I found the answer to be $\begin{pmatrix} 9&9 \\10&8 \end{pmatrix}$, however I lost 2 points (out of 10) to the question--which is the reason for my confusion.

I found the determinant to be -8, which is 3 in $ℤ_{11}$ and therefore, in calculating the inverse, $\left(\frac13\right)$ = 4 in $ℤ_{11}$. While I was computing the inverse, I had this intermediary matrix $M^{-1} = \begin{pmatrix} 20&{-24} \\{-12}&8 \end{pmatrix}$ which I then converted to mod 11 to obtain my final answer. My professor circled the -12 and -24 in the matrix and asked "What is '24' inverse?" This was where I lost points, and I'm completely unsure why. I talked to my professor about this, and I came away thinking I understood but upon further thought I'm still unsure. His reasoning was along the lines of -12 and -24 are additive inverses...but then I get lost. I don't understand why my method of computation is "incorrect." If anything is unclear, or if anything is wrong with the formatting (this is my first post on this website), please let me know and I'll do my best to elaborate on/fix anything! Thanks.

(Also, if it helps I can post my entire computation with every step in detail)

  • $\begingroup$ determinant $\frac{1}{3}$? :O $\endgroup$ – user87543 Feb 1 '14 at 5:03
  • 1
    $\begingroup$ I'm pretty sure he means $\frac{1}{3}$ as in the inverse of $3$ in $\mathbb{Z}_{11}$ which is 4. $\endgroup$ – Robert Wolfe Feb 1 '14 at 5:09
  • $\begingroup$ @Bryan but the determinant is $-8\equiv3 \text { mod }11$... $\endgroup$ – user87543 Feb 1 '14 at 5:12
  • $\begingroup$ Yes, the determinant is actually -8 which is 3 in $ℤ_{11}$. Sorry, I said the wrong words for what I was thinking, I'll edit. $\endgroup$ – Tadpole Feb 1 '14 at 5:13
  • $\begingroup$ I mean the determinant of his inverse. $\endgroup$ – Robert Wolfe Feb 1 '14 at 5:13

Apparently your prof wanted you to write all the elements as part of the "usual", basic set of representatives modulo $\;11\;$ , i.e. $\;\Bbb Z_{11}=\{0,1,2,3,...,10\}\pmod{11}\;$, the devil knows why , and then


$$\;\;\text{(this is what's meant by "additive inverses")}$$

so that your matrix would look as


  • $\begingroup$ Thank you for the response! It's clear to me now where I erred. Thanks so much. $\endgroup$ – Tadpole Feb 1 '14 at 5:18

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