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I want to find $A^{-1} \pmod{26}$ for

$A=\begin{bmatrix}10&3\\5&3\end{bmatrix}$

and I did the conventional $\frac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$ and found the inverse of the fraction mod $26$, cool, then reduced mod $26$ the matrix.

I obtained:

$$\begin{bmatrix}21&5\\17&18\end{bmatrix}$$

Cool - but the wolframalpha calculator obtained the transpose of above? What the?

Who is right?

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  • $\begingroup$ sometimes WA treats rows/columns in a way that is not clear, leading often to transpose issues if not dealt with properly $\endgroup$ – danimal Apr 1 '15 at 10:33
  • $\begingroup$ You are right, wolfram alpha is wrong! $\endgroup$ – user3184807 Apr 1 '15 at 10:34
  • $\begingroup$ Trying both answers shows yours is right, and the transpose doesn't work. $\endgroup$ – coffeemath Apr 1 '15 at 10:36
  • $\begingroup$ Yay. Silly wolfram alpha :). Smart humans! $\endgroup$ – Wolfram Apr 1 '15 at 10:38
  • $\begingroup$ Strangely enough, "Inverse[{{10,3},{5,3}},Modulus->26]" doesn't work. $\endgroup$ – Batman Apr 1 '15 at 11:04
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Be careful that Wolfram addresses your $a,b,c,d$ to $1,2,3,4$, so you should have written your $a,b,c,d$ in the corresponding order to the entries $1,2,3,4$, which gives the matrix $$ \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix} $$ It is not handy, but that is how this beta widget works.

See this;

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  • $\begingroup$ Oh yes I see, so it was silly humans afterall :P [the programmers] $\endgroup$ – Wolfram Apr 1 '15 at 11:28

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