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I got the following question:
Find, if possible, the inverse of the matrix:
$\begin{bmatrix}3, -1\\2, -2\end{bmatrix} $

and I did the following:
$\begin{bmatrix}3, -1\\2, -2\end{bmatrix}^{-1} = \frac{1}{ad-bc}\begin{bmatrix}-2, 1\\-2, 3\end{bmatrix} = \frac{1}{4} \begin{bmatrix}-2, 1\\-2, 3\end{bmatrix}$

Test: $( A * A^{-1} = I)$
$\begin{bmatrix}3, -1\\2, -2\end{bmatrix} *\begin{bmatrix}-0.5,0.25\\-0.5, 0.75\end{bmatrix} = \begin{bmatrix}-1, 0\\0, -1\end{bmatrix}$

Is this correct?

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    Your matrix's determinant has the wrong sign: it should be $;\color{red}-\frac14;$ – DonAntonio Apr 06 '14 at 12:42
  • Oh wow! Maybe it's time to eat something... Thanks! –  Apr 06 '14 at 12:45
  • In the above comment the determinant was meant to be $;\color{red}-4;$ , of course... – DonAntonio Apr 06 '14 at 12:48
  • @amWhy That's a typo. Thanks though. Ctrl-C + Ctrl-V is dangerous –  Apr 06 '14 at 12:53
  • @Pontus (hehehe) Yes, I know the danger! – amWhy Apr 06 '14 at 12:54
  • Have you found your answer? If so, why not help others? could you please change to a Q&A style post? (Remove the answer from the question, and post an answer yourself - Read more on meta) Help build another great stackexchange knowledge base :) – TastySpaceApple Apr 06 '14 at 21:11
  • Well no one has actually answered the question, can the identity matrix be negative? I only got an answer to my specific question but not the actual topic of the question. –  Apr 06 '14 at 21:56

2 Answers2

4

No, the identity matrix cannot be negative. If your check yields $AA^{-1} = -I$ then something must have gone wrong.

-1

La faute est dans le calcul du déterminant, on a : ad-bc=-4 et non 4

The error is in the calculation of the determinant. We have $ad-bc=-4$ and not $4$.

Daniel R
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Mohamed
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