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General Formula for the inverse of a 3×3 Matrix

This came about from some lunchtime fun a couple of days ago – we had an empty whiteboard and a boardpen: it was the logical thing to do.

\textrm{let } M =\begin{pmatrix}a&b&c\\ d&e&f\\ g&h&i \end{pmatrix}.
Now produce the matrix of minors

\begin{pmatrix} ei-fh&di-fg&dh-eg\\ bi-ch&ai-cg&ah-bg\\ bf-ce&af-cd&ae-bd \end{pmatrix}.

Use the alternating law of signs to produce the matrix of cofactors M'

M' = \begin{pmatrix} ei-fh&fg-di&dh-eg\\ ch-bi&ai-cg&bg-ah\\ bf-ce&cd-af&ae-bd \end{pmatrix}

Transpose M'

{M'}^T = \begin{pmatrix} ei-fh&ch-bi&bf-ce\\ fg-di&ai-cg&cd-af\\ dh-eg&bg-ah&ae-bd \end{pmatrix}

\textrm{det }M = a(ei-fh) - b(di-fg) + c(dh-eg)

The grand formula:
M^{-1} =
\frac{1}{a(ei-fh) - b(di-fg) + c(dh-eg)} \begin{pmatrix} ei-fh&ch-bi&bf-ce\\ fg-di&ai-cg&cd-af\\ dh-eg&bg-ah&ae-bd \end{pmatrix}

I suggest you print it out and put it on your bedroom ceiling…

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This entry was posted on Friday 18th July, 2008 at 5:16 pm and is filed under General with tags , , , , , , . You can follow any responses to this entry through the RSS 2.0 feed You can leave a response, or trackback from your own site.

3 Responses to “General Formula for the inverse of a 3×3 Matrix”

  1. brian Says:
    Tuesday 29th July, 2008 at 4:53 am

    This has some errors. The determinant is wrong; also the first row should have be

    (ei – fh) (ch – ib) (bf – ce)

    Reply
  2. jebavarde Says:
    Tuesday 29th July, 2008 at 8:07 pm

    Sorry, made some typos when copying it out. Will correct.
    Edit: think that’s all corrected now. Oops :P

    Reply
  3. Jared Says:
    Tuesday 26th May, 2009 at 7:46 pm

    THANKS :)

    this site helped me so much in making a matrix program

    haven’t seen a site that gives a single formula like this

    the only other way would have been programming through that whole process of finding the determinant and minor cofactors and dividing and and and … well, that would have been so tedious :P

    SO THANKS :D

    Reply

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