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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General Formula for the inverse of a 3×3 Matrix

9 thoughts on “General Formula for the inverse of a 3×3 Matrix

  1. Jared says:

    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

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