I know the normal process is to do row operations to transform the matrix to get the identity matrix and then apply the same row operations in the identity matrix to get the inverse. But this process seems so tedious and I'd rather not waste so much time and energy doing the row operations to get it.

Is there some sort of general formula or shortcut to obtain the inverse similar to how the inverse of a 2x2's inverse is found? Thanks.

  • 1
    $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – Travis Apr 20 '15 at 4:52
  • $\begingroup$ If the matrix has a nonzero determinant $|A|$ (as an invertible matrix must), the entries of the inverse can be expressed by the usual ratios of (signed) cofactors to $|A|$. So you wind up computing nine $2\times 2$ determinants in addition to $|A|$. Is that easier or not? $\endgroup$ – hardmath Apr 20 '15 at 4:53
  • $\begingroup$ If you carry out the calculations one time with symbols instead of numbers, then you have a result that can be reused for any 3x3 matrix $\endgroup$ – Nick Alger Apr 20 '15 at 4:54
  • $\begingroup$ Why do you need to find the inverse of a $3 \times 3$ matrix anyway? $\endgroup$ – littleO Apr 20 '15 at 4:56
  • $\begingroup$ I've always used the formula that is given by Jessica K in answer $\endgroup$ – iostream007 Apr 20 '15 at 6:51

One way is to recall that, for $\operatorname{det}(A)\neq 0$,

$$A^{-1} = \frac{1}{\operatorname{det}(A)} \operatorname{adj}(A)$$

and use the Rule of Sarrus to calculate $\operatorname{det}(A)$. Whether that is easier or not is debatable, but the formula for a $2\times 2$ matrix is basically this.

  • $\begingroup$ If $a_{22} \neq 0$, then the calculation of $\det(A)$ may as well be done by Dodgson's method of condensation since the required $2\times 2$ determinants are already at hand. If $a_{22} = 0$, then a workaround is possible, if slightly more awkward, assuming $A$ is indeed invertible. $\endgroup$ – hardmath Apr 20 '15 at 4:58

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