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Should the formula for the inverse of a 2x2 matrix be obvious?

$$( \operatorname{diag}(a,d))^{-1} = \operatorname{diag}\left( \frac1a, \frac1d \right) = \frac1{ad} \operatorname{diag}(d,a) $$ That answers at least part of the question.
Michael Hardy's user avatar
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Should the formula for the inverse of a 2x2 matrix be obvious?

For $\mathbf{A}$ near zero we have $$ (1-\mathbf{A})^{-1}\approx 1+\mathbf{A} $$ so it has to negate the off-diagonals. (If you want to get all fancy about it you could notice that we use $\exp$ to ...
Dan Piponi's user avatar
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