Clever calculation of the adjugate of a matrix

Asked Feb 01 '23 at 11:08

Active Feb 01 '23 at 11:26

Viewed 27 times

I would like to calculate $\text{adj}({A})$ the adjugate of a square matrix $A$. Matrix $A$ is singular. Is there a clever way of doing so, instead of the direct approach which is computationally expensive?

If $A$ was invertible, then one could calculate the inverse $A^{-1}$, and the adjugate would be

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

But here, $A$ is not invertible.

edited Feb 01 '23 at 11:26

asked Feb 01 '23 at 11:08

Hosam Hajeer

21,978

4

Adjoint is an archaic term, nowadays this is called the adjugate. – Luke Collins Feb 01 '23 at 11:10
Also what do you mean "just take the inverse"? Are you concerned with algorithmic complexity? – Luke Collins Feb 01 '23 at 11:11
If $A$ is invertible, then I think the simplest and most efficient way to compute the adjugate would be by computing the inverse using Gauss-Jordan elimination, and then multiplying by the determinant of $A$. If there is a better way, I'd be interested to know. – Hosam Hajeer Feb 01 '23 at 11:28
1

Yes. It does. Thank you @AnneBauval – Hosam Hajeer Feb 01 '23 at 11:53

Clever calculation of the adjugate of a matrix

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