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Let's say we have matrix ${\bf A}$. Its inverse matrix is given by

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

I know that dividing by the determinant makes sense because we are getting rid of the area done by the transformation but what is going on with the adjoint matrix?

Rodrigo de Azevedo
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SirMrpirateroberts
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1 Answers1

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So we know that $AA^{-1}=I$. Also, $A[\operatorname{adj}(A)] = [\det(A)]I$ (see here).

Therefore,

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

Rodrigo de Azevedo
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Annika
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  • Right, the adjugate matrix of A is defined as a matrix of all "minors" of the matrix A, i.e., the determinants of submatrixes of A without one particular row and column. Because of the Laplace expansion (https://en.wikipedia.org/wiki/Laplace_expansion) the determinant of the whole A can be expanded in terms of these minors, which results in $A adj(A)$ being $det(A)$ times the identity matrix $I$. – Nadav Har'El Jul 14 '23 at 08:53