Prove that $ \det \left( A^{-1} \right) = \frac{1}{\det(A)} $

Question

If I have a non-singular matrix $\bf A$, how can I prove the following?

$$ \det \left( {\bf A}^{-1} \right) = \frac{1}{\det({\bf A})} $$

I know that ${\bf A} {\bf A}^{-1} = {\bf I}$, but I am not sure what to do with that knowledge.

Yes, but I am not sure what that help me with. Having a hard time putting 2 and 2 together. — CuriousFellow, Feb 03 '14 at 04:59
The equation about determinants holds for all $A$ and $B$, so it holds if $B$ is something particular. — , Feb 03 '14 at 05:01
Since $\det(AB)=\det A\det B$, letting $B=A^{-1}$ gives $\det I=1=\det A\det(A^{-1})$, so $\det(A^{-1})=\frac{1}{\det A}$. — , Feb 03 '14 at 05:43
See also: http://math.stackexchange.com/questions/1121697/determinant-of-the-inverse-matrix — Martin Sleziak, Jan 27 '15 at 13:34

score 34 · Accepted Answer · edited May 16 '18 at 19:41

34

first of all we know that

$$\det(A \cdot B)=\det(A)\times\det(B)$$

also we know that

$$A\times A^{-1}=I$$

we know that

$$\det(A \cdot A^{-1})=\det(I)$$

or

$$\det(A)\times\det(A^{-1})=\det(I)$$

Can you continue from this? Ask yourself: what is $\det(I)$?) Take the example of the $3\times 3$ identity matrix:

$$ I = \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$

$$ \det(I) = 1$$

So,

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

or

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

edited May 16 '18 at 19:41

Alex Kritchevsky

answered Feb 03 '14 at 05:03

dato datuashvili

2

I assume the determinant of an identity matrix is 1, right? If so, then I understand this. Sorry, I just began learning Linear Algebra. – CuriousFellow Feb 03 '14 at 05:06
of course for nXn identity matrix has determinant one – dato datuashvili Feb 03 '14 at 05:15
@CuriousFellow for diagonal matrices, the determinant is simply the product along the diagonal – John D Jan 12 '22 at 16:51

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