-1

I am unsure on how to prove the following problem:

If a $2 \times 2$ matrix $A$ is non-singular, prove $|A| = \frac{1}{|A^{-1}|}$

I know that $|A|\cdot|A^{-1}| = I$ but i’m not sure where to go from there.

Any help on proving this would be appreciated.

Stephen Donovan
  • 5,163
Jayden
  • 9

1 Answers1

1

In fact, since $A\cdot A^{-1}=I$, taking the determinant of both sides we get: $|A \cdot A^{-1}|=|I|$.

Because determinant is multiplicative (this means $|A|\cdot|B|=|A\cdot B|$, for every matrix $A,B$), $|A\cdot A^{-1}|=|A|\cdot|A^{-1}|.$ Also, $|I|=1$ (As a diagonal matrix, its determinant is exactly the product of the main diagonal elements). From here, we obtain:

$$|A|\cdot|A^{-1}|=1 \Rightarrow |A|=\frac{1}{|A^{-1}|}.$$

And this is available for any $n \times n$ matrix, not just $2 \times 2$.

Paul Rebenciuc
  • 727