If $A$ is an invertible matrix of order 2, then $|A^*|$ is (where $A^*$ is inverse of matrix $A$)

Question

I am having some formula without proof (u can find it by ur self) $\def\adj{\operatorname{adj}}$

$A^* = 1/|A| \cdot \adj A $

and

$|\adj A| = |A|^{n-1}$ where $n$ is order of matrix

By using these formulae

$$\begin{align} |A^*| &= 1/|A| \cdot |\adj A|\qquad \text {as $|A|$ is constant}\\ &= 1/|A| \cdot |A|^{n-1}\\ &= |A|^{n-2} \end{align} $$ Now here $n$ is 2 So finally we gotta

$|A^*| = |A|^0 = 1$ But why it is not 1. Plz help me.

Answer 1

Note when $M$ has order $n$, we have $|kM| = k^n \cdot|M|$, not $k\cdot|M|.$ Therefore, $$|A^*| = 1/|A|^n \cdot |\text{adj } A| = |A|^{-1}$$

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For future reference: OP has used $A^*$ for the inverse of $A$ so I have used this notation. This should not be confused with the common notation of the conjugate transpose.