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

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.

Note when $$M$$ has order $$n$$, we have $$|kM| = k^n \cdot|M|$$, not $$k\cdot|M|.$$ Therefore, $$|A^*| = 1/|A|^n \cdot |\text{adj } A| = |A|^{-1}$$
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.
• @5Dots You have written $$\text{|A*| = 1/|A| × |adj A| (as |A| is constant)}$$ My whole point is that this is wrong unless $n=1$. Therefore, when you ask the only question: "But why it is not 1...... " it's because you misapplied the determinant. Mar 9, 2020 at 5:08
• @5Dots As best I can tell, you're mixing up two formulas. The following both are true: $$\text{inv}(A) = 1/|A| \cdot \text{adj}(A) \\ |\text{inv}(A)| = 1/|A|^n \cdot |\text{adj}(A)|$$ Mar 9, 2020 at 5:54