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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.

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1 Answer 1

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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.

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  • $\begingroup$ @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. $\endgroup$ Mar 9, 2020 at 5:08
  • $\begingroup$ @5Dots As best I can tell, you're mixing up two formulas. The following both are true: $$\text{inv}(A) = 1/|A| \cdot \text{adj}(A) \\ |\text{inv}(A)| = 1/|A|^n \cdot |\text{adj}(A)|$$ $\endgroup$ Mar 9, 2020 at 5:54
  • $\begingroup$ Ok thanks i got this, sorry for the trouble man. Appreciates a lot $\endgroup$
    – 5 Dots
    Mar 9, 2020 at 9:29

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