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Using the fact that $AC^T = (\det A)I$ where $A$ is some arbitrary matrix and $C$ is the cofactor matrix for $A$, how can I prove that $\det C = (\det A)^{n-1}$

I really don't know how to progress on this. Tried doing some random operations on both equations to see if I can make a link somewhere and: $AC^T = \det A \implies C^T = A^{-1}\det A \implies \det C^T = (\det A)A^{-1}$

$\det C = (\det A)^{n-1} \implies \det C^T = (\det A)^{n-1}$

To be honest, I'm not even sure if the last operation I applied on the first line and the operation I applied on the second line are even legal operations.

Suggestions are welcome.

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1 Answer 1

up vote 1 down vote accepted

$$\det(A)\det(C)=\det(A)\det(C^T)=\det(AC^T)=\det((\det A)I)=(\det A)^n$$ If $\det(A)\neq0$, then divide both sides by $\det(A)$.
If $\det(A)=0$ then it follows easily from the definition of cofactor matrix that $\det(C)=0$.

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