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If have a matrix $A$ that all I know about is its size and what its determinant is? For example a $4\times4$ matrix with a determinant of $3$. How can I find the determinant of the cofactor matrix $cof(A)$.

What I have done is I created a lower triangular matrix where the product of the diagonals will give me $3$. So do I just find the cofactor matrix of the matrix I came up with and find its determinant. Would that be correct?

V. Vancak
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A.Mo
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1 Answers1

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For your $n \times n$ matrix $A$,

  • is $\det(A) \neq 0$,
  • is $n > 1$,
  • do you accept the formulas $A^{-1} = cof(A)^T / \det(A)$,
  • $\det(A B) = \det(A) \, \det(B)$ for two $n \times n$ martices $A$ and $B$ and
  • $\det(B^{T}) = \det(B)$ for any matrix $B$ ?

If so, we have

$\det(A) I_n = cof(A)^T A$.

Taking the determinant on both sides, we obtain

$\det(A)^n = \det(cof(A)^T) \, \det(A)$ and so

$\det(cof(A)) = \det(cof(A)^T) = \det(A)^{n-1}$.

Neo
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    It's worth noting that the result is still true when $\det A=0$ (and even, under suitable conventions, when $n=1$). – blargoner Dec 17 '22 at 17:24