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Given that

$ U = \begin{bmatrix}1&4&6\\0&2&5\\0&0&3\end{bmatrix} $

find

$ det(U^{-1}) $

I just have one question.. would I get the answer if I simply inversed U and found its determinant? or do i need to first inverse the U matrix, get the determinant, and finally divide it by det(U)?

In any case, I looked up two different methods of inverting a 3-by-3 matrix..

  1. creating a matrix of cofactors and finding its determinants and dividing it by det(U) = 1/6

the second method:

  1. Row reduction but not dividing by det(U) = 1/6...

So in other words, I get 1/6 as the determinant of inverse U when I divide it by det(U) on the one hand, but also, I get 1/6 as determinant of inverse U when I do not divide it by det(U)...

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    $\begingroup$ Determinant of inverse is inverse of determinant, for any invertible matrix $\endgroup$ – J. W. Tanner Nov 17 '20 at 6:02
  • $\begingroup$ but what if the matrix is an upper triangular matrix? meaning the determinant is the product of the main diagonal entries... does that property still apply? $\endgroup$ – swordlordswamplord Nov 17 '20 at 6:03
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    $\begingroup$ The property is true for any invertible matrix, so it still applies with whatever other restriction you want to put on it. $\endgroup$ – Robert Israel Nov 17 '20 at 6:08
  • $\begingroup$ the other comments appear to be telling me that the answer is 1... but it seems like u said.. the inverse of the determinant $\endgroup$ – swordlordswamplord Nov 19 '20 at 0:31
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It can be shown that $det(AB)=det(A)det(B)$. In particular, if $A\cdot A^{-1}=I$, you get that $det(A\cdot A^{-1})=det(A)det(A^{-1})=det(I)=1$. Hence $det(A^{-1})=det(A)^{-1}$.

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$U = \begin{bmatrix}1&4&6\\0&2&5\\0&0&3\end{bmatrix}. \tag 1$

The determinant is a multiplicative function of its argument; thus

$\det(U) \det(U^{-1}) = \det(UU^{-1}) = \det(I) = 1; \tag 2$

since $U$ is upper triangular, its eigenvalues are $1$, $2$, and $3$, and we have

$\det(U) = 1 \cdot 2 \cdot 3 = 6; \tag3$

therefore in accord with (2)

$\det(U^{-1}) = (\det(U))^{-1} = \dfrac{1}{6}. \tag 4$

This is the same answer as is obtained by first computing $U^{-1}$ and taking the determinant of the result.

To proceed the "long way", and actually find $U^{-1}$ explicitly, one might follow the method outlined here.

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    $\begingroup$ @J. W. Tanner: thanks once again for the small edit, and damn the apostrophes! Cheers!!! $\endgroup$ – Robert Lewis Nov 17 '20 at 6:20

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