# What is the determinant of an inversed Matrix where the matrix is an upper triangular matrix?

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)...

• Determinant of inverse is inverse of determinant, for any invertible matrix – J. W. Tanner Nov 17 '20 at 6:02
• 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? – swordlordswamplord Nov 17 '20 at 6:03
• The property is true for any invertible matrix, so it still applies with whatever other restriction you want to put on it. – Robert Israel Nov 17 '20 at 6:08
• the other comments appear to be telling me that the answer is 1... but it seems like u said.. the inverse of the determinant – swordlordswamplord Nov 19 '20 at 0:31

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 $$det(A^{-1})=det(A)^{-1}$$.

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

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.

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