What is the Determinant of an Invertible 3x3 Matrix?

lubricarret
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Homework Statement



Let A be an invertible 3x3 matrix. Suppose it is known that:
A =
[u v w
3 3 -2
x y z]
and that adj(A) =
[a 3 b
-1 1 2
c -2 d]
Find det(A)

(answer without any unknown variables)

Homework Equations



The Attempt at a Solution



I found A^(-1) to be equal to
(1/det(A)) * adj(A)

So, then re-arranging the formula;
I*det(A) = A*adj(A)
So then det(A) =
[u v w
3 3 -2
x y z]
*
[a 3 b
-1 1 2
c -2 d]

I know the solution to this problem is det(A) = 16. Therefore it must be that
[3 3 -2] * [3 1 2] = det(A)

But, what I am confused about is:
Why is the det equal only to the position (2,2) of the matrix A*adj(A)? As the solution is taken as the position a_(2,2)...

Thanks!
 
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If I*det(A)=A*adj(A) then you can evaluate any diagonal element to figure out det(A). a_(2,2) just happens to be the one you can figure out that doesn't have any unknowns in it. No big mysteries here.
 
Thanks again Dick for clearing this up for me!
 
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