LetAandBbe square matrices of the same dimensions. Suppose that

det(AB2) = 1

det(AB) = 3

Show that

AandBare invertible and find their determinants.

I'm not even sure where to start with this!

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- Jan 14th 2011, 10:41 AMAritaAlgebraLet
*A*and*B*be square matrices of the same dimensions. Suppose that

det(*AB*2) = 1

det(*AB*) = 3

Show that

*A*and*B*are invertible and find their determinants.

I'm not even sure where to start with this!

- Jan 14th 2011, 11:02 AMpickslides
What is B2 here?

- Jan 14th 2011, 11:12 AMArita
The 2 is suppose to represent squared. So it would be det(AB(squared))=1

- Jan 14th 2011, 11:26 AMpickslides
Would it be helpful to say $\displaystyle \det(AB)= \det(A)\det(B)$ ?

- Jan 14th 2011, 11:29 AMHallsofIvy
Start with these facts:

1) det(AB)= det(A)det(B)

2) A is invertible if and only det(A) is not 0.

3) xy= 0 if and only if either x= 0 or y= 0. So that if xy is not 0 then neither x nor y is 0.

det(AB)= 3 which is not 0 so neither det(A) nor det(B)= 0 and both A and B are invertible.

$\displaystyle det(AB^2)= det(AB)det(B)$ and you are told both det(AB) and det(AB^2). - Jan 14th 2011, 11:34 AMArita
So far what I have is det(AB)=det(A)det(B)=3 and det(AB(squared))= det(A)det(B(squared))

(square root of 1) = det(A)(square root (det(B)))

1 = det(A)(square root (det(B)))

1 = 1/3(square root of 9)

1 = 1/3(3)

1 = 1

det(AB)= det(A)det(B)

3 = 1/3(9)

3 = 3

But I'm not sure if this is right? - Jan 14th 2011, 11:40 AMpickslides
Now you're on the right track! Hall's hint was the killer here!

$\displaystyle \det(AB^2)= \det(AB)\det(B)$

$\displaystyle 1= 3\det(B)$

$\displaystyle \det(B)= \frac{1}{3}$

now find $\displaystyle \det(A)$ given $\displaystyle \det(A)\det(B)=3$ - Jan 14th 2011, 11:48 AMArita
So det(AB^2) = det(AB)det(B)?

I thought it would be det(AB^2) = det(A)det(B)^2