Question

Find the 3x3 matrix B such that:

B^6 = -I_3

It says I need to use the determinant anyone able to give me a hint on how to start this problem?

I realize:
|I| = 1,
thus,
|A^6| = -1^3|I| = -1

Since A^6 = -I it is safe to say A is invertible.
Not sure how that is helpful though.

Answer 1

\[B^6 = -I_3\]
sorry it should be this.

Answer 2

Absolutely no idea...so sorry :( Can you help with my algebra II problem? http://openstudy.com/study#/updates/53ae01d2e4b0a819ab143fea

Answer 3

Obviously, if \(B=I_3\), then \(B^6=I_3\) as well. Is there anything preventing you from using complex scalars? If not, you could use \(B=\sqrt{-1}I_3\), then \(B^6=-I_3\).

Answer 4

I cant use complex scalers and it is B^6 = -I not B^6 = I

Answer 5

Says I need to use the determinant to solve it :\

Answer 6

Right, I was just pointing out that raising the identity to any power yields the identity, *unless* you multiply the identity by \(i\).

Well, it was worth a shot. I haven't come across a problem like this before, and I can't seem to find any examples online addressing this sort of question. Sorry

Answer 7

A^6 = -I

|A| |A| |A| |A| |A| |A| = -I

Assuming, -I = A, which is the only option in regards to this question
|AA| |AA| |AA| = -I
|1| |1| |1| =/= -1

Answer 8

(-I)(-I) = (I)

Answer 9

I feel like that isn't sufficient proof though

Answer 10

sorry that should be,

1*1*1 =/= -1

Answer 11

I feel like even with a matrix with an alternating negative it will always come out positive.

Answer 12

when it is to an even power

Answer 13

This is 100% speculation on my part, but... could this have something to do with eigenvalues?
\[\begin{align*}B^6&=-I_3\\
B^6+I_3&=0\\
\text{det}(B^6+I_3)&=0\\
\text{det}(B^6-\lambda I_3)&=0&\text{with }\lambda=-1
\end{align*}\]
If there's any sense to what I'm saying, it would seem all you need to do is construct a matrix with \(-1\) as an eigenvalue, or *the* eigenvalue with multiplicity 3. So if you could find the entries, such that
\[|B^6-\lambda I|=\begin{vmatrix}
b_{1,1}-\lambda&b_{1,2}&b_{1,3}\\
b_{2,1}&b_{2,2}-\lambda&b_{2,3}\\
b_{3,1}&b_{3,2}&b_{3,3}-\lambda
\end{vmatrix}=0\]
for \(\lambda=-1\), you should (could?) be set.

Answer 14

That seems to make the most sense

Answer 15

Wouldnt you still end up with the matrix -I?

Answer 16

Like I said, what I'm suggesting is by no means foolproof. I'm not even sure one could argue for its validity. Just thought I'd try my hand at the problem :/

Answer 17

Well I do appreciate your help, this question kind of sucks :| I would just argue that a matrix A does not exist such that A^6 = -I

But aren't there weird matrices that turn into identity matrix after raised to a certain power

Answer 18

I would think so, but when it comes down to providing an example, no such matrix seems to exist :P

Answer 19

Actually, yes, I can think of one:
\[\begin{pmatrix}0&1\\1&0\end{pmatrix}\]
Square this and you get the identity.

I agree with what you said earlier: I don't see how raising anything to an even power would yield a negative result.

Answer 20

I'm pretty sure I can show that any matrix with a negative determinant, raised to an even power would have a positive determinant and since the determinant of A^6 = -1 then no matrix exists

Answer 21

I think that is my proof :/

Answer 22

On a whim, I plugged a generic matrix into WA:

http://www.wolframalpha.com/input/?i=MatrixPower%5B%7B%7Ba%2Cb%2Cc%7D%2C%7Bd%2Ce%2Cf%7D%2C%7Bg%2Ch%2Ci%7D%7D%2C6%5D

You have to be able to find constants \(a\) through \(f\) so that each entry in the matrix agrees with \(-I_3\). I just don't see anyone solving a 9 equation system like that manually.

Answer 23

If there is such a matrix \(B\), finding it must involve some trick I'm not aware of or haven't learned.

Answer 24

yeah I have no idea either, this is an introductory course so I dont know, it cant be that advanced.

Answer 25

For a sketch of your proof against such a \(B\), you could use
\[B^6=(B^2)^3\]
It might be simpler to show there's no way to get the negative identity when you square a matrix, and so there can't possibly be way to get the negative identity when you raise to the sixth power.

Answer 26

You dont think I can just use the determent?

Answer 27

to show this?

Answer 28

like I did above?

Answer 29

If you think it works I'll vouch for it :)

Answer 30

thanks for the help it is much appreciated

Answer 31

\[

\det(B^6)= [\det(B)]^6
\]Consider \[
\det( -I_3) = (-1)^3\det(I_3) = -1
\]If we simply tried to solve for the determinant \(b=\det(B)\), we get: \[
[\det(B)]^6 = b^6= -1
\]We have a complex determinant. We can't get a complex determinant by multiplying and adding real numbers (which is how the determinant is computed).