# Linear algebra : calculating the determinant

1. Jan 22, 2009

### fluidistic

-1. The problem statement, all variables and given/known data
Let A and B be 2x2 matrices such that A²B=3I and A^T*B³=-I, where A^T is the transposed matrix of A, and I is the identity matrix.
Calculate det(A).
0. The attempt at a solution
I know I don't know how to approach correctly the problem, but I've tried something.
I know that the determinant of A squared times the determinant of B is equal to 9 and that minus the determinant of A times the determinant of B cubed is worth 1. From this, I can't reach the answer. And I've no idea about a different approach.
If you have an idea, I'd be glad to be its tester.

Last edited: Jan 22, 2009
2. Jan 22, 2009

### gabbagabbahey

What exactly is the question?

3. Jan 22, 2009

### fluidistic

Oops you're right. The question is to find out det (A).
I edit the first post to make it clearer.

4. Jan 22, 2009

### gabbagabbahey

Okay, in that case you are on the right track ; basically you have 2 equations:

$$(\det A)^2\det B=9$$ and $$(\det A)(\det B)^3=1$$

with two unknowns; $\det A$ and $\det B$

How would you normally go about solving a system of two (non-linear) equations and two unknowns? Try solving the first for $\det B$ and substituting it into the second.

5. Jan 22, 2009

### fluidistic

That's what I've tried to do, but I gave up because I made an error.
Well now you gave me confidence and I could reach a result : det (A)=$$\sqrt [4] {9^3}$$. Thank you very much!

6. Jan 22, 2009

### gabbagabbahey

Unfortunately you still have an error; you should get $$\det A =9^{3/5}$$

7. Jan 22, 2009

### fluidistic

Oh well, that's maybe a bit too late for me right now. I just saw my error! Unforgivable, I wrote that $$(det(A)^2)^3=det(A)^5$$ instead of $$det(A)^6$$.
I got it now. Thank you!