## maggieL 2 years ago 1) if A is a 3x3 matrix satisfying det(2AA^T)=8, then detA=±1 T or F?

1. hartnn

hi! still need help with this ?

2. maggieL

3. hartnn

ok, we use the property that if there's a nXn matrix B, \(|aB|= a^n|B|\) which means if we take out the constant, it will be raised to n'th power

4. hartnn

5. hartnn

where |...| is for determinant...

6. maggieL

errr......

7. maggieL

don't get it

8. hartnn

here, since A is 3X3, n= 3 \(|2AA^T|=2^3|AA^T|\) from the property i mentioned....try to get this and ask doubts if any..

9. maggieL

any more hints? I'm really bad at it:(

10. hartnn

try to understand that step first...next step are very simple, only this first step is bummer

11. hartnn

when i take constant out of the determinant (here, constant = 2) it gets raised to the power of 'n' where a matrix is of order nXn (here n=3) thats how we get 2^3 outside got this ?

12. maggieL

yea this part I get. I don't get why the whole thing equals 8

13. hartnn

that whole thing =8 is GIVEN part. we need to find whether det (A) = +/- 1 or not lets go to next steps, since \(|XY|=|X||Y|\) we will have \(2^3|AA^T|=8|A||A^T|\) got this step too ?

14. maggieL

yepp

15. hartnn

lastly , determinant of any matrix is same as determinant of its transpose, \(so, |A|=|A^T|\) so, \(8|A||A|=8 \implies |A|^2=1\) what can you say about det(A)from here ?

16. hartnn

does det(A) = +1 or -1 ?

17. maggieL

ohh I get it now! thanks! I got stuck in another T/F question if you don't mind helping as well?

18. hartnn

sure :)

19. maggieL

posted!