Question

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

Answer 1

hi! still need help with this ?

Answer 2

yes please!

Answer 3

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

Answer 4

so what about \(|2AA^T|=....?\)

Answer 5

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

Answer 6

errr......

Answer 7

don't get it

Answer 8

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..

Answer 9

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

Answer 10

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

Answer 11

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 ?

Answer 12

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

Answer 13

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 ?

Answer 14

yepp

Answer 15

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 ?

Answer 16

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

Answer 17

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

Answer 18

sure :)

Answer 19

posted!