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

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hi! still need help with this ?
yes please!
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

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Other answers:

so what about \(|2AA^T|=....?\)
where |...| is for determinant...
don't get it
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..
any more hints? I'm really bad at it:(
try to understand that step step are very simple, only this first step is bummer
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 ?
yea this part I get. I don't get why the whole thing equals 8
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 ?
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 ?
does det(A) = +1 or -1 ?
ohh I get it now! thanks! I got stuck in another T/F question if you don't mind helping as well?
sure :)

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