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maggieL

  • one year ago

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

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  1. hartnn
    • one year ago
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    hi! still need help with this ?

  2. maggieL
    • one year ago
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    yes please!

  3. hartnn
    • one year ago
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    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
    • one year ago
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    so what about \(|2AA^T|=....?\)

  5. hartnn
    • one year ago
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    where |...| is for determinant...

  6. maggieL
    • one year ago
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    errr......

  7. maggieL
    • one year ago
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    don't get it

  8. hartnn
    • one year ago
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    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
    • one year ago
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    any more hints? I'm really bad at it:(

  10. hartnn
    • one year ago
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    try to understand that step first...next step are very simple, only this first step is bummer

  11. hartnn
    • one year ago
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    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
    • one year ago
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    yea this part I get. I don't get why the whole thing equals 8

  13. hartnn
    • one year ago
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    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
    • one year ago
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    yepp

  15. hartnn
    • one year ago
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    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
    • one year ago
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    does det(A) = +1 or -1 ?

  17. maggieL
    • one year ago
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    ohh I get it now! thanks! I got stuck in another T/F question if you don't mind helping as well?

  18. hartnn
    • one year ago
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    sure :)

  19. maggieL
    • one year ago
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    posted!

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