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Determinant question

  1. Jan 21, 2009 #1
    i am given 2 matrices

    A=
    a1 b1 c1
    a2 b2 c1
    a3 b3 c3

    B=
    -c2 3c1 -c3
    b2 -3b1 b3
    -5a2 15a1 -5a3


    and also given is: 5detA+detB=10

    what i need to fin is det(3A2B-1)

    what i did to help me was

    det(3)=27
    det(A2)=detA*detA
    det(B-1)=1/detB

    i see that if i perform 2 "swaps" on B, once between R1 and R3 , and then between C1 and C2, these actions wont change detB.
    now i can divide my new R1 by 5, and my new c1 by 3 and transpose B to get a new B* , detB=(5*3)detB*

    now my new matrix B* is ALMOST identical to A, except for the (-) signs before some of its numbers, which are drivin me mad,
    how can i get rid of them, if i can get rid of them i can find detA.

    any ideas? have all my steps beeen legal??
     
  2. jcsd
  3. Jan 21, 2009 #2
    a1 -a2 -a3
    -b1 b2 b3
    c1 -c2 -c3

    Find "patterns" in the negative signs so you can multiply certain rows or columns by -1 to get rid of them.
     
  4. Jan 21, 2009 #3
    thanks, got it.. so...
    what i had was

    B*=
    a1 -b1 c1
    -a2 b2 -c2
    -a3 b3 -c3

    so i'll multiply det(B*)*(-1)*(-1)*(-1)=detA
    so detB* =-detA

    detB=(5*3)detB*
    detB=-15detA


    5detA+detB=10
    5detA-15detA=10
    detA=-1

    now to solve the question
    det(3A^2B^-1)=27*(-1)*(-1)*(1/-15)
    =-27/15

    does this all look right??
     
  5. Jan 22, 2009 #4
    Right up until to the very end! Since det(A)=-1 and det(B) = -15det(A), we have det(B) = +15.

    Apart from that - well done, Dell!
     
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