# Determinant question

1. Jan 21, 2009

### Dell

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. Jan 21, 2009

### mutton

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.

3. Jan 21, 2009

### Dell

thanks, got it.. so...

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??

4. Jan 22, 2009

### Unco

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!