# Multipying Determinants

1. Nov 1, 2013

### bowlbase

1. The problem statement, all variables and given/known data
This is a practice problem where the solutions are given.
Both are 3x3 matrices.

det A=-2 and det B=1
find the following:
1)det(A6)
2) det(B-1A3B3AT)
3) det(4(AT)2(B-1)4)
4) det((2BT)-1)

2. Relevant equations

3. The attempt at a solution
I get the first two easily enough, 64 and 16. However, I'm not getting anything near the 256 and 1/8 (answers in order).

I guess it is the multiples of 4 and 2 that are screwing me up. I know that det AT=det A and det A-1=det $\frac{1}{A}$

I thought maybe making two matrices such that the diagonals multiply to -2 and 1 respectively. So, 1*1*-2 and 1*1*1. Then multiplying each row by 4. Then: 4*4*-8 and 4*4*4.

So the new det would be A=128 and B=64. And then placing them back in the 3rd equation but this clearly doesnt work as 644 is much greater than 1282.

However, if I left it as 1282 divided by 64 then I get the correct answer. I don't know if this is a fluke or not but it seems to me that not taking 64 to the 4th power is counter to the equation.

thanks for the help.

2. Nov 1, 2013

### Dick

If A is an nxn matrix and c is a constant, then det(cA)=c^n*det(A).

3. Nov 1, 2013

### bowlbase

Thank you so much! I guess we hadn't covered that yet.