Multiplying Determinants Practice Problems

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Homework Statement


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)

Homework Equations





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 doesn't 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.
 
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bowlbase said:

Homework Statement


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)

Homework Equations





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 doesn't 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.

If A is an nxn matrix and c is a constant, then det(cA)=c^n*det(A).
 
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Thank you so much! I guess we hadn't covered that yet.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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