Multiplying Determinants Practice Problems

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SUMMARY

This discussion focuses on calculating determinants of 3x3 matrices A and B, where det A = -2 and det B = 1. The solutions to the problems posed include det(A^6) = 64, det(B^(-1)A^3B^3A^T) = 16, det(4(A^T)^2(B^(-1))^4) = 256, and det((2B^T)^(-1)) = 1/8. Key insights include the application of the determinant properties, specifically that det(cA) = c^n * det(A), which clarifies the confusion around the multiplicative constants in the determinant calculations.

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  • Understanding of matrix determinants
  • Familiarity with properties of determinants, including det(AB) = det(A)det(B)
  • Knowledge of matrix transposition and its effect on determinants
  • Basic algebraic manipulation skills
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  • Study the properties of determinants in linear algebra
  • Learn about the effects of scalar multiplication on determinants
  • Explore the concept of matrix inverses and their determinants
  • Practice additional determinant problems involving 3x3 matrices
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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.
 

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