Why Can det|P| Be Both +3 and -3?

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Homework Help Overview

The discussion revolves around the properties of determinants in linear algebra, specifically addressing why the determinant of a matrix P can yield both positive and negative values when considering its absolute value. Participants are exploring the implications of the notation used in the context of determinants.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the distinction between the matrix P and the scalar p, as well as the meaning of the notation |P| versus det(P). There is confusion regarding the interpretation of absolute values in the context of determinants.

Discussion Status

Several participants are actively engaging in clarifying the notation and the underlying concepts. There is an ongoing exploration of the relationship between the determinant and its absolute value, with some participants providing examples to illustrate their points. The discussion reflects a mix of understanding and confusion, with no clear consensus reached yet.

Contextual Notes

Participants note the significance of uppercase and lowercase letters in mathematical notation, which may be contributing to misunderstandings. There is also mention of a potential error in a textbook regarding the notation used for determinants.

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For part b:
Could anyone why it is + or - 3? I really don't understand why there would be two solutions as det|P| as it would just be the absolute value of P, meaning just +ve?
 
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synkk said:
2uonakz.png


ir8f1w.png

For part b:
Could anyone why it is + or - 3? I really don't understand why there would be two solutions as det|P| as it would just be the absolute value of P, meaning just +ve?
It's "±" because your equation has the absolute value of the determinant of p .
 
synkk said:
2uonakz.png


ir8f1w.png

For part b:
Could anyone why it is + or - 3? I really don't understand why there would be two solutions as det|P| as it would just be the absolute value of P, meaning just +ve?
You seem to be confusing P (the matrix) and p (a number). Uppercase and lowercase are significant here, and you seem to be ignoring the difference.

The notation |P| doesn't mean "absolute value" of P; it is the determinant of P, also written as det(P).

What you show as |det p| makes no sense, because you're not taking the deteriminant of the number p - you want the determinant of the matrix P.
 
SammyS said:
It's "±" because your equation has the absolute value of the determinant of p .

Why?


Mark44 said:
You seem to be confusing P (the matrix) and p (a number). Uppercase and lowercase are significant here, and you seem to be ignoring the difference.

The notation |P| doesn't mean "absolute value" of P; it is the determinant of P, also written as det(P).

What you show as |det p| makes no sense, because you're not taking the deteriminant of the number p - you want the determinant of the matrix P.

What is shown in the OP is what the book has shown, not me.

Thanks for your input so far.
 
synkk said:
What is shown in the OP is what the book has shown, not me.
OK, then that's an error in the book. Apparently the author got confused between P and p.

What is written as |det p| should be written as |det P| or |det(P)|.
Since |det (P)| = 3, then det(P) = ± 3.
synkk said:
Thanks for your input so far.
 
SammyS said:
It's "±" because your equation has the absolute value of the determinant of p .

synkk said:
Why?

The solution to

4|x|=12

is

x = ±3 .

Do you agree?
 
SammyS said:
The solution to

4|x|=12

is

x = ±3 .

Do you agree?

No I've never learned this before, could you explain it please?
 
synkk said:
No I've never learned this before, could you explain it please?

the solution to 4|x|=12 is x=±3 BECAUSE if you plug x back in it works for both 3 and -3 ie. if you only list one of these as a solution then you have not completely solved the equation.

Therefore if you have 4|det P|=12 then |det P| = 3 meaning that det P can be either 3 or -3

make sense?
 
The solutions to |y| = k, where k > 0 are y = k or y = -k.
 
  • #10
d2j2003 said:
the solution to 4|x|=12 is x=±3 BECAUSE if you plug x back in it works for both 3 and -3 ie. if you only list one of these as a solution then you have not completely solved the equation.

Therefore if you have 4|det P|=12 then |det P| = 3 meaning that det P can be either 3 or -3

make sense?

yes, thank you and to everyone else.
 

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