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

  • Thread starter Thread starter synkk
  • Start date Start date
  • Tags Tags
    Matrices
AI Thread Summary
The discussion centers on the confusion regarding the determinant of a matrix P, specifically why it can be both +3 and -3. The key point is that the notation |P| represents the determinant of matrix P, not its absolute value, leading to the conclusion that if |det(P)| = 3, then det(P) can indeed be ±3. Participants clarify that the distinction between uppercase P (matrix) and lowercase p (number) is crucial for understanding the problem. An example is provided to illustrate that equations involving absolute values yield two solutions, reinforcing the concept. The conversation concludes with a resolution of the initial confusion, affirming the correct interpretation of the determinant.
synkk
Messages
216
Reaction score
0
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?
 
Physics news on Phys.org
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.
 
Back
Top