# Homework Help: Matrices simple question

1. Jun 5, 2012

### synkk

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?

2. Jun 5, 2012

### SammyS

Staff Emeritus
It's "±" because your equation has the absolute value of the determinant of p .

3. Jun 5, 2012

### Staff: Mentor

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.

4. Jun 5, 2012

### synkk

Why?

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

Thanks for your input so far.

5. Jun 5, 2012

### Staff: Mentor

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.

6. Jun 5, 2012

### SammyS

Staff Emeritus
The solution to

4|x|=12

is

x = ±3 .

Do you agree?

7. Jun 5, 2012

### synkk

No i've never learnt this before, could you explain it please?

8. Jun 5, 2012

### d2j2003

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?

9. Jun 5, 2012

### Staff: Mentor

The solutions to |y| = k, where k > 0 are y = k or y = -k.

10. Jun 6, 2012

### synkk

yes, thank you and to everyone else.