Product of Two Terms: Determinant Evaluation and Vanishing Values in Terms of p

[Nylex]

Hi, I' not sure if I've done this question correctly so I just want someone to tell me where I've gone wrong (if I have).

Evaluate the following determinant as a product of two terms. Hence find, in terms of p the values of x for which it vanishes.

Grr, I can't seem to use LaTeX properly so I'll just "draw" the determinant:

x x 0
0 x p
-2x -p x

$$= x \left\vert \begin{array}{cc} x & p\\-p & x \end{array}\right\vert - x \left\vert \begin{array}{cc} 0 & p\\-2x & x \end{array}\right\vert$$

= x(x^2 + p^2) - x(2xp)
=x^3 + p^2.x - 2x^2.p
- x(x^2 - 2px + p^2) = 0

x = {-(-2p) +- [(-2p)^2 - 4.1.p^2]^1/2}/2

x = [2p +- (4p^2 - 4p^2)^1/2]/2

x = p

It doesn't look right to me :/.

 

[poolwin2001]

Hi, I' not sure if I've done this question correctly so I just want someone to tell me where I've gone wrong (if I have).

Evaluate the following determinant as a product of two terms. Hence find, in terms of p the values of x for which it vanishes.

Grr, I can't seem to use LaTeX properly so I'll just "draw" the determinant:

x x 0
0 x p
-2x -p x

$$= x \left\vert \begin{array}{cc} x & p\\-p & x \end{array}\right\vert - x \left\vert \begin{array}{cc} 0 & p\\-2x & x \end{array}\right\vert$$

= x(x^2 + p^2) - x(2xp)
=x^3 + p^2.x - 2x^2.p
- x(x^2 - 2px + p^2) = 0

x = {-(-2p) +- [(-2p)^2 - 4.1.p^2]^1/2}/2

x = [2p +- (4p^2 - 4p^2)^1/2]/2

x = p

It doesn't look right to me :/.

This is correct x(x-p)^2=0
==>x=0 or x=p

 

[Nylex]

This is correct x(x-p)^2=0
==>x=0 or x=p

LOL, can't believe I didn't see that . Thanks a lot .