3x3 Matrix (Lorentz force law)- determinant

Using the Lorentz force law for my problem, I get the following matrix.

I ex ey ez I

I 1 . -1 . 0 I

I 1 . 0 . -1 I

I have evaluated the ey term as a -ve ( ey((1X-1)-(0x1)) = ey(-1-0) = -1ey)

= ex((-1x-1)-(0x0)) + ey((1X-1)-(0x1)) + ez ((1x0)-(-1x1))

= e_{x} - e_{y} + e_{z }Another person evaluates this term as +ve. so could someone point out the obvious mistake I made, please, I can't see it.

The set problem is

An electron in a magnetic field **B**=2.0T(**e**_{x}-**e**_{z}) has velocity v=(2.5x10^{7} ms-1 (**e**_{x}-**e**_{y})

a) calculate magnetic force on electron at that instant

Now, F = -e 2(2.5x10^{7}) (**e**_{x}-**e**_{z})* (**e**_{x}-**e**_{y})

so

F = -8x10-12 (e_{x} - e_{y} + e_{z})

Re: 3x3 Matrix (Lorentz force law)- determinant

You have done the cross product incorrectly. The j coefficient is the **negative** of the minor.

Re: 3x3 Matrix (Lorentz force law)- determinant

AH! . . . . . I WAS missing something after all.

Ta!

Re: 3x3 Matrix (Lorentz force law)- determinant

The cross product of vectors <a, b, c> and <p, q, r> is

$\displaystyle \left|\begin{array}{ccc}i & j & k \\ a & b & c \\ p & q & r\end{array}\right|$

expanding that on the first row it is

(br- cq)i**-** (ar- cp)j+ (aq- bp)k.