# I must be making a simple mistake here.

• Mar 25th 2010, 07:33 AM
qwerty11
I must be making a simple mistake here.
Ok the problem is,
Given:

|r s t|
|u v w| = 4
|x y z|

Evaluate:

|-x -y -z|
|r s t|
|u v w|

My steps at finding the solution are:
1.
|r s t|
(-)|-x -y -z|
|u v w|

2.
|r s t|
(-)(-)|u v w|
|-x -y -z|

3.
|r s t|
(-)(-)(-1)|u v w|
|x y z|

4. (-)(-)(-1)(4)= -4

Which for some reason is wrong because I am supposed to show that it equals 4 not -4. What am I doing wrong?
• Mar 25th 2010, 08:34 AM
Tinyboss
It's two row swaps, and a row multiplication by -1, so your answer looks right to me.
• Mar 25th 2010, 09:10 AM
tonio
Quote:

Originally Posted by qwerty11
Ok the problem is,
Given:

|r s t|
|u v w| = 4
|x y z|

Evaluate:

|-x -y -z|
|r s t|
|u v w|

My steps at finding the solution are:
1.
|r s t|
(-)|-x -y -z|
|u v w|

2.
|r s t|
(-)(-)|u v w|
|-x -y -z|

3.
|r s t|
(-)(-)(-1)|u v w|
|x y z|

4. (-)(-)(-1)(4)= -4

Which for some reason is wrong because I am supposed to show that it equals 4 not -4. What am I doing wrong?

How did you get to $\begin{pmatrix}\!\!-x&\!\!-y&\!\!-z\\r&s&t\\u&v&w\end{pmatrix}$ from the original matrix ?

First, you interchanged the 2nd and 3rd rows, and thus the determinant is multiplied by -1, and then you interchanged the 1st and 2nd rows and again the det. gets multiplied by -1, so we still have the original determinant. But now you multiplied the first row by -1 and thus the det. of the new matrix equals $-4$ ...
Thus you're right: it is -4

Tonio