I need major help :(

• Oct 3rd 2006, 07:38 PM
kgpretty
I need major help :(
Questions can be found here:
http://www.freewebs.com/kgpretty/math.JPG

I attempted #4.. can you use 4 .. determinants? to solve the equation like:
Code:

```x|k  l|  -a|l  m|  x|k  l|  -a|l  m|  |m  n|    |n  o|    |o  k|    |n  o|```
For #7, I've think I've got that covered. Just tell me.. would the augmented matrix be:
|log1 log1 log1|0|
to represent the first row?

or |1 1 1|0|, where log is factorised out?

For all the other questions.. I simply don't know what to do
• Oct 3rd 2006, 07:55 PM
ThePerfectHacker
Quote:

Originally Posted by kgpretty

I attempted #4..

No only 3.
Expand along the first row:
Code:

```x [x b]-a[a b]+b[a x]   [b x]  [a x]    [a b]  =0```
• Oct 3rd 2006, 08:20 PM
ThePerfectHacker
Do my excersice.
Show that the determinant below is always zero.
----
Once you have done that you can use the fact, x,y,z>0
log(2x)=log2 +log x
log(2y)=log2 +log y
log(2z)=log2 +log z
log(3x)=log3+log x
log(3y)=log3+log y
log(3z)=log3+log z
Let,
a=log x
b=log y
c=log z
Then you end with,
Code:

```a                b            c a+log 2  b+log 2  c+log 2 a+log 3  b+log 3  c+log 3```
Then use that result.
• Oct 4th 2006, 09:06 AM
kgpretty
Quote:

Originally Posted by ThePerfectHacker
No only 3.
Expand along the first row:
Code:

```x [x b]-a[a b]+b[a x]   [b x]  [a x]    [a b]  =0```

When I expand along the first row, the result isn't equal to 1.
From the first expansion, you get:
x^3 - b^2x

There's no x^3 in the other expansions to cancel the x^3 already there. How can I be equal to 0?
• Oct 4th 2006, 09:22 AM
ThePerfectHacker
Quote:

Originally Posted by kgpretty
When I expand along the first row, the result isn't equal to 1.
From the first expansion, you get:
x^3 - b^2x

There's no x^3 in the other expansions to cancel the x^3 already there. How can I be equal to 0?

Expand it each determinant.
x(x^2-b^2)-a(ax-ab)+b(ab-ax)=0
Thus,
x^3-xb^2-a^2x+a^2b+ab^2-abx=0
Thus,
x^3-x(b^2-ab+a^2)+a^2b+ab^2=0
• Oct 4th 2006, 05:31 PM
kgpretty
Quote:

Originally Posted by ThePerfectHacker
Expand it each determinant.
x(x^2-b^2)-a(ax-ab)+b(ab-ax)=0
Thus,
x^3-xb^2-a^2x+a^2b+ab^2-abx=0
Thus,
x^3-x(b^2-ab+a^2)+a^2b+ab^2=0

I'm sorry, I don't get it.
How does x^3-x(b^2-ab+a^2)+a^2b+ab^2 equal to 0?
• Oct 4th 2006, 05:37 PM
ThePerfectHacker
Quote:

Originally Posted by kgpretty
I'm sorry, I don't get it.
How does x^3-x(b^2-ab+a^2)+a^2b+ab^2 equal to 0?

Because the left hand side (determiant) was expaned into this equation. Now the problem says to solve when it is equal to zero.