Expert: Steve Holleran - 12/28/2007

Question
QUESTION: 1.can you please solve for y using cramer's rule
3x-2y-4z=1
2x+5y-2z=-3
5x+3y-3z=10

2. can you find the sum of this infinite geometric series-

 ͚
 Ó (2/3)^k
k-1

ANSWER: Hi Alisa,

1.  Okay, to use cramer's rule to solve for y, you want to use the following setup:

         y = D(y) / D   

where D is the determinant of the x-y-z coefficients:

         3    -2    -4

 D =     2     5     -2

         5     3     -3


and D(y) is the same except with the constants on the right side of the = signs in the column for the y's:



         3    1    -4

 D(y)=    2    -3    -2

         5    10    -3

I don't know how you've been taught to evaluate these, but I do them by what's called the "basket" method, and its really hard to describe--it really has to be drawn, and I don't know how to do that here.

Anyway, it comes out like this:

D(y) = (27 - 10 -80) - (6- - 60 - 6) = -57

D =    (-45 + 20 - 24) - (-100 + 12 - 18) = 57

so then y = -1.


2.  I can't make out the symbols here--maybe you could write out what the series is?

Steve

---------- FOLLOW-UP ----------

QUESTION: i'm going to try to explain the second quest. without the symbols-

sigma with an infinity symbol above, a k-1 below, and times (2/3)^k

Answer
Hi Alisa,

I still think something's wrong here.  Are you SURE that below the sigma it doesn't have k = 1?

Then you would have the sum, k going from 1 to inf, of (2/3)^k.  This is a geometric series with a common ratio of 2/3, so using the formula to sum an inf geom series

 S(inf) = a / (1-r)  where a is the first term and r is the common ratio, you have :

    S(inf) = (2/3)/[1-2/3] = (2/3) / (1/3) = 2.


I hope this is the way its supposed to be.
Let me know if not.

Steve