Expert: Paul Klarreich - 12/26/2007

Question
1. PLEASE SOLVE BY CRAMER'S RULE-
3X-2Y-4Z=1
2X+5Y-2Z=-3
5X+3Y-3Z=10

2. SOLVE FOR X-
2COS^2X-COSX=1

Answer
Questioner:   ALISA
Category:  Advanced Math
Private:  No
 
Subject:  PLEASE HELP
Question:  1. PLEASE SOLVE BY CRAMER'S RULE-
3X-2Y-4Z=1
2X+5Y-2Z=-3
5X+3Y-3Z=10

2. SOLVE FOR X-
2COS^2X-COSX=1
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1. PLEASE SOLVE BY CRAMER'S RULE-

>> Please, don't shout.

1. Cramer's rule gives you:
    3  -2  -4
D =  2   5  -2 [a determinant]
    5   3   -3

D = -45 +20 -24 +18 -12 +100 = -81 + 138 = 57

Now

    
     1  -2  -4
    -3   5  -2 [D with the x-column replaced by RHS]
    10   3   -3
x  = -------------
       D = 57

Same for y and z.  I'll leave the hackwork to you.  I assume you have already learned how to evaluate a determinant.

..........................................


2 Cos^2X-Cos X=1

Use this 'abbreviation trick':

Write C for cos x, just temporarily.  Now your equation is:
2C^2 - C = 1
2C^2 - C - 1 = 0

(2C + 1)(C - 1) = 0

Now 'disabbreviate':  {OK,OK, so I made up that word.]

cos x = -1/2  and cos x = 1

The first gives you a principal value of 120 deg, and another value of 240 deg, since cosine is negative in Quad II, III.

The second gives you x = 0.