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Omnoculus
Posts:
2
From:
New York State
Registered:
2/18/10


A textbook problem misprint?
Posted:
Feb 18, 2010 5:42 PM


I have this textbook "Mathematics for 3D Game Programming & Computer Graphics" and Chapter on matrices starts from a linear algebra perspective. They have this example problem of a linear system that that go through and reduce to canonical form using an algorithm. I have been going over it again and again and I cannot come up with the same answer they get in the book for one operation in a particular element. The weird thing is the rest of the operations end up with the right solution for the system.
I'll repro the work here:
Given system: 3x + 2y  3z = 13 4x  3y + 6z = 7 x  z = 5
Augmented matrix:
 3 2 3  13   4 3 6  7   1 0 1  5 
First swap rows one and two and multiply new row 1 by 1/4:
 1 3/4 3/2  7/4   3 2 3  13   1 0 1  5 
Next make row 2 = 3 * (row 1) + row 2 and make row 3 = 1 * (row 1) + row 3 :
 1 3/4 3/2  7/4   0 17/4 15/2  73/4   0 3/4 5/2  27/4 
Multiply row 2 by 4/17:
 1 3/4 3/2  7/4   0 1 30/17  73/17   0 3/4 5/2  27/4 
Here's the step I have a problem with  2/3R2 + R1 > R1 and 3/4R2 + R3 > R3
 1 0 3/17  25/17   0 1 30/17  73/17   0 0 20/17  60/17 
My problem is with the 25/17 in element 1,4 The operation there should be ( 2/3 * 73/17 ) + 7/4 unless I am wrong. But then how could the answer be negative? The answer I get is 941/204 but using this does not ultimately solve the system. I'll show work below but the last steps with their data:
Multiply Row 3 by 17/20:
 1 0 3/17  25/17   0 1 30/17  73/17   0 0 1  3 
Finally 3/17R3 + R1 > R1 and 30/17R3 + R2 > R2 :
 1 0 0  2  x = 2  0 1 0  1  y = 1  0 0 1  3  z = 3
These values plugin to the equations correctly. So what am I missing? For 2/3R2 + R1 > R1 get the following:
(2/3 * 73/17) + 7/4 = 146/51 + 7/4 = 584/204 + 357/204 = 941/204
Any help is appreciated



