Posted by YANETH on Tuesday, November 1, 2011 at 6:34pm.
x+y+z=14
z=4x
x+2z=18
=================
x+y+4x = 14 or 5x+y = 14
x+8x = 18
or x = 2
so
y = 14 - 10 = 4
z = 4*2=8
3 unknowns, three equations
I did it with substitution, however can form augmented matrix and use Gauss Jordan:
+1 1 1 14
-4 0 1 0
+1 0 2 18
then plug and chug
post it.
matix solution
|1 1 1 | [x y z]'= [14 0 18]'
|4 0 1 |
|1 0 2 |
H*[x y z]' = [14 0 18]' (' stands for transpose; in this case makes row vector a column vector)
Answer
inv(H)*H*[x y z]'= inv(H)*[14 0 18]'
[x y z] = inv(H)*[14 0 18]'
inverting a 3x3 matrix
inv(H)=1/|H|*adj(H) |H|=det(H)
adj(H)=
|+(0-0) -(-8-1) +(0-0)|'
|-(2-0) -(2-1) -(0-1)|
|+(1-0) -(1+4) +(0+4)|
|0 9 0|'
|-2 -1 1|
|1 -5 4|
=|0 -2 1|
|9 -1 -5|
|0 1 4|
det(H)= use the second column
since it has 2 zeros it makes
it easy.
det(H)=-1(-8-1)+0*(2-1)-0*(1+4)=9
inv(H)=1/9*|0 -2 1|
|9 -1 -5|
|0 1 4|
[x y z]'= inv(H)*[14 0 18]'
=[18/9 (9*14-5*18)/9 (18*4)/9]'
=[ 2 4 8 ]'
hope this helps
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