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December 21, 2014

December 21, 2014

Posted by **YANETH** on Tuesday, November 1, 2011 at 6:34pm.

The sum of 3 numbers is 14

the largest is 4 times the smallest

the sum of the smallest and twice the largest is 18

what are the 3 numbers? show work?

A. how many unknowns you have?

-how many equations need to be solve for that many unknown?

B. what are the equations?

C. used the system of equations using matrices.

- College algebra -
**Damon**, Tuesday, November 1, 2011 at 6:56pmx+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

- College algebra -
**Henry**, Thursday, November 3, 2011 at 10:11pmpost it.

- College algebra -
**JandBsDad**, Friday, November 4, 2011 at 1:46ammatix 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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