magnum
 
Reply Mon 3 May, 2004 07:01 am
i have this equations ---->

x=4z
x=3y
y=z+4

i want to know how you can solve this system with a matrix????

note: it's solvable without a matrix but in the answers of x,y,z i'm not interested
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Type: Discussion • Score: 1 • Views: 831 • Replies: 13
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magnum
 
  1  
Reply Mon 3 May, 2004 02:32 pm
isn't there anybody on this topic who can help me?????
0 Replies
 
BoGoWo
 
  1  
Reply Mon 3 May, 2004 02:43 pm
x = 48,
y = 16,
z = 12,

what's a matrix?
0 Replies
 
magnum
 
  1  
Reply Tue 4 May, 2004 02:13 am
matrix is a system wich helps you solve equations with multiple unknown variables. The equations above were also solvable without a matrix but if it gets more complicate then a matrix would really help you finding an answer
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magnum
 
  1  
Reply Tue 4 May, 2004 02:17 am
i don't think you can solve these equations without a matrix

x- y- z = 2
3x- 3y+ 2z = 16
2x- y+ z = 9
0 Replies
 
satt fs
 
  1  
Reply Tue 4 May, 2004 03:47 am
x = 3
y = -1
z = 2
0 Replies
 
magnum
 
  1  
Reply Tue 4 May, 2004 04:06 am
ok

but how did you come to this answer??

like i said before i am not interested in answer only in the way of generating the answers
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satt fs
 
  1  
Reply Tue 4 May, 2004 04:17 am
Sorry, but a matrix is not displayed neatly for a casual handling on this page.

(If a size of a matrix is small as in the case of 3x3, then calculating the inverse is good for an exercise. Otherwise, there are multiple algorithms for a linear equation.)
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magnum
 
  1  
Reply Tue 4 May, 2004 04:20 am
ok and you used a matrix to solve this or did you use another way to handle this problem???
0 Replies
 
satt fs
 
  1  
Reply Tue 4 May, 2004 04:32 am
I used a matrix.
(You can offer an equation of a larger size. For equations of higher ranks, I would use a Java program.)
0 Replies
 
Brandon9000
 
  1  
Reply Tue 4 May, 2004 07:49 am
magnum wrote:
i don't think you can solve these equations without a matrix

x- y- z = 2
3x- 3y+ 2z = 16
2x- y+ z = 9

You could use a matrix, but you can always solve n equations in n unknowns algebraically. In this case, n = 3.

For instance, from the first of these equations, x = 2 + y + z.
Substituting into the second equation:
3(2+y+z) +3y +2z = 16
6 +3y +3z +3y +2z = 16
6 + 6y + 5z = 16
6y + 5z = 10
6y = 10 - 5z
y = 5/3 - (5/6)z

And remember you already have x = 2 + y + z = 2 + [5/3 - (5/6)z] + z = 11/3 + z/6

So, you have x and y both in terms of z, and you can substitute those formulas into the 3rd equation.
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magnum
 
  1  
Reply Fri 7 May, 2004 02:04 pm
well ok but if you have like 8 variables with 8 equations you cannot easily solve algebraically. That's why a matrix is handy but i still not know how it works Confused
0 Replies
 
Brandon9000
 
  1  
Reply Fri 7 May, 2004 02:09 pm
magnum wrote:
well ok but if you have like 8 variables with 8 equations you cannot easily solve algebraically. That's why a matrix is handy but i still not know how it works Confused

True. I just don't remember the algorithm, but I'm sure you can find it in some math preparation book, like a Schaum's outline, in your local book store. It would probably be in some book titled "College Mathematics," or "Algebra," or "Pre-calculus."
0 Replies
 
fachatta
 
  1  
Reply Fri 28 May, 2004 03:12 pm
the equations can be written as:

x + 0Y -4Z = 0
x - 3Y + 0Z = 0
0X + y - Z = 4

As a matrix:

1 0 -4 0
1 -3 0 0
0 1 -1 4

subtract the first row from the second and the matrix becomes

1 0 -4 0
0 -3 4 0
0 1 -1 4

divide the second row by -3

1 0 -4 0
0 1 -4/3 0
0 1 -1 4

subtract the second row from the third row

1 0 -4 0
0 1 -4/3 0
0 0 1/3 4

multiply the third row by 3

1 0 -4 0
0 1 -4/3 0
0 0 1 12

add 4/3 times row 3 to row 2

1 0 -4 0
0 1 0 16
0 0 1 12

add 4 times row 3 to row 1

1 0 0 48
0 1 0 16
0 0 1 12

the matrix is now fully reduced, and the answer is

x = 48
y = 16
z = 12

this is the only solution to this problem
0 Replies
 
 

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