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Question 1181896: solve the following using gaussian elimination method 2x + y + 3z = 4 x + y + 2z = 0 2x + 4y + 6z = -8
Answer by greenestamps(13209)   (Show Source):
You can put this solution on YOUR website!
It is rather pointless to have someone do this for you, because (1) you don't learn how to do it yourself; and (2) 100 different people will use probably at least 90 different paths to get the answer.
Nevertheless, I will go through one POSSIBLE sequence of steps for solving this example, with explanations of why I chose that path. If you want to learn anything about how to do this, you should then solve the problem using a different path.
The initial matrix is this:
The matrix we want to end up with is this:
The matrix in that form will tell us that the solution to the system is x=A, y=B, and z=C.
At each step, there are three things you can do:
(a) switch the order of the rows;
(b) multiply a row by a nonzero constant;
(c) add a multiple of one row to another row
In terms of the objectives in choosing a path to the answer, the sequence of steps you want to accomplish is this:
(1) get a 1 in position (1,1)
(2) use the 1 in position (1,1) to get 0's in the rest of column 1
(3) get a 1 in position (2,2)
(4) use the 1 in position (2,2) to get 0's in the rest of column 2
(5) get a 1 in position (3,3)
Following, then, is ONE POSSIBLE path to the solution.
As a preliminary step, seeing that the numbers in row 3 are all multiples of 2, I would use (b) to make the numbers in row 3 smaller:
Step (1): getting a 1 in position (1,1)
I could of course use (b) to multiply row 1 by 1/2 -- but that would introduce fractions, which will surely complicate the process. Since rows 2 and 3 both have 1's in column 1, I can use (a) to accomplish step (1). I can rearrange the rows in any order I want; I chose this:
Step (2): getting 0's in positions (2,1) and (3,1)
I use (c) with the 1 in (1,1) to do this: multiply row 1 by -1 and add to row 2 to get the 0 in (2,1); multiply row 1 by -2 and add to row 3 to get the 0 in (3,1).
Step (3): getting a 1 in position (2,2)
By the path I have chosen, I already have that....
Step (4): getting 0's in positions (1,2) and (3,2)
Again use (c) with the 1 in position (2,2) to do this: multiply row 2 by -1 and add to row 1 to get the 0 in position (1,2); multiply row 2 by 1 and add to row 3 to get the 0 in position (3,2).
In this problem, we can't do step (5) -- getting a 1 in position (3,3) -- because row 3 is all 0's.
That row of all 0's means there is no unique solution to the system of equations; there are infinitely many solutions.
We have done all we can do to the matrix. Translating the rows of the matrix into equations, the matrix in the final form tells us this:
We use this to set up a parametric description of the infinite number of solutions to the system:
z=t
x+z=4 --> x = 4-z --> x = 4-t
y+z=-4 --> y = -4-z --> y = -4-t
The parametric solution to the system of equations is (x,y,z) = (4-t, -4,-t,t)
Substitute any number for t to get a solution to the system of equations:
t=0 --> (4,-4,0)
t=1 --> (3,-5,1)
t=-3 --> (7,-1,-3)
etc...
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