|
Question 48491This question is from textbook College Algebra
: The system of linear equations has a unique solution. Find the solution using Gaussian elimination or Gauss-Jordan elimination.
x + y + 6z = 3
x + y + 3z = 3
x + 2y+ 4z = 7
I know this has to do with combining rows and eliminating variables until only one is left, then back-substituting into the system to find the remaining variables, but I honestly don't know where to begin the elimination.
Thank you so very much for all your help!
This question is from textbook College Algebra
Answer by AnlytcPhil(1806)   (Show Source):
You can put this solution on YOUR website! x + y + 6z = 3
x + y + 3z = 3
x + 2y + 4z = 7
I will assume you have had matrices. If not,
post again and I will show you how to solve
it without matrices:
The augmented matrix is
[1 1 6 | 3]
[1 1 3 | 3]
[1 2 4 | 7]
We need to get 0's in the three lower left
positions, that is, in the positions below
the upper left to lower-right diagonal:
To get a 0 where the 1 is in row 2 column 1 is,
we multiply row 1 temporarily by -1 and add it
to 1 times row 2. This is easy to do mentally
if you write -1 to the left of row 1 and 1 left
of row 2:
-1[1 1 6 | 3]
1[1 1 3 | 3]
[1 2 4 | 7]
[1 1 6 | 3]
[0 0 -3 | 0]
[1 2 4 | 7]
To get a 0 where the 1 is in row 3 column 1 is,
we multiply row 1 temporarily by -1 and add it
to 1 times row 3. This is easy to do mentally
if you write -1 to the left of row 1 and 1 left
of row 3:
-1[1 1 6 | 3]
[0 0 -3 | 0]
1[1 2 4 | 7]
[1 1 6 | 3]
[0 0 -3 | 0]
[0 1 -2 | 4]
To get a 0 where the 1 is in row 3 column 2 is,
we merely need to swap rows 2 and 3
[1 1 6 | 3]
[0 1 -2 | 4]
[0 0 -3 | 0]
Now we need to get 1's on the upper left to
lower right diagonal. Two of the diagonal
elements are already 1's. To get a 1 where
the -3 is, we multiply row 3 by -1/3
[1 1 6 | 3]
[0 1 -2 | 4]
-1/3[0 0 -3 | 0]
[1 1 6 | 3]
[0 1 -2 | 4]
[0 0 1 | 0]
This is the system:
1x + 1y + 6z = 3
0x + 1y - 2z = 4
0x + 0y + 1z = 0
or
x + y + 6z = 3
y - 2z = 4
z = 0
The bottom equation tells us that
z = 0
Substitute z = 0 into the 2nd equation
x + y + 6z = 3
y - 2z = 4
y - 2·0 = 4
y = 4
Substitute z = 0 and y = 4 into the 1st
equation
x + 4 + 6·0 = 3
x + 4 + 0 = 3
x = -1
So the solution is (x, y, z) = (-1, 4, 0)
Edwin
|
|
|
| |
