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Question 153430This question is from textbook Advanced Mathematical Concepts
: I need help in completing the problem. I need help on starting the problem after I set it up as an augmented matrix. It would be great if I could get it step-by-step: 2x+6y+8z=5; -2x+9y-12z=-1; 4x+6y-4z=3. Thanks a lot!
This question is from textbook Advanced Mathematical Concepts
Found 2 solutions by stanbon, Edwin McCravy:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 2x+6y+8z =5
-2x+9y-12z=-1
4x+6y-4z =3
--------------------
+2...+6...+8....+5
-2...+9...-12...-1
+4...+6...-4....+3
-------------------------
Add 1st to 2nd to get:
+2...+6...+8....+5
+0...+15..-4....+4
+4...+6...-4....+3
------------------------
Add -2 times the 1st to the 3rd to get:
+2...+6...+8....+5
+0...+15..-4....+4
+0...-6..-20....-7
-----------------------
Add 3rd to 1st to get:
+2...+0...-12...-2
+0...+15..-4....+4
+0...-6..-20....-7
--------------------
Divide thru equation 3 by -6:
+2...+0...-12...-2
+0...+15..-4....+4
+0...+1..+10/3..+7/6
--------------------
Add -15 time 3rd to the 2nd to get:
+2...+0...-12...-2
+0...+0...-54...-13.5
+0...+1..+10/3..-7/6
------------------------
Solve the 2nd equation for z:
-54z = -13.5
z = 0.25 or 1/4
---------------------
Substitute z = 1/4 into the 3rd equation to solve for "y":
y +(10/3)(1/4) = -7/6
y + (5/6) = -7/6
y = -2
------------------
Substitute z = 1/4 into the 1st equation to solve for "x"
2x -12*(1/4) = -2
2x - 3 = -2
2x = 1
x = 1/2
----------------
Final solution:
x = 1/2
y = -2
z = 1/4
===============
Cheers,
Stan H.
Answer by Edwin McCravy(20060)  (Show Source):
You can put this solution on YOUR website! Stanbon's solution has a mistake.
Edwin's Solution:
[ 2 6 8 | 5]
[-2 9 -12 | -1]
[ 4 6 -4 | 3]
We must get 0's where the -2,
the 4, and the 6 are now. We
make three 0's in the lower
left hand corner
The top row will never change.
The other two will. To get a 0
where the -2 is, We multiply 1
times the first row and add it
to 1 times the 2nd row. It's a
good idea to put what you're
going to multiply a row by out
to the left of the row, even
when it is just 1, like this:
1[ 2 6 8 | 5]
1[-2 9 -12 | -1]
[ 4 6 -4 | 3]
The next matrix is then
[ 2 6 8 | 5]
[ 0 15 -4 | 4]
[ 4 6 -4 | 3]
To get a 0 where the 4 is,
We multiply -2 times the first
row and add it to 1 times the
3rd row.
-2[ 2 6 8 | 5]
[ 0 15 -4 | 4]
1[ 4 6 -4 | 3]
[ 2 6 8 | 5]
[ 0 15 -4 | 4]
[ 0 -6 -20 | -7]
To get a 0 where the -6 is,
we notice that the least common
multiple of 15 and 6 is 30. So,
We multiply 2 times the second
row and add it to 5 times the
3rd row.
[ 2 6 8 | 5]
2[ 0 15 -4 | 4]
5[ 0 -6 -20 | -7]
We get this:
[ 2 6 8| 5]
[ 0 15 -4| 4]
[ 0 0 -108|-27]
Now that we have three 0's in the
lower left corner, we go back to
a system of equations:
Start at the bottom, and go back up.
This is called back-substitution.
Start with the bottom equation, solve for
z:
Substitute  for  in the
second equation:
Finally sibstitute both for
and  for 
So the solution is
(x,y,z)=( ,,)
Edwin
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