This is an under-determined system because there are fewer equations
than there are variables.
2x - y + 3z = 5
-x + 4y + 4z = -1
Put in 1's
2x - 1y + 3z = 5
-1x + 4y + 4z = -1
Write the system as a matrix [by eliminating the letters, putting | for
equal signs and enclosing in brackrts]:
[ 2 -1 3 | 5]
[-1 4 4 | -1]
We want to use row operations
to make it look like this:
[ 1 0 # | #]
[ 0 1 # | #]
where there are numbers where the #'s are
but 1's and 0's as indicated:
[ 2 -1 3 | 5]
[-1 4 4 | -1]
Multiply the 2nd row by 2
[ 2 -1 3 | 5]
[-2 8 8 | -2]
Add the 1st row to the 2nd row to
get a 0 in the lower left corner:
[ 2 -1 3 | 5]
[ 0 7 11 | 3]
Multiply the 1st row by 7
[14 -7 21 |35]
[ 0 7 11 | 3]
Add the 2nd row to the 1st row to
get a 0 where the -7 is:
[14 0 32 |38]
[ 0 7 11 | 3]
Get a 1 where the 14 is by dividing the
1st row through by 14
Get a 1 where the 7 is by dividing the
2nd row through by 7
[ 1 0 | ]
[ 0 1 | ]
Reduce the fractions:
[ 1 0 | ]
[ 0 1 | ]
Write the matrix as a system of equations by putting
the letters and the equal signs back in:
1x + 0y + z =
0x + 1y + z =
Simplifying:
x + z =
y + z =
Solving for x and y
x = - z
y = - z
Now you can choose any arbitrary number for z to equal,
say z = a
x = - z
y = - z
z = a
And write the general solution this way, letting z = a:
x = - a
y = - a
z = a
Or as an ordered triple:
(x, y, z) = ( - a, - a, a)
That's the general solution:
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Then if you want some sample solutions, you can pick various
values for a. If you pick a= 0, you have this solution:
(x, y, z) = (, , 0)
which has fractions.
But by picking the right values for a=z you can get integer solutions.
Here are some particular solutions:
(x,y,z) = (5,2,-1)
(x,y,z) = (21,13,-8)
(x,y,z) = (37,24,-15)
(x,y,z) = (53,35,-22)
(x,y,z) = (69,46,-29)
(x,y,z) = (85,57,-36)
(x,y,z) = (101,68,-43)
(x,y,z) = (117,79,-50)
Edwin