Download Systems of Linear Equations

Three Equations Containing Three Variables
a11x  a12 y  a13 z  b1
a21x  a22 y  a23 z  b2
a31x  a32 y  a33 z  b3
The first two cases are called consistent since there are
solutions. The last case is called inconsistent.
The solution will be one of three cases:
1. Exactly one solution, a point (x, y, z)
2. A dependent system with infinitely many solutions
3. No solution
With two equations and two variables, the graphs were
lines and the solution (if there was one) was where the
lines intersected. Graphs of three variable equations are
planes. Let’s look at different possibilities. Remember the
solution would be where all three planes all intersect.
Planes intersect at a point: consistent with
one solution
Planes intersect in a line: consistent
system called dependent with an infinite
number of solutions
Three parallel planes: no intersection so
system called inconsistent with no solution
No common intersection of all three
planes: inconsistent with no solution
We will be doing a “monster” elimination. (This just means
it’s like elimination that you learned with two equations and
variables but it’s now “monster” because the problems are
bigger and meaner and uglier).
2x  y  z  4
 3 x  2 y  2 z  10
x  2 y  3z  7
coefficient is a 1 here so will easy to work with
Your first strategy would be to choose one equation to
keep that has all 3 variables, but then use that equation to
“eliminate” a variable out of the other two. I’m going to
choose the last equation to “keep” because it has just x.
2x  y  z  4
 3 x  2 y  2 z  10
x  2 y  3z  7
5 here
y  5for
z
10
keep over
later
use
x  2 y  3z  7
Now use the third equation multiplied through by whatever it
takes to eliminate the x term from the first equation and add
these two equations together.
In this case when added to eliminate the x’s you’d need a −2.
 2( x  2 y  3z )  7  2
 2 x  4 y  6 z  14
2x  y  z  4
put this equation up with
the other one we kept
5 y  5 z  10
2x  y  z  4
 3 x  2 y  2 z  10
x  2 y  3z  7
5 y  5 z  10
x  2 y  3z  7
Now use the third equation multiplied through by whatever
it takes to eliminate the x term from the second equation
and add these two equations together.
In this case when added to eliminate the x’s you’d need a 3.
3x  2 y  3z   7  3
3x  6 y  9 z  21
 3x  2 y  2 z  10
we won’t “keep” this equation, but
we’ll use it together with the one we
“kept” with y and z in it to eliminate
the y’s.
 4 y  7 z  11
2x  y  z  4
 3 x  2 y  2 z  10
x  2 y  3z  7
5 here
y  5for
z
10
keep over
later
use
z 1
x  2 y  3z  7
So we’ll now eliminate y’s from the 2 equations in y and z
that we’ve obtained by multiplying the first by 4 and the
second by 5
5 y  5 z  10
 4 y  7 z  11
20 y  20 z  40
 20 y  35 z  55
we can add this to the one’s we’ve
kept up in the corner
15z  15
z 1
2x  y  z  4
 3 x  2 y  2 z  10
x  2 y  3z  7
x  2 y  3z  7
5 here
y  5for
z
10
keep over
later
use
z 1
Now we are ready to take the equations in the corner and
“back substitute” using the equation at the bottom and
substituting it into the equation above to find y.
5 y  51  10
5 y  5
y  1
Now we know both y and z we can sub
them in the first equation and find x
x  21  31  7
x5  7
x2
These planes intersect at a point,
namely the point (2 , −1 , 1).
The equations then have this
unique solution. This is the ONLY
x, y and z that make all 3
equations true.
The planes intersect in a point.
 x  2y  z  5
y  z  10
Let’s do another one:
2x  3y  z  0
 x  2y  z  5
3x  4 y  z  1
I’m
going
to “keep”
this onewe
since
If we
multiply
the equation
keptit
will be easy to use to eliminate x’s
by 3
2 and add it to the last
first equation
equation
from
others.
we can
eliminate x’s.
 2 x  4 y  2 z  10
2x  3y  z  0
we’ll keep y  z  10
this one
 3x  6 y  3z  15
3x  4 y  z  1
2 y  2 z  16
Now we’ll use the 2 equations we have with y and z to
eliminate the y’s.
The system is inconsistent,
no common intersection between the three planes.
2x  3y  z  0
 x  2y  z  5
3x  4 y  z  1
y  z  10
 x  2y  z  5
y  z  10
This means the equations are
inconsistent and have no solution.
The planes don’t have a common
intersection and there is not any
(x, y, z) that make all 3 equations true.
2 y  2 z  16
we’ll multiply the first
equation by -2 and add these
together
 2 y  2 z  20
2 y  2 z  16
0  4
Oops---we eliminated the y’s alright but the z’s ended
up being eliminated too and we got a false equation.
x  y  2z  1
 3one
y since
3z  it0
I’m going to “keep” this
Let’s do another one:
x  y  2z  1
2x  y  z  2
4 x  y  5z  4
will be easy to use to eliminate x’s
If we
multiply
thethe
equation
wewe
kept
If we
multiply
equation
kept
from
others.
by by
−4 −2
and
add
it to
thethe
last
equation
and
add
it to
second
weequation
can eliminate
x’s.
we can
eliminate x’s.
 2 x  2 y  4 z  2
2x  y  z  2
we’ll keep  3 y  3 z  0
this one
 4 x  4 y  8 z  4
4 x  y  5z  4
 3 y  3z  0
Now we’ll use the 2 equations we have with y and z to
eliminate the y’s.
x  y  2z  1
 3 y  3z  0
x  y  2z  1
2x  y  z  2
4 x  y  5z  4
 3 y  3z  0
 3 y  3z  0
This means the equations are
consistent and have infinitely many
solutions. The planes intersect in a
line. To find points on the line we can
solve the 2 equations we saved for x
and y in terms of z.
multiply the first equation by −1
and add the equations to
eliminate y.
3 y  3z  0
 3 y  3z  0
00
Oops---we eliminated the y’s alright but the z’s ended up
being eliminated too but this time we got a true equation.
x  y  2z  1
2x  y  z  2
4 x  y  5z  4
x  y  2z  1
 3 y  3z  0
z t
First we just put z = t since it can be any real number.
Now solve for y in terms of z.
Now sub it −t for y in first equation and solve for x in
terms of t.
The solution is (1 − t , −t , t) where t is any real number.
For example: Let z be 1. Then (0 , −1 , 1) would be a solution.
Notice is works in all 3 equations.
But so would the point
you get when z = 2 or 3
any other real
 3 y  3z  0
x  z  2z  1 or
number so there are
infinitely many
x  1 t
y  t
solutions.
The three planes intersect in the line.