Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
Introduction
Math 1038 Lect 1: Technique of solving
system of linear equations
System of linear
equations of two
unknowns
Substitution
Elimination
System of linear
equations of three
unknowns
Charles Li
The Chinese University of Hong Kong
Substitution
Elimination
More examples
System of linear
equations with
parameters (new)
Warning: The notes may be updated later. Check online
for the latest version.
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
Introduction
Last updated: September 3, 2022
System of linear
equations of two
unknowns
Substitution
Reference:
1. Strang : Linear Algebra and Its Applications Section
1.3
Elimination
System of linear
equations of three
unknowns
Substitution
Elimination
2. Beezer, A first course in Linear algebra. Ver 3.5
More examples
Downloadable at
System of linear
equations with
http://linear.ups.edu/download.html
parameters (new)
Print version can be downloaded at
http://linear.ups.edu/download/fcla-3.50-print.pdf
Beezer, Ver 3.5 Sect SSLE (print version p7 - p14)
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
Introduction
System of linear
equations of two
unknowns
Substitution
Elimination
Introduction
System of linear
equations of three
unknowns
Substitution
Elimination
More examples
System of linear
equations with
parameters (new)
In this section we give examples of systems of linear
equations and solve one example.
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
Example 1
Solve the system of equations
2x1 + 3x2 = 3
Introduction
System of linear
equations of two
unknowns
Substitution
x1 − x2 = 4
Answer: Equation 1 − 2 × equation 2
Elimination
System of linear
equations of three
unknowns
Substitution
Elimination
5x2 = −5.
So x2 = −1. Substitute x2 = −1 into equation 1:
2x1 + 3(−1) = 3
x1 = 3.
So x1 = 3 and x2 = −1 is a solution.
More examples
System of linear
equations with
parameters (new)
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
Introduction
Checking:
2(3) + 3(−1) = 3
(3) − (−1) = 4.
In fact it is the unique solution.
System of linear
equations of two
unknowns
Substitution
Elimination
System of linear
equations of three
unknowns
Substitution
Main goal: One of the main goals of this course is to
solve systems of linear equations with more variables and
more equations.
Elimination
More examples
System of linear
equations with
parameters (new)
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Example 2
Charles Li
Introduction
x1 + 2x2 + 2x3 = 4
x1 + 3x2 + 3x3 = 5
2x1 + 6x2 + 5x3 = 6
System of linear
equations of two
unknowns
Substitution
Elimination
System of linear
equations of three
unknowns
Substitution
Example 3
Elimination
More examples
x1 + 2x2 + x4 = 7
x1 + x2 + x3 − x4 = 3
3x1 + x2 + 5x3 − 7x4 = 1
System of linear
equations with
parameters (new)
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
Introduction
System of linear
equations of two
unknowns
Substitution
Elimination
System of linear equations of two unknowns:
Substitution
System of linear
equations of three
unknowns
Substitution
Elimination
More examples
System of linear
equations with
parameters (new)
You should all be very familiar with the procedure of
solving a system equations of two unknowns.
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
Example 4
Introduction
3x + 4y = 2
(1)
4x + 5y = 3
(2)
System of linear
equations of two
unknowns
Substitution
Elimination
System of linear
equations of three
unknowns
Use (2), we can solve for y in terms of x:
Substitution
3 4
y = − x.
5 5
Substitution this into (1):
3 4
3x + 4( − x) = 2
5 5
12 x
− =2
5
5
x = 2.
Elimination
(3)
More examples
System of linear
equations with
parameters (new)
Substituting x = 2 into (3), we can solve for y:
y=
3 4
− × 2 = −1.
5 5
So the solution is x = 2, y = −1.
Remark: There are other ways to use substitution, for
example
1. Solve x by (2) in terms of y and substitute it into (1)
2. Solve y by (1) in terms of x and substitute it into (2)
3. Solve x by (1) in terms of y and substitute it into (2)
But you cannot get the solution by
1. Solve y by (2) in terms of x and substitute it into (2)
(No substitution back to the original equation)
2. Solve y by (1) in terms of x and substitute it into (1)
3. Solve x by (2) in terms of y and substitute it into (2)
4. Solve x by (1) in terms of y and substitute it into (1)
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
Introduction
System of linear
equations of two
unknowns
Substitution
Elimination
System of linear
equations of three
unknowns
Substitution
Elimination
More examples
System of linear
equations with
parameters (new)
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
Introduction
System of linear
equations of two
unknowns
Substitution
Elimination
System of linear equations of two unknowns:
Elimination
System of linear
equations of three
unknowns
Substitution
Elimination
More examples
System of linear
equations with
parameters (new)
Example 5
Again, let’s solve the system of linear equations from the
previous example.
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
Introduction
3x + 4y = 2
(4)
4x + 5y = 3
(5)
System of linear
equations of two
unknowns
Substitution
Elimination
Consider (4) − 34 (5).
−)
System of linear
equations of three
unknowns
3x
3x
+
+
4y
15
4 y
1
4y
=
=
=
2
9
4
− 14
Thus y = −1. Substituting it into (4):
3x + 4(−1) = 2
x = 2.
So we obtain the solution x = 2, y = −1.
Substitution
Elimination
More examples
System of linear
equations with
parameters (new)
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
Introduction
Remark:
1. The number 34 is so chosen such that the coefficient
of x is eliminated.
2. At some point, we still need to use substitution to get
the solution.
System of linear
equations of two
unknowns
Substitution
Elimination
System of linear
equations of three
unknowns
Substitution
Elimination
More examples
System of linear
equations with
parameters (new)
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
Introduction
System of linear
equations of two
unknowns
Substitution
Elimination
System of linear equations of three unknowns:
Substitution
System of linear
equations of three
unknowns
Substitution
Elimination
More examples
System of linear
equations with
parameters (new)
(Cold joke) There were two men trying to decide
what to do for a living. They went to see a counselor, and he decided that they had good problem
solving skills.
He tried a test to narrow the area of specialty.
He put each man in a room with a stove, a table, and a pot of water on the table. He said ”Boil
the water”. Both men moved the pot from the table to the stove and turned on the burner to boil
the water. Next, he put them into a room with
a stove, a table, and a pot of water on the floor.
Again, he said ”Boil the water”. The first man put
the pot on the stove and turned on the burner.
The counselor told him to be an Engineer, because he could solve each problem individually.
The second man moved the pot from the floor to
the table, and then moved the pot from the table
to the stove and turned on the burner.
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
Introduction
System of linear
equations of two
unknowns
Substitution
Elimination
System of linear
equations of three
unknowns
Substitution
Elimination
More examples
System of linear
equations with
parameters (new)
The counselor told him to be a mathematician because
he reduced the problem to a previously solved
problem.
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
Introduction
Example 6
System of linear
equations of two
unknowns
Solve the following system of linear equations
Substitution
x + 2y + 2z = 4
(1)
x + 3y + 3z = 5
(2)
2x + 6y + 5z = 6
(3)
Elimination
System of linear
equations of three
unknowns
Substitution
Elimination
System of linear
equations with
parameters (new)
Find x in terms of y, z by (1) :
x = 4 − 2y − 2z.
Substituting (4) into (2), we obtain
(4 − 2y − 2z) + 3y + 3z = 5
More examples
(4)
i.e.,
y + z = 1.
(5)
Substituting (4) into (3), we obtain
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
2(4 − 2y − 2z) + 6y + 5z = 6,
Introduction
System of linear
equations of two
unknowns
i.e.,
2y + z = −2
(6)
The equations are reduced to solving a linear system of
equations with two unknowns:
(
y +z =1
2y + z = −2
Solve y in terms of z by (5):
y =1−z
Then substitute y = 1 − z into (6):
2(1 − z) + z = −2
(7)
Substitution
Elimination
System of linear
equations of three
unknowns
Substitution
Elimination
More examples
System of linear
equations with
parameters (new)
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
i.e.
Introduction
z = 4.
By (7)
System of linear
equations of two
unknowns
Substitution
Elimination
y = 1 − z = −3.
Substitute y = −3, z = 4 into (4),
System of linear
equations of three
unknowns
Substitution
Elimination
x = 4 − 2y − 2z = 4 − 2 × (−3) − 2 × 4 = 2.
Hence x = 2, y = −3, z = 4 is a solution.
More examples
System of linear
equations with
parameters (new)
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
Introduction
System of linear
equations of two
unknowns
Substitution
Elimination
System of linear equations of three unknowns:
Elimination
System of linear
equations of three
unknowns
Substitution
Elimination
More examples
System of linear
equations with
parameters (new)
Using substitution all the way to solve linear equations is
not the best way. Instead, we can use elimination to
simplify the system of linear equations first.
Example 7
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
Introduction
We solve the following system by a sequence of equation
operations.
System of linear
equations of two
unknowns
x + 2y + 2z = 4
System of linear
equations of three
unknowns
Substitution
Elimination
x + 3y + 3z = 5
2x + 6y + 5z = 6
−1 times equation 1, add to equation 2:
x + 2y + 2z = 4
0x + 1y + 1z = 1
2x + 6y + 5z = 6
Substitution
Elimination
More examples
System of linear
equations with
parameters (new)
−2 times equation 1, add to equation 3:
x + 2y + 2z = 4
0x + 1y + 1z = 1
0x + 2y + 1z = −2
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
Introduction
System of linear
equations of two
unknowns
Substitution
Elimination
−2 times equation 2, add to equation 3:
x + 2y + 2z = 4
System of linear
equations of three
unknowns
Substitution
Elimination
0x + 1y + 1z = 1
More examples
0x + 0y − 1z = −4
System of linear
equations with
parameters (new)
−1 times equation 3:
x + 2y + 2z = 4
0x + 1y + 1z = 1
0x + 0y + 1z = 4
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
Introduction
which can be written more clearly as
System of linear
equations of two
unknowns
Substitution
x + 2y + 2z = 4
y +z =1
z=4
The third equation requires that z = 4 to be true.
Making this substitution into equation 2 we arrive at
y = −3,
and finally, substituting these values of y and z into the
first equation, we find that x = 2.
Elimination
System of linear
equations of three
unknowns
Substitution
Elimination
More examples
System of linear
equations with
parameters (new)
Remark: We can add several more eliminations to solve
x, y, z without substitution:
x + 2y + 2z = 4
0x + 1y + 1z = 1
0x + 0y + 1z = 4
−1 times equation 3, add to equation 2 and −2 times
equation 3, add to equation 1
x + 2y + 0z = −4
0x + 1y + 0z = −3
0x + 0y + 1z = 4
−2 times equation 2, add to equation 1
x + 0y + 0z = 2
0x + 1y + 0z = −3
0x + 0y + 1z = 4
So x = 2, y = −3, z = 4 is a solution.
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
Introduction
System of linear
equations of two
unknowns
Substitution
Elimination
System of linear
equations of three
unknowns
Substitution
Elimination
More examples
System of linear
equations with
parameters (new)
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
Introduction
System of linear
equations of two
unknowns
Substitution
Elimination
More examples
System of linear
equations of three
unknowns
Substitution
Elimination
More examples
System of linear
equations with
parameters (new)
Example 8
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
x1 − 5x2 + 3x3 = 1
2x1 − 4x2 + x3 = 0
x1 + x2 − 2x3 = −1
Introduction
System of linear
equations of two
unknowns
Substitution
Elimination
−2 times equation 1, add to equation 2:
x1 − 5x2 + 3x3 = 1
0x1 + 6x2 − 5x3 = −2
x1 + x2 − 2x3 = −1
−1 times equation 1, add to equation 3:
x1 − 5x2 + 3x3 = 1
0x1 + 6x2 − 5x3 = −2
0x1 + 6x2 − 5x3 = −2
System of linear
equations of three
unknowns
Substitution
Elimination
More examples
System of linear
equations with
parameters (new)
−1 times equation 2, add to equation 3:
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
x1 − 5x2 + 3x3 = 1
0x1 + 6x2 − 5x3 = −2
0x1 + 0x2 + 0x3 = 0
Introduction
System of linear
equations of two
unknowns
Substitution
Elimination
5
6 times equation 2, add to equation 1:
2
7
x1 + 0x2 − x3 = −
6
3
0x1 + 6x2 − 5x3 = −2
0x1 + 0x2 + 0x3 = 0
We can express x1 , x2 in terms of x3 :
2 7
x1 = − + x3
3 6
1 5
x2 = − + x3
3 6
System of linear
equations of three
unknowns
Substitution
Elimination
More examples
System of linear
equations with
parameters (new)
The solution set is
2 7
1 5
{(− + a, − + a, a) | a real numbers.}
3 6
3 6
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
Introduction
Example 9
System of linear
equations of two
unknowns
Substitution
Elimination
x1 − 5x2 + 3x3 = 1
2x1 − 4x2 + x3 = 0
x1 + x2 − 2x3 = −2
−2 times equation 1, add to equation 2:
x1 − 5x2 + 3x3 = 1
0x1 + 6x2 − 5x3 = −2
x1 + x2 − 2x3 = −2
System of linear
equations of three
unknowns
Substitution
Elimination
More examples
System of linear
equations with
parameters (new)
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
−1 times equation 1, add to equation 3:
x1 − 5x2 + 3x3 = 1
Introduction
System of linear
equations of two
unknowns
0x1 + 6x2 − 5x3 = −2
Substitution
0x1 + 6x2 − 5x3 = −3
System of linear
equations of three
unknowns
Elimination
Substitution
−1 times equation 2, add to equation 3:
Elimination
More examples
x1 − 5x2 + 3x3 = 1
0x1 + 6x2 − 5x3 = −2
0x1 + 0x2 + 0x3 = −1
The last equation, 0 = −1 has no solution. So the system
of linear equations has no solution.
System of linear
equations with
parameters (new)
Example 10
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
x1 + 2x2 + 0x3 + x4 = 7
x1 + x2 + x3 − x4 = 3
3x1 + x2 + 5x3 − 7x4 = 1
Introduction
System of linear
equations of two
unknowns
Substitution
Elimination
−1 times equation 1, add to equation 2:
x1 + 2x2 + 0x3 + x4 = 7
0x1 − x2 + x3 − 2x4 = −4
3x1 + x2 + 5x3 − 7x4 = 1
−3 times equation 1, add to equation 3:
x1 + 2x2 + 0x3 + x4 = 7
0x1 − x2 + x3 − 2x4 = −4
0x1 − 5x2 + 5x3 − 10x4 = −20
System of linear
equations of three
unknowns
Substitution
Elimination
More examples
System of linear
equations with
parameters (new)
−5 times equation 2, add to equation 3:
x1 + 2x2 + 0x3 + x4 = 7
0x1 − x2 + x3 − 2x4 = −4
0x1 + 0x2 + 0x3 + 0x4 = 0
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
Introduction
System of linear
equations of two
unknowns
Substitution
Elimination
−1 times equation 2:
System of linear
equations of three
unknowns
x1 + 2x2 + 0x3 + x4 = 7
Substitution
0x1 + x2 − x3 + 2x4 = 4
More examples
0x1 + 0x2 + 0x3 + 0x4 = 0
−2 times equation 2, add to equation 1:
x1 + 0x2 + 2x3 − 3x4 = −1
0x1 + x2 − x3 + 2x4 = 4
0x1 + 0x2 + 0x3 + 0x4 = 0
Elimination
System of linear
equations with
parameters (new)
which can be written more clearly as
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
x1 + 2x3 − 3x4 = −1
x2 − x3 + 2x4 = 4
0=0
Introduction
System of linear
equations of two
unknowns
Substitution
Elimination
The last equation 0 = 0 is always true, so we can ignore it
and only consider the first two equations.
We can analyze the second equation without
consideration of the variable x1 . It would appear that
there is considerable latitude in how we can choose x2 ,
x3 , x4 and make this equation true. Let us choose x3 and
x4 to be anything we please, say x3 = a and x4 = b. Now
we can take these arbitrary values for x3 and x4 ,
substitute them in equation 1, to obtain
x1 + 2a − 3b = −1
x1 = −1 − 2a + 3b
System of linear
equations of three
unknowns
Substitution
Elimination
More examples
System of linear
equations with
parameters (new)
Similarly, equation 2 becomes
x2 − a + 2b = 4
x2 = 4 + a − 2b
So our arbitrary choices of values for x3 and x4 (a and b)
translate into specific values of x1 and x2 . Now we can
easily and quickly find many more (infinitely more).
Suppose we choose a = 5 and b = −2, then we compute
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
Introduction
System of linear
equations of two
unknowns
Substitution
Elimination
System of linear
equations of three
unknowns
Substitution
x1 = −1 − 2(5) + 3(−2) = −17
x2 = 4 + 5 − 2(−2) = 13
and you can verify that
(x1 , x2 , x3 , x4 ) = (−17, 13, 5, −2) makes all three
equations true. The entire solution set is written as
{(−1 − 2a + 3b, 4 + a − 2b, a, b) | a, b real numbers}.
Elimination
More examples
System of linear
equations with
parameters (new)
Example 11
Solve the following system of linear equations:
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
x2 + x3 + 2x4 + 2x5 = 2
x1 + 2x2 + 3x3 + 2x4 + 3x5 = 4
−2x1 − x2 − 3x3 + 3x4 + x5 = 3
Swap equation 1 and equation 2:
x1 + 2x2 + 3x3 + 2x4 + 3x5 = 4
x2 + x3 + 2x4 + 2x5 = 2
−2x1 − x2 − 3x3 + 3x4 + x5 = 3
2 times equation 1, and add it to equation 3:
x1 + 2x2 + 3x3 + 2x4 + 3x5 = 4
x2 + x3 + 2x4 + 2x5 = 2
3x2 + 3x3 + 7x4 + 7x5 = 11
Introduction
System of linear
equations of two
unknowns
Substitution
Elimination
System of linear
equations of three
unknowns
Substitution
Elimination
More examples
System of linear
equations with
parameters (new)
-3 times equation 2 and add it to equation 3:
x1 + 2x2 + 3x3 + 2x4 + 3x5 = 4
x2 + x3 + 2x4 + 2x5 = 2
x4 + x5 = 5
Now the system of linear equations looks like an ”inverted
stair”. We can then solve the system of linear equations
by substitution (a better method well be given later). By
the last equation:
x4 = 5 − x5 .
Solve x2 in terms of other variables by equation 2:
x2 = 2 − x3 − 2x4 − 2x5
= 2 − x3 − 2(5 − x5 ) − 2x5
= −8 − x3
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
Introduction
System of linear
equations of two
unknowns
Substitution
Elimination
System of linear
equations of three
unknowns
Substitution
Elimination
More examples
System of linear
equations with
parameters (new)
Solve x1 in terms of other variables by equation 1:
x1 = 4 − 2x2 − 3x3 − 2x4 − 3x5
= 4 − 2(−8 − x3 ) − 3x3 − 2(5 − x5 ) − 3x5
= 10 − x3 − x5 .
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
Introduction
x3 , x5 can be taken as any values. Set x3 = a, x5 = b, the
solution set can be given by
System of linear
equations of two
unknowns
Substitution
Elimination
{(10 − a − b, −8 − a, a, 5 − b, b) | a, b real numbers}.
A better method
Instead of subsitution, we use more elimination:
x1 + 2x2 + 3x3 + 2x4 + 3x5 = 4
x2 + x3 + 2x4 + 2x5 = 2
x4 + x5 = 5
-2 times equation 3 and add it to equation 2:
x1 + 2x2 + 3x3 + 2x4 + 3x5 =
x2 + x3
4
= −8
x4 + x5 =
5
System of linear
equations of three
unknowns
Substitution
Elimination
More examples
System of linear
equations with
parameters (new)
-2 times equation 3 and add it to equation 1:
x1 + 2x2 + 3x3 +
+ x5 = −6
= −8
x2 + x3
x4 + x5 =
5
-2 times equation 2 and add it to equation 1:
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
Introduction
System of linear
equations of two
unknowns
Substitution
x1 +
+ x3
x2 + x3
+ x5 = 10
= −8
x4 + x5 =
5
Notice the following:
1. The system of equations looks like an ”inverted”
stairs.
2. The leftmost variables in the equations are x1 , x2 and
x4 .
3. Only the first equation has variable x1 .
4. Only the second equation has variable x2 .
5. Only the third equation has variable x4 .
Elimination
System of linear
equations of three
unknowns
Substitution
Elimination
More examples
System of linear
equations with
parameters (new)
Move x3 , x5 to another side.
x1 = 10 − x3 − x5
x2 = −8 − x3
x4 = 5 − x5
The right hand sides have x3 , x5 as variables only and
x3 , x5 can be taken as any values. Set x3 = a, x5 = b, the
solution set can be given by
{(10 − a − b, −8 − a, a, 5 − b, b) | a, b real numbers}.
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
Introduction
System of linear
equations of two
unknowns
Substitution
Elimination
System of linear
equations of three
unknowns
Substitution
Elimination
More examples
Observations:
1. some systems of linear equations have exactly one
solution.
2. some systems of linear equations have no solution.
3. some systems of linear equations have infinite many
solutions. The solutions can be expressed in terms
of 1 or 2 (or even more) variables.
System of linear
equations with
parameters (new)
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
Introduction
System of linear
equations of two
unknowns
Substitution
Elimination
System of linear equations with parameters (new)
System of linear
equations of three
unknowns
Substitution
Elimination
More examples
System of linear
equations with
parameters (new)
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Example 12
Let α be a real number. Denote the following system of
linear equation by (Sα ).
Charles Li
Introduction
System of linear
equations of two
unknowns
x1 + x2 + (1 + α)x3 = 1 + α
−αx1
−αx3 = 1 − α
(1 − α)x1 + x2 + (2 − α)x3 = 3 − α
Substitution
2
Elimination
System of linear
equations of three
unknowns
Substitution
1. Suppose the system (Sα ) has a solution. What are
the possible values of α.
2. For each such value of α that the system (Sα ) has a
solution, write down the solution set.
x1 + x2 + (1 + α)x3 = 1 + α
−αx1
−αx3 = 1 − α
(1 − α)x1 + x2 + (2 − α)x3 = 3 − α2
Elimination
More examples
System of linear
equations with
parameters (new)
α× equation 1 + equation 2, −(1 − α)× equation 1 +
equation 3
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
x1 + x2 +
(1 + α)x3 = 1 + α
αx2 +
α 2 x3 = 1 + α 2
2
αx2 + (1 − α + α )x3 = 2
Introduction
System of linear
equations of two
unknowns
Substitution
Elimination
−equation 2 + equation 3
System of linear
equations of three
unknowns
x1 + x2 + (1 + α)x3 = 1 + α
αx2 +
2
α x3 = 1 + α
Substitution
Elimination
2
+ (1 − α)x3 = 1 − α2
When α = 0, the second equation is
0 = 1.
It has no solution. So suppose α 6= 0.
More examples
System of linear
equations with
parameters (new)
When α = 1, the system of equations is
x1 + x2 + 2x3 = 2
x2 + x3 = 2
+
0=0
Charles Li
Introduction
System of linear
equations of two
unknowns
−equation 2 + equation 1.
x1
Math 1038 Lect 1:
Technique of
solving system of
linear equations
+ x3 = 0
x2 + x3 = 2
+ 0=0
Substitution
Elimination
System of linear
equations of three
unknowns
Substitution
Elimination
The solution set is therefore
{(−a, 2 − a, a) | a is a real number}.
For α 6= 0, 1 by the third equation, we have
x3 = 1 + α.
By the second equation
αx2 = 1 + α2 − α2 x3 = 1 + α2 − α2 (1 + α) = 1 − α3 .
More examples
System of linear
equations with
parameters (new)
So
x2 = α−1 − α2 .
By the first equation
Math 1038 Lect 1:
Technique of
solving system of
linear equations
Charles Li
x1 = 1 + α − x2 − (1 + α)x3 = −α−1 − α.
Introduction
System of linear
equations of two
unknowns
Conclusion
1. When α 6= 0, the system of linear equations has
solution.
2. When α = 1, The solution set is
Substitution
Elimination
System of linear
equations of three
unknowns
Substitution
Elimination
More examples
{(−a, 2 − a, a) | a is a real number}.
When α 6= 0, 1, the system has a unique solution.
x1 = −α−1 − α.
x2 = α−1 − α2 .
x3 = 1 + α.
System of linear
equations with
parameters (new)