GTS 112 Linear Algebra
Homework 1
Due Date: Friday February 17, 2017 before 4:45PM
Disclaimer: This is my attempt to help you learn better. Do not use this as your main study material. There might be mistakes/typos.
Homework Assignment
(1) Use geometry (draw lines) to find solutions (if exist) of the linear systems below. What is your
conclusion?
3x + y
=
x − 2y
=
x+y
Solution:
−8
1
10
st solve each equation for
5
3
x 2
2
4
13
- x +
3
3
graphs of the equations are
wn at the right.
=
−3
-10
10
-10
intersection: x = 2.41
y = 1.12
(2.41, 1.12)
tiply each equation
byhas10
This system
a unique solution with x3= −2, y = 3
then solve for y:
1
24
(2) Use geometry (draw lines) to find solutions (if exist) of the linear systems below. What is your
x +
conclusion?
35
35
-5
5
17
1
x x − 2y = −6
26
13
2x + y = 8
graphs of these equations
-3
almost indistinguishable.
= −2
intersection:x +x 2y= -3.31
y = -2.24
Solution
(-3.31, -2.24)
2y = -6 (L1)
y
x – 2y = –6
+ y = 8
(L2)
10
x + 2y = –2
2y = -2 (L3)
(2,4)
(–4,1)
L1 and L2 intersect:
x
10
–1 0
x - 2y = -6
(1)
(6,–4)
2x + y = 8
(2)
Multiply (2) by 2 and add to (1):
–1 0
2x + y = 8
5x = 10
x = 2
hasget
no solutions.
Substitute xHence,
= 2 this
in system
(1) to
2 - 2y = -6
-2y = -8
y = 4
Solution: x = 2, y = 4
1
L1 and L3 intersect:
(C) L2 and L3 intersect:
x - 2y = -6
(3)
2x + y = 8
(5)
x + 2y = -2
(4)
x + 2y = -2 (6)
Add (3) and (4):
Multiply (6) by -2 and add
2x = -8
to (5)
x = -4
-3y = 12
Substitute x = -4 in (3)
y = -4
(3) Use the back substitution technique to find a solution set of the following system.
x + 2y + z
=
0
3y + 2z
=
−z
=
−8
1
Solution: From the third equation, we have z = −1. Use this to substitute into the second equation.
3y + 2z
3y + 2(−1)
3y
y
= −8
=
−8
= −6
= −2
Finally substitute the values of y and z into the first equation.
x + 2y + z
=
0
x + 2(−2) + (−1)
=
0
x
=
5
This system has one unique solution which is x = 5, y = −2, z = −1.
2
Example 1. Solve the following system by using the Gauss-Jordan elimination method.


+z =5
x + yto
Use the Gauss-Jordan Elimination technique
solve the linear systems in problem (4)-(6). Determine
2x linear
+ 3y +equations
5z = 8 has a solution. If there is one solution, find
whether or not the system of the following


4x
+
5z
=
2
out what it is. If there are an infinite number of solutions,
write down any two of them.
(4)
Solution: The augmented matrix of the system is the following.
 x1 + x2 + x3 = 5
1 1 1 5
2x1 + 3x2 + 5x3 = 8
2 3 5 8
4 4x
01 +
5 5x
23 = 2
Solution:
We will now perform row operations until we obtain a matrix in reduced row echelon form.




1 1 1 5
5
1 1 1
R2 −2R1
 2 3 5 8 −
−−−−→  0 1 3 −2 
4 0 5 2
4 0 5
2

5
1
1 1
R −4R1
 0
1 3 −2 
−−3−−−→
0 −4 1 −18


5
1 1 1
−−−−−→  0 1 3 −2 
0 0 13 −26
R3 +4R2


5
1 1 1
−−−→  0 1 3 −2 
0 0 1 −2

1
R
13 3


5
1 1 1
R −3R3
 0 1 0
4 
−−2−−−→
0 0 1 −2


7
1 1 0
R −R3
 0 1 0
4 
−−1−−→
0 0 1 −2

1 0 0
3
4 
−−−−→  0 1 0
0 0 1 −2
R1 −R2

From this final matrix, we can read the solution of the system. It is
This system has one unique solution which is x1 = 3, x2 = 4, x3 = −2. It is a consistent system.
x = 3, y = 4, z = −2.
2
3
(5)
4y + z
=
2
2x + 6y − 2z
=
3
=
4
4x + 8y − 5z
Solution:
We will now perform row operations until we obtain a matrix in



0 4
1 2
2 6 −2
R1 ↔R2
 2 6 −2 3  −−−−−→  0 4
1
4 8 −5 4
4 8 −5

3
2
6 −2
4
1
2 
−−−−−→  0
0 −4 −1 −2
R3 −2R1

reduced row echelon form.

3
2 
4

2 6 −2 3
R +R2
 0 4
1 2 
−−3−−→
0 0
0 0


2 6
−−−→  0 1
0 0
1
R
4 2

3
−2
1/4 1/2 
0
0

2 0 −7/2 0
1/4 1/2 
−−−−−→  0 1
0 0
0
0
R1 −6R2


1 0 −7/4 0
1/4 1/2 
−−−→  0 1
0 0
0
0
1
R
2 1

This last matrix is in reduced row echelon form so we can stop. It corresponds to the augmented
We matrix
have; of the following system.
(
x − 7z = 0
x − 7/4z4 = 0
y + 1z = 1
y + 1/4z4 = 2 1/2
We can express the solutions of this system as 0 = 0
This means that
x = 47 z,
y=
1
2
− 14 z.
Since there is no specific value for z, it can
x be=chosen
7/4zarbitrarily. This means that there are infinitely
many solutions for this system. We can yrepresent
solutions by using a parameter t as follows.
= 1/2all−the
1/4z
7
x=
yhas
= 12infinitely
− 14 t, zmany
= t solutions. If z = 0, then x = 7/4,
The unknowns x, y depend on z. This
system
4 t,
y = 1/2, z = 0 is a solution of this system. If z = 4, then x = 7, y = −1/2, z = 4 is another solution of
value of the parameter t gives us a solution of the system. For example,
thisAny
system.
t = 4 gives the solution (x, y, z) = (7, − 21 , 4)
t = −2 gives the solution (x, y, z) = (− 72 , 1, −2).
4
4
Example 2. Solve the following system by using the Gauss-Jordan elimination method.


x + 2y − 3z = 2
6x + 3y − 9z = 6


7x + 14y − 21z = 13
(6)
Solution: The augmented matrix of the system is the following.
 u + 2v − 3w = 2
2 6
2 −−3
6u1+ 3v
9w =
 6 3 −9 6 
7u + 14v − 21w = 13
7 14 −21 13
Solution:
Let’s now perform row operations on this augmented matrix.




1 2 −3 2
2
1
2 −3
R2 −6R1
 6 3 −9 6  −−−−−→  0 −9
9 −6 
7 14 −21 13
7 14 −21 13

2
1
2 −3
R −7R1
 0 −9
9 −6 
−−3−−−→
0
0
0 −1

We obtain a row whose elements are all zeros except the last one on the right. Therefore, we conclude
Thethat
lastthe
rowsystem
gives us
= -1 which
impossible.i.e.,
Therefore
this
system has no solutions. It is inconsistent.
of 0equations
is is
inconsistent,
it has no
solutions.
(7) Use Gauss-Jordan Elimination to find the inverse of the following matrices (if exists)


Example 3. Solve the following system by−5
using−2the −2
Gauss-Jordan elimination method.
 2
1
0 

1
0
1

4y
+
z
=
2

2x + 6y − 2z = 3
Solution


4x + 8y − 5z = 4
 -5 -2 -2  1
 1
0
0
0
1
Solution:
is the
following.
 The augmentedmatrix of the system

1
0  0
1
0 ~  2
0
45.  2
1
 1
1 2 -2 -2
0
1  0
0
1  0 4  -5
 2 6 −2(-2)
3 R +
R1 ↔ R3
1
4 8 −5 4
1

~ 0
0
0
1  0
0

1 -2  0
1

-2
3
1
0
2R2 + R3 → R3
1

-2 
5
 0

 0
 1
R2 →
5R1 + R3 →
1
0
1  0


1 -2  0
~ 0
0
0 -1  1
(-1)R3 → R3
1
0
1  0
0
1



1 -2  0
1 -2  ~
~ 0
0
0
1  -1 -2 -1 
2R3 + R2 → R2
(-1)R3 + R1 → R1
 1
23

Thus, the inverse is  -2 -3
 -1 -2
 2 1

47.  1 1
 -1 -1




R1 ↔
1
0
0
1 0
0 1
0 0
R2
1

0
0
0
1
0
0
0
1
 1

 -2
 -1
0
1
0
R2
R3
1

0
0
0
1
2
1

-2 
1
2
-3
-2
2

-4 
-1 
2

-4  .
-1 
 1 1 0  0 1 0
1 1
0





0  ~  2 1 1  1 0 0  ~  0 -1
 -1 -1 0  0 0 1 
0 0
1
(-2)R1 + R2 → R2
R1 +
R → R3
5 3
0
1
0
 0 1

 1 -2
 0 1
0

0
1
From this matrix, we conclude that the inverse does not exist.
 -1 -2

49.  4 3
 4 0
1 2

~ 4 3
 1 0 0


 0 1 0  (-1) R1 → R1
 0 0 1
-2  -1 0 0  (-4)R + R → R


1
2
2
0  0 1 0
2
0
4
(8) Use Gauss-Jordan Elimination to find the inverse of the following matrices (if exists)


1 −1 0
 2 −1 1 
0 −1 1
Solution:
(9) Find the determinant of the following matrix. Is it invertible?


2 −1
3 0 −1
 1
0
1 0
2 


 1 −1
1
0
4 


 3
2 −5 1
1 
3
0
4 0 −3
Solution
2
1
1
3
3
−1
3
0
1
−1
1
2 −5
0
4
0 −1 0
2 0
4 =
1
1 0 −3 2 −1 3 −1 1
0 1
2 1 −1 1
4 3
0 4 −3 2
1 1
2 4 + (−1)(−1)3+2 1
= (−1)(−1)1+2 1 1
3
3 4 −3 = −2 + 4
=
(1)(−1)4+4 2
(10) Find the determinant of the following matrix. Is it invertible?

0 −10
3
1
0
0
 4
12 −7 −5
15
2

 2 −6 −2
0
8 −4

 1 −3 14
0 −5
3

 −7 −5
1
0 −10
0
−3
9
3
0 −12
6
3
1
4
−1
2
−3








Solution: The determinant is zero because row 6 is (-3/2) times row 3. Therefore this matrix is not
invertible.
6