الحل الجديد TMA-MT132-Fall-2023-20 كتابةةة_044857_062404
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King Khalid University *
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Course
132
Subject
Mathematics
Date
Nov 24, 2024
Type
docx
Pages
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MT132 - M132
: Linear Algebra
Tutor-Marked
Assignment
(TMA) Fall 23/24
Cut-Off Date: Based on the Published Deadline. Total Marks: 60 marks turned to 15 marks Contents
Warnings and Declaration…………………………………….………………………………
......
1
Question 1 ……………….…………………………………. ………………………………...…..2
Question 2 ………………………………………………………………………………….…..…..3
Question 3 ………………………………………………………………………………….…..…..4
Question 4 ………………………………………………………………………………….…..…..5
Marking Criteria ...…………..………………………………………………………….………..…6
Plagiarism Warning:
As per AOU rules and regulations, all students are required to submit their own TMA work and avoid plagiarism. The AOU has implemented sophisticated techniques for plagiarism detection. You must provide all references in case you use and quote another person's work in your TMA. You will be penalized for any act of plagiarism as per the AOU's rules and regulations.
Declaration of No Plagiarism by Student (to be signed and submitted by student with TMA work):
I hereby declare that this submitted TMA work is a result of my own efforts and I have not plagiarized any other person's work. I have provided all references of information that I have used and quoted in my TMA work.
Name of Student:… Haya Dossary Signature:………
ID… 20417394 Date:
…………………………………………………
MT132 – M132 / TMA Page 1
of 19
2023/2024 Fall
The TMA covers only chapters 1 and 2. It consists of four questions; each question
is worth 15 marks. You should give the details of your solutions and not just the
final results.
Q−1: [5×3 marks] Step
1
5
(
a
)
TRUE
:
Giventwoequationsis
−
¿
x
+
2
y
+
kz
=
6
3
x
+
6
y
+
8
z
=
4
Now findthe valueof k for infinitely many solutionfor thelinear system .
Explanation
:
System hasinfinitely many solutionif
−
a
1
a
2
=
b
1
b
2
=
c
1
c
2
a
1
=
1,
a
2
=
3,
b
1
=
2,
b
2
=
6,
c
1
=
k ,c
2
=
8
1
3
=
2
6
=
k
8
by
1
st
∧
2
nd
−
¿
1
3
=
k
8
k
=
8
3
Step
2
5
(
b
)
FALSE
:
Justification
−
Giventwomatrices
−
¿
A
=
|
a
+
b
2
|
B
=
|
12
|
|
−
13
||
2
a
+
3
b
3
|
Given A
=
Bthen
−
¿
|
a
+
b
2
|
=
|
12
|
|
−
13
||
2
a
+
3
b
3
|
comparisonof
¿
−
¿
a
+
b
=
1
−−−−−−−−−−−−−−−
(
i
)
2
a
+
3
b
=−
1
−−−−−−−−−−−
(
ii
)
now solvethese equationsby Substitutemethod
−
¿
a
=
1
−
b
put thevalue
∈
equation
(
ii
)
−
¿
2
(
1
−
b
)
+
3
b
=−
1
2
−
2
b
+
3
b
=−
1
b
=−
3
put thevalue of b
∈
equation
(
i
)
−
¿
a
=
1
−
(
−
3
)
MT132 – M132 / TMA Page 2
of 19
2023/2024 Fall
a
=
4
If matrices A
=
Bthen a
=
4
∧
b
=−
3.
Step
3
5
(
c
)
FALSE
:
If AC
=
BC ,then A ≠Bis true.
Justification
:
let wetake anexample
−
¿
let A
=
|
10
|
, B
=
|
00
|
,C
=
|
00
|
|
00
||
20
||
3 4
|
since
−
¿
AB
=
|
10
||
00
|
=
|
00
|
|
00
||
20
||
00
|
AC
=
|
10
||
00
|
=
|
00
|
|
00
||
3 4
||
00
|
Here AB
=
AC but B≠C .
Step
4
5
(
d
)
FALSE
:
Not necessarily.If Aisinvertible
∧
AB
=
0,
it doesnot imply that B
=
0.
Itonly impliesthat Bisthe zero
¿
∈
otherwor
zero
¿¿
Justification
:
let wetake anexample
−
¿
let A
=
|
01
|
, B
=
|
−
10
|
|
00
||
00
|
since
−
¿
AB
=
|
01
||
−
10
|
=
|
00
|
|
00
||
00
||
00
|
Here AB
=
0
but A≠
0
∨
B≠
0
Step
5
5
(
e
)
FALSE
:
No ,that statement is incorrect .Thetranspose of the product of twomatrices AB is equal
¿
the product of their trans
This property holds truefor matriceswhere the multiplicationoperationisdefined .
(
AB
)
T
=
B
T
A
T
MT132 – M132 / TMA Page 3
of 19
2023/2024 Fall
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Q−2:
[10+5 marks]
a)
Consider the linear system :
{
x
+
y
=
2
y
+
z
=
2
x
+
z
=
2
ax
+
by
+
cz
=
0
,
For which values of a,b
and c
does the linear system has
(i)
A unique solution.
(ii)
Infinitely many solutions.
(iii)
No solution.
b)
Show that A
=
[
1
2
0
3
−
1
2
−
2
3
−
2
]
has no inverse.
Q−2:
[10+5 marks]
x
+
y
=
2
y
+
z
=
2
x
+
z
=
2
ax
+
by
+
cz
=
0
Theaugmented
¿
1
1
0
2
0
1
1
2
1
0
1
2
a
y
c
0
Now the raw operation of the above MATRIX
1
1
0
2
0
1
1
2
1
0
1
2
a
y
c
0
R
3
−
R
1
, R
4
−
a R
1
→
1
1
0
2
0
1
1
2
0
−
1
1
0
0
−
a
+
b
c
−
2
a
R
1
+
R
3
,R
2
+
R
3
, R
4
−
(
b
+
a
)
R
1
→
1
0
−
1
0
0
1
1
2
0
0
2
2
0
0
a
−
b
+
c
−
2
b
i)
Hereb
−
a
=
0
,c≠
0
a,bare arbitary
⟹
a
=
bthenthe system hasunique solotuion
ii)
b
−
a
=
0
,c
=
0
a
=
0
⟹
a
=
b
=
c
=
0
thenthe system hasinfinitelymany solotuion
iii)
if b
−
a
=
0
,c
=
0
a≠
0
thenthe system hasno soluation
b
¿
A
=
1
2
0
3
−
1
2
−
2
3
−
2
MT132 – M132 / TMA Page 4
of 19
2023/2024 Fall
det
(
A
)
=
1
(
12
−
6
)
−
2
(
−
6
+
4
)
=−
4
−
2
(
−
2
)
=−
4
+
4
=
0
A
−
1
=
adj
(
A
)
de t
(
A
)
If det
(
A
)=
0
thenit isundefine ,noinverse
¿
MT132 – M132 / TMA Page 5
of 19
2023/2024 Fall
Q−2:
[5+5+5 marks]
a)
Find scalars a
and b
such that Z
=
aX
+
bY
where X
=
[
5
4
−
3
]
,Y
=
[
8
1
−
1
]
, and Z
=
[
−
14
−
22
16
]
.
b)
Let A
−
1
=
[
1
1
−
1
−
3
2
−
1
3
−
3
−
2
]
, B
=
[
0
−
1
0
0
2
−
1
−
2
0
1
]
, and C
=
[
1
−
1
0
]
.
(i)
Find the matrix A
.
(ii)
Find a matrix X
, such that D
T
X
=
C
where D
=
A B
−
1
.
A)
Z
=
aX
+
bY
[
¿
−
14
−
22
16
]
=
a
[
¿
5
4
−
3
]
+
b
[
¿
8
1
−
1
]
Thecorresponding
¿
[
5
8
−
14
4
1
−
22
−
3
−
1
16
]
Multipleeachelement of R
1
by
1
/
5
¿
make theentry at
1,1
a
1
[
1
8
/
5
−
14
/
5
4
1
−
22
−
3
−
1
16
]
Performtherow operation R
2
=
R
2
−
4
R
1
¿
maketheentry at
3,1
a
0
[
1
8
5
−
14
5
0
−
27
5
−
54
5
−
3
−
1
16
]
[
1
8
5
−
14
5
0
−
27
5
−
54
5
0
19
/
5
38
/
5
]
Muyiple MT132 – M132 / TMA Page 6
of 19
2023/2024 Fall
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[
1
8
5
−
14
5
0
1
2
0
19
/
5
38
/
5
]
Perform the row operation [
1
8
5
−
14
5
0
1
2
0
0
0
]
Perform [
1
0
−
6
0
1
2
0
0
0
]
Therefore
a= -6 and b =2
(b)
i)
The matrix A
−
1
=
[
1
1
−
1
−
3
2
−
1
3
−
3
−
2
]
A
(
¿¿
−
1
)
−
1
=
A
¿
Findthe determinant .
-19
Since the determinant is non-zero, the inverse exists.
MT132 – M132 / TMA Page 7
of 19
2023/2024 Fall
Set up a Matrix where the left half is the original matrix and the right half is its identity matrix.
Find the reduced row echelon form.
The right half of the reduced row echelon form is the inverse.
(ii)
Now
And
Find the determinant.
MT132 – M132 / TMA Page 8
of 19
2023/2024 Fall
Since the determinant is non-zero, the inverse exists.
Set up a Matrix where the left half is the original matrix and the right half is its identity matrix.
Find the reduced row echelon form.
The right half of the reduced row echelon form is the inverse.
Now Two matrices can be multiplied if and only if the number of columns in the first matrix is equal
to the number of rows in the second matrix. In this case, the first matrix is MT132 – M132 / TMA Page 9
of 19
2023/2024 Fall
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And the second matrix is .
Multiply each row in the first matrix by each column in the second matrix.
Simplify each element of the matrix by multiplying out all the expressions.
Transpose of the matrix by switching its rows to columns.
MT132 – M132 / TMA Page 10
of 19
2023/2024 Fall
Swap With To put a nonzero entry at .
Multiply each element of By To make the entry at A .
MT132 – M132 / TMA Page 11
of 19
2023/2024 Fall
Perform the row operation To make the entry at A .
Swap With To put a nonzero entry at .
Multiply each element of By To make the entry at A .
Perform the row operation MT132 – M132 / TMA Page 12
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2023/2024 Fall
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To make the entry at A .
Perform the row operation To make the entry at A .
Perform the row operation To make the entry at A .
Therefore
Explanation:
All parts of question-2 have solved.
MT132 – M132 / TMA Page 13
of 19
2023/2024 Fall
Q−3:
[7+8 marks]
Let
A
=
[
1
−
1
2
4
1
0
1
6
2
−
3
5
4
3
2
−
1
1
]
and B
=
[
1
0
1
−
2
]
a)
Find the determinant of A
.
b)
Solve the linear system AX
=
B
.
a) Find the determinant of A
Choose the row or column with the most 0 elements. If there are no 0 elements choose any row or column. Multiply every element in row 2 by its cofactor and add.
det
(
A
)
=−
1
[
−
1
2
4
−
3
5
4
2
−
1
1
]
+
0
[
−
1
2
4
−
3
5
4
2
−
1
1
]
−
1
[
−
1
−
1
4
2
−
3
4
3
2
1
]
+
6
[
1
−
1
2
2
−
3
5
3
2
−
1
]
Multiply 0 by −
1
[
−
1
2
4
−
3
5
4
2
−
1
1
]
+
0
[
−
1
2
4
−
3
5
4
2
−
1
1
]
−
1
[
−
1
−
1
4
2
−
3
4
3
2
1
]
+
6
[
1
−
1
2
2
−
3
5
3
2
−
1
]
MT132 – M132 / TMA Page 14
of 19
2023/2024 Fall
Evaluate [
−
1
2
4
−
3
5
4
2
−
1
1
]
−1
−15+0−
1
[
−
1
−
1
4
2
−
3
4
3
2
1
]
+
6
[
1
−
1
2
2
−
3
5
3
2
−
1
]
Evaluate [
−
1
−
1
4
2
−
3
4
3
2
1
]
−
1
⋅
−
15
+
0
−
1
⋅
31
+
6
[
1
−
1
2
2
−
3
5
3
2
−
1
]
Evaluate
[
1
−
1
2
2
−
3
5
3
2
−
1
]
−
1
⋅
−
15
+
0
−
1
⋅
31
+
6.2
Simplify the determinant.
det
(
A
)=−
4
b) Solve the linear system AX=B
Let X=
[
¿
W
X
y
z
]
1
−
1
2
4
1
0
1
6
2
−
3
5
4
3
2
−
1
1
[
¿
W
X
y
z
]
= [
1
0
1
−
2
]
Write the system as a matrix.
MT132 – M132 / TMA Page 15
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2023/2024 Fall
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[
1
−
1
2
4
1
0
1
6
2
−
3
5
4
3
2
−
1
1
¿
1
0
1
−
2
]
Perform the row operation R2=R2−R1
to make the entry at 2,1
a 0
.
[
1
−
1
2
4
0
1
−
1
2
2
−
3
5
4
3
2
−
1
1
¿
1
−
1
1
−
2
]
Perform the row operation R3=R3−2R1
to make the entry at 3,1
a 0
.
[
1
−
1
2
4
0
1
−
1
2
0
−
1
1
−
4
3
2
−
1
1
¿
1
−
1
−
1
−
2
]
Perform the row operation R4=R4−3R1
to make the entry at 4,1
a 0
.
[
1
−
1
2
4
0
1
−
1
2
0
−
1
1
−
4
0
5
−
7
−
11
¿
1
−
1
−
1
−
5
]
Perform the row operation R3=R3+R2
to make the entry at 3,2
a 0
.
[
1
−
1
2
4
0
1
−
1
2
0
0
0
−
2
0
5
−
7
−
11
¿
1
−
1
−
2
−
5
]
Perform the row operation R4=R4−5R2
to make the entry at 4,2
a 0
.
[
1
−
1
2
4
0
1
−
1
2
0
0
0
−
2
0
0
−
2
−
21
¿
1
−
1
−
2
0
]
Swap R4
with R3
to put a nonzero entry at 3,3
.
[
1
−
1
2
4
0
1
−
1
2
0
0
−
2
−
21
0
0
0
−
2
¿
1
−
1
0
−
2
]
Multiply each element of R3
by -1/2 to make the entry at 3,3
a 1
.
MT132 – M132 / TMA Page 16
of 19
2023/2024 Fall
[
1
−
1
2
4
1
0
1
−
1
2
−
1
0
0
−
1
−
21
2
0
0
0
0
−
2
−
2
]
Multiply each element of R4
by −12
to make the entry at 4,4
a 1
.
[
1
−
1
2
4
1
0
1
−
1
2
−
1
0
0
1
21
2
0
0
0
0
1
1
]
Performtherow operation R
3
=
R
3
−
212
R
4
¿
makethe entryat
3,4
a
0
.
Performtherow operation R
2
=
R
2
−
2
R
4
¿
makethe entry at
2,4
a
0
Performtherow operation R
1
=
R
1
−
4
R
4
¿
makethe entry at
1,4
a
0
.
Performtherow operation R
2
=
R
2
+
R
3
¿
make theentry at
2,3
a
0
.
Performtherow operation R
1
=
R
1
−
2
R
3
¿
makethe entry at
1,3
a
0
.
Performtherow operation R
1
=
R
1
+
R
2
¿
makethe entry at
1,2
a
0
.
Thereducedrow echelon form.
[
1
0
0
0
29
2
0
1
0
0
−
27
2
0
0
1
0
−
21
2
0
0
0
1
1
]
Usetheresult
¿
declarethe final solution
¿
the system of equations .
w
=
9
2
x
=
−
27
2
y
=
−
21
2
z
=
1
Q−4: [7+8 marks] a)
Determine whether the following set of vectors in R
4
is linearly independent or
linearly dependent.
S
=
{
(
5,3,0,6
)
,
(
4,6,4,12
)
,
(
0,2
,
−
3,4
)
,
(
0,1
,
−
2,2
)
}
.
MT132 – M132 / TMA Page 17
of 19
2023/2024 Fall
b)
Write the vector u
=(
9
,
−
1
,
−
11
)
as a linear combination of the vectors
v
1
=
(
5
,
−
3,1
)
,v
2
=(
4,0
,
−
3
)
and v
3
=(
1
,
−
1,2
)
.
2
V
1
+
v
2
−
5
V
3
=
U
a)
Step Given.. {
(
5,3,0,6
)
,
(
4,6,4,12
)
,
(
0,2
,
−
3,4
)
, (
0,1
,
−
2,2
)
A = 5
4
0
0
3
6
2
1
0
4
−
3
−
2
6
12
4
2
Performing matrix column operation
C
1
— C
3
−
2
C
4
A
=
5
4
0
0
3
6
0
1
0
4
1
−
2
6
12
0
2
A
=
[
5
4
0
3
6
1
6
12
2
]
R3— R3-2R2
A
=
[
5
4
0
3
6
1
0
0
0
]
.
Since, All the elements in row3 are zero Hence, The determinant of the matrix A is zero
Det A = 0
The vectors are linearly dependent
b)
U
=(
9
,
−
1
,
−
11
)
MT132 – M132 / TMA Page 18
of 19
2023/2024 Fall
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V
1
=
(
5
,
−
3,1
)
,V
2
=
(
4,0
,
−
3
)
∧
V
3
=
(
5
,
−
3,1
)
,
Consider C
1
V
1
+
¿
C2V2+ C3V3 = U
C
1
(
5
,
−
3,1
)
+
¿
C2
(
4,0
,
−
3
)
+ C3
(
5
,
−
3,1
)
= (
9
,
−
1
,
−
11
)
[
1
−
3
2
−
3
0
−
1
5
4
1
]
[
¿
C
1
C
2
C
3
]
= [
¿
11
−
1
9
]
Performing matrix column operation
R
2
—
>
R
2
+
3
R
1
,R
3
— R
3
−
5
R
1
[
1
−
3
2
0
−
9
5
0
19
−
9
]
[
¿
C
1
C
2
C
3
]
= [
¿
11
−
34
64
]
R
3
— R
3
+
2
R
2
[
1
−
3
2
0
−
9
5
0
1
1
]
[
¿
C
1
C
2
C
3
]
= [
¿
11
−
34
−
4
]
R
1
—
>
R
1
+
3
R
3
, R
2
—
¿
R
2
−
9
R
3
[
1
0
5
0
0
14
0
1
2
]
[
¿
C
1
C
2
C
3
]
= [
¿
−
23
−
70
−
4
]
[
1
0
5
0
0
1
0
1
2
]
[
¿
C
1
C
2
C
3
]
= [
¿
−
23
−
5
−
4
]
R
1
—
>
R
1
−
5
R
2
, R
3
—
¿
R
3
−
R
2
[
1
0
0
0
0
1
0
1
0
]
[
¿
C
1
C
2
C
3
]
= [
¿
2
−
5
1
]
C
1
=
2
,C
2
=
1,
C
3
=−
5
End of the answer MT132 – M132 / TMA Page 19
of 19
2023/2024 Fall