The document summarizes properties and techniques for computing determinants of matrices. It defines determinants and provides examples of computing 2x2 and 3x3 determinants. It discusses properties of determinants including how determinants change when multiplying a matrix by a scalar or performing row operations. It also defines what makes a matrix singular versus nonsingular.
This presentation describes Matrices and Determinants in detail including all the relevant definitions with examples, various concepts and the practice problems.
This presentation describes Matrices and Determinants in detail including all the relevant definitions with examples, various concepts and the practice problems.
Math 2318 - Test 3In this test we will try something differe.docxandreecapon
Math 2318 - Test 3
In this test we will try something different. The answers are provided, your job is to show the work in how to get that
solution. On problem 1 only A is a vector space. You will show why it is a vector space but you will also show why B
and C are not vector spaces. On question 2 only V is a vector space. You will show why it is a vector space and you
will also show why W and U are not vector spaces.
Solve the problem.
1) Determine which of the following sets is a subspace of Pn for an appropriate value of n.
A: All polynomials of the form p(t) = a + bt2, where a and b are in ℛ
B: All polynomials of degree exactly 4, with real coefficients
C: All polynomials of degree at most 4, with positive coefficients
A) A and B B) C only C) A only D) B only
1)
2) Determine which of the following sets is a vector space.
V is the line y = x in the xy-plane: V = x
y
: y = x
W is the union of the first and second quadrants in the xy-plane: W = x
y
: y ≥ 0
U is the line y = x + 1 in the xy-plane: U = x
y
: y = x + 1
A) U only B) V only C) W only D) U and V
2)
Find a matrix A such that W = Col A.
3) W =
3r - t
4r - s + 3t
s + 3t
r - 5s + t
: r, s, t in ℛ
A)
0 3 -1
4 -1 3
0 1 3
1 -5 1
B)
3 0 -1
4 -1 3
0 1 3
1 -5 1
C)
3 -1
4 3
1 3
1 -5
D)
3 4 0 1
0 -1 1 -5
-1 3 3 1
3)
Determine if the vector u is in the column space of matrix A and whether it is in the null space of A.
4) u =
5
-3
5
, A =
1 -3 4
-1 0 -5
3 -3 6
A) In Col A and in Nul A B) In Col A, not in Nul A
C) Not in Col A, in Nul A D) Not in Col A, not in Nul A
4)
Use coordinate vectors to determine whether the given polynomials are linearly dependent in P2. Let B be the standard
basis of the space P2 of polynomials, that is, let B = 1, t, t2 .
5) 1 + 2t, 3 + 6t2, 1 + 3t + 4t2
A) Linearly dependent B) Linearly independent
5)
Find the dimensions of the null space and the column space of the given matrix.
6) A = 1 -5 -4 3 0
-2 3 -1 -4 1
A) dim Nul A = 2, dim Col A = 3 B) dim Nul A = 4, dim Col A = 1
C) dim Nul A = 3, dim Col A = 2 D) dim Nul A = 3, dim Col A = 3
6)
1
Solve the problem.
7) Let H =
a + 3b + 4d
c + d
-3a - 9b + 4c - 8d
-c - d
: a, b, c, d in ℛ
Find the dimension of the subspace H.
A) dim H = 3 B) dim H = 1 C) dim H = 4 D) dim H = 2
7)
Assume that the matrix A is row equivalent to B. Find a basis for the row space of the matrix A.
8) A =
1 3 -4 0 1
2 4 -5 5 -2
1 -5 0 -3 2
-3 -1 8 3 -4
, B =
1 3 -4 0 1
0 -2 3 5 -4
0 0 -8 -23 17
0 0 0 0 0
A) {(1, 3, -4, 0, 1), (0, -2, 3, 5, -4), (0, 0, -8, -23, 17), (0, 0, 0, 0, 0)}
B) {(1, 3, -4, 0, 1), (0, -2, 3, 5, -4), (0, 0, -8, -23, 17)}
C) {(1, 3, -4, 0, 1), (2, 4, -5, 5), -2, (1, ...
Math 2318 - Test 3In this test we will try something differe.docxandreecapon
Math 2318 - Test 3
In this test we will try something different. The answers are provided, your job is to show the work in how to get that
solution. On problem 1 only A is a vector space. You will show why it is a vector space but you will also show why B
and C are not vector spaces. On question 2 only V is a vector space. You will show why it is a vector space and you
will also show why W and U are not vector spaces.
Solve the problem.
1) Determine which of the following sets is a subspace of Pn for an appropriate value of n.
A: All polynomials of the form p(t) = a + bt2, where a and b are in ℛ
B: All polynomials of degree exactly 4, with real coefficients
C: All polynomials of degree at most 4, with positive coefficients
A) A and B B) C only C) A only D) B only
1)
2) Determine which of the following sets is a vector space.
V is the line y = x in the xy-plane: V = x
y
: y = x
W is the union of the first and second quadrants in the xy-plane: W = x
y
: y ≥ 0
U is the line y = x + 1 in the xy-plane: U = x
y
: y = x + 1
A) U only B) V only C) W only D) U and V
2)
Find a matrix A such that W = Col A.
3) W =
3r - t
4r - s + 3t
s + 3t
r - 5s + t
: r, s, t in ℛ
A)
0 3 -1
4 -1 3
0 1 3
1 -5 1
B)
3 0 -1
4 -1 3
0 1 3
1 -5 1
C)
3 -1
4 3
1 3
1 -5
D)
3 4 0 1
0 -1 1 -5
-1 3 3 1
3)
Determine if the vector u is in the column space of matrix A and whether it is in the null space of A.
4) u =
5
-3
5
, A =
1 -3 4
-1 0 -5
3 -3 6
A) In Col A and in Nul A B) In Col A, not in Nul A
C) Not in Col A, in Nul A D) Not in Col A, not in Nul A
4)
Use coordinate vectors to determine whether the given polynomials are linearly dependent in P2. Let B be the standard
basis of the space P2 of polynomials, that is, let B = 1, t, t2 .
5) 1 + 2t, 3 + 6t2, 1 + 3t + 4t2
A) Linearly dependent B) Linearly independent
5)
Find the dimensions of the null space and the column space of the given matrix.
6) A = 1 -5 -4 3 0
-2 3 -1 -4 1
A) dim Nul A = 2, dim Col A = 3 B) dim Nul A = 4, dim Col A = 1
C) dim Nul A = 3, dim Col A = 2 D) dim Nul A = 3, dim Col A = 3
6)
1
Solve the problem.
7) Let H =
a + 3b + 4d
c + d
-3a - 9b + 4c - 8d
-c - d
: a, b, c, d in ℛ
Find the dimension of the subspace H.
A) dim H = 3 B) dim H = 1 C) dim H = 4 D) dim H = 2
7)
Assume that the matrix A is row equivalent to B. Find a basis for the row space of the matrix A.
8) A =
1 3 -4 0 1
2 4 -5 5 -2
1 -5 0 -3 2
-3 -1 8 3 -4
, B =
1 3 -4 0 1
0 -2 3 5 -4
0 0 -8 -23 17
0 0 0 0 0
A) {(1, 3, -4, 0, 1), (0, -2, 3, 5, -4), (0, 0, -8, -23, 17), (0, 0, 0, 0, 0)}
B) {(1, 3, -4, 0, 1), (0, -2, 3, 5, -4), (0, 0, -8, -23, 17)}
C) {(1, 3, -4, 0, 1), (2, 4, -5, 5), -2, (1, ...
How to Create Map Views in the Odoo 17 ERPCeline George
The map views are useful for providing a geographical representation of data. They allow users to visualize and analyze the data in a more intuitive manner.
The Roman Empire A Historical Colossus.pdfkaushalkr1407
The Roman Empire, a vast and enduring power, stands as one of history's most remarkable civilizations, leaving an indelible imprint on the world. It emerged from the Roman Republic, transitioning into an imperial powerhouse under the leadership of Augustus Caesar in 27 BCE. This transformation marked the beginning of an era defined by unprecedented territorial expansion, architectural marvels, and profound cultural influence.
The empire's roots lie in the city of Rome, founded, according to legend, by Romulus in 753 BCE. Over centuries, Rome evolved from a small settlement to a formidable republic, characterized by a complex political system with elected officials and checks on power. However, internal strife, class conflicts, and military ambitions paved the way for the end of the Republic. Julius Caesar’s dictatorship and subsequent assassination in 44 BCE created a power vacuum, leading to a civil war. Octavian, later Augustus, emerged victorious, heralding the Roman Empire’s birth.
Under Augustus, the empire experienced the Pax Romana, a 200-year period of relative peace and stability. Augustus reformed the military, established efficient administrative systems, and initiated grand construction projects. The empire's borders expanded, encompassing territories from Britain to Egypt and from Spain to the Euphrates. Roman legions, renowned for their discipline and engineering prowess, secured and maintained these vast territories, building roads, fortifications, and cities that facilitated control and integration.
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Roman architecture and engineering achievements were monumental. They perfected the arch, vault, and dome, constructing enduring structures like the Colosseum, Pantheon, and aqueducts. These engineering marvels not only showcased Roman ingenuity but also served practical purposes, from public entertainment to water supply.
How to Make a Field invisible in Odoo 17Celine George
It is possible to hide or invisible some fields in odoo. Commonly using “invisible” attribute in the field definition to invisible the fields. This slide will show how to make a field invisible in odoo 17.
This is a presentation by Dada Robert in a Your Skill Boost masterclass organised by the Excellence Foundation for South Sudan (EFSS) on Saturday, the 25th and Sunday, the 26th of May 2024.
He discussed the concept of quality improvement, emphasizing its applicability to various aspects of life, including personal, project, and program improvements. He defined quality as doing the right thing at the right time in the right way to achieve the best possible results and discussed the concept of the "gap" between what we know and what we do, and how this gap represents the areas we need to improve. He explained the scientific approach to quality improvement, which involves systematic performance analysis, testing and learning, and implementing change ideas. He also highlighted the importance of client focus and a team approach to quality improvement.
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
We all have good and bad thoughts from time to time and situation to situation. We are bombarded daily with spiraling thoughts(both negative and positive) creating all-consuming feel , making us difficult to manage with associated suffering. Good thoughts are like our Mob Signal (Positive thought) amidst noise(negative thought) in the atmosphere. Negative thoughts like noise outweigh positive thoughts. These thoughts often create unwanted confusion, trouble, stress and frustration in our mind as well as chaos in our physical world. Negative thoughts are also known as “distorted thinking”.
Instructions for Submissions thorugh G- Classroom.pptxJheel Barad
This presentation provides a briefing on how to upload submissions and documents in Google Classroom. It was prepared as part of an orientation for new Sainik School in-service teacher trainees. As a training officer, my goal is to ensure that you are comfortable and proficient with this essential tool for managing assignments and fostering student engagement.
Synthetic Fiber Construction in lab .pptxPavel ( NSTU)
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
2. Ch03_2
3.1 Introduction to Determinants
Definition
The determinant of a 2 2 matrix A is denoted |A| and is given
by
Observe that the determinant of a 2 2 matrix is given by the
different of the products of the two diagonals of the matrix.
The notation det(A) is also used for the determinant of A.
21
12
22
11
22
21
12
11
a
a
a
a
a
a
a
a
Example 1
1
3
4
2
A
14
12
2
))
3
(
4
(
)
1
2
(
1
3
4
2
)
det(
A
3. Ch03_3
Definition
Let A be a square matrix.
The minor of the element aij is denoted Mij and is the determinant
of the matrix that remains after deleting row i and column j of A.
The cofactor of aij is denoted Cij and is given by
Cij = (–1)i+j Mij
Note that Cij = Mij or Mij .
5. Ch03_5
Definition
The determinant of a square matrix is the sum of the products
of the elements of the first row and their cofactors.
These equations are called cofactor expansions of |A|.
n
nC
a
C
a
C
a
C
a
A
n
n
A
C
a
C
a
C
a
C
a
A
A
C
a
C
a
C
a
A
A
1
1
13
13
12
12
11
11
14
14
13
13
12
12
11
11
13
13
12
12
11
11
,
is
If
4,
4
is
If
3,
3
is
If
6. Ch03_6
Example 3
Evaluate the determinant of the following matrix A.
1
2
4
1
0
3
1
2
1
A
Solution
2
4
0
3
)
1
)(
1
(
1
4
1
3
)
1
(
2
1
2
1
0
)
1
(
1 4
3
2
13
13
12
12
11
11
C
a
C
a
C
a
A
6
6
2
2
)]
4
0
(
)
2
3
[(
)]
4
1
(
)
1
3
[(
2
)]
2
1
(
)
1
0
[(
7. Ch03_7
Theorem 3.1
The determinant of a square matrix is the sum of the products of
the elements of any row or column and their cofactors.
ith row expansion:
jth column expansion: nj
nj
j
j
j
j
in
in
i
i
i
i
C
a
C
a
C
a
A
C
a
C
a
C
a
A
2
2
1
1
2
2
1
1
Example 4
Find the determinant of the following matrix using the second row.
1
2
4
1
0
3
1
2
1
A
Solution
6
6
0
12
)]
4
2
(
)
2
1
[(
1
)]
4
1
(
)
1
1
[(
0
)]
2
1
(
)
1
2
[(
3
2
4
2
1
1
1
4
1
1
0
1
2
1
2
3
23
23
22
22
21
21
C
a
C
a
C
a
A
8. Ch03_8
Example 5
Evaluate the determinant of the following 4 4 matrix.
3
0
1
0
5
3
2
7
2
0
1
0
4
0
1
2
Solution
6
)
2
3
(
6
3
1
2
1
)
2
(
3
3
1
0
2
1
0
4
1
2
3
)
(
0
)
(
3
)
(
0
)
(
0 43
33
23
13
43
43
33
33
23
23
13
13
C
C
C
C
C
a
C
a
C
a
C
a
A
9. Ch03_9
Example 6
Solve the following equation for the variable x.
7
2
1
1
x
x
x
Solution
Expand the determinant to get the equation
7
)
1
)(
1
(
)
2
(
x
x
x
Proceed to simplify this equation and solve for x.
3
or
2
0
)
3
)(
2
(
0
6
7
1
2
2
2
x
x
x
x
x
x
x
x
There are two solutions to this equation, x = – 2 or 3.
10. Ch03_10
Computing Determinants of 2 2
and 3 3 Matrices
22
21
12
11
a
a
a
a
A 21
12
22
11
A a
a
a
a
33
32
31
23
22
21
13
12
11
a
a
a
a
a
a
a
a
a
A
32
31
22
21
12
11
33
32
31
23
22
21
13
12
11
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
left)
right to
from
products
(diagonal
right)
left to
from
products
(diagonal
A
33
21
12
32
23
11
31
22
13
32
21
13
31
23
12
33
22
11
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
12. Ch03_12
Let A be an n n matrix and c be a nonzero scalar.
(a) If then |B| = c|A|.
(b) If then |B| = –|A|.
(c) If then |B| = |A|.
3.2 Properties of Determinants
Theorem 3.2
Proof (a)
|A| = ak1Ck1 + ak2Ck2 + … + aknCkn
|B| = cak1Ck1 + cak2Ck2 + … + caknCkn
|B| = c|A|.
B
A
cRk
B
A
Rj
Ri
B
A
cRj
Ri
14. Ch03_14
Example 2
If |A| = 12 is known.
Evaluate the determinants of the following matrices.
,
10
4
2
5
2
0
3
4
1
A
16
4
0
5
2
0
3
4
1
)
c
(
5
2
0
10
4
2
3
4
1
(b)
10
12
2
5
6
0
3
12
1
)
a
( 3
2
1 B
B
B
Solution
(a) Thus |B1| = 3|A| = 36.
(b) Thus |B2| = – |A| = –12.
(c) Thus |B3| = |A| = 12.
1
2
3
B
A
C
2
3
2
B
A
R
R
3
1
2
3
B
A
R
R
15. Ch03_15
Theorem 3.3
Let A be a square matrix. A is singular if
(a) all the elements of a row (column) are zero.
(b) two rows (columns) are equal.
(c) two rows (columns) are proportional. (i.e., Ri=cRj)
Proof
(a) Let all elements of the kth row of A be zero.
0
0
0
0 2
1
2
2
1
1
kn
k
k
kn
kn
k
k
k
k C
C
C
C
a
C
a
C
a
A
(c) If Ri=cRj, then , row i of B is [0 0 … 0].
|A|=|B|=0
B
A
cRj
Ri
Definition
A square matrix A is said to be singular if |A|=0.
A is nonsingular if |A|0.
16. Ch03_16
Example 3
Show that the following matrices are singular.
8
4
2
4
2
1
3
1
2
(b)
9
0
4
1
0
3
7
0
2
)
(a B
A
Solution
(a) All the elements in column 2 of A are zero. Thus |A| = 0.
(b) Row 2 and row 3 are proportional. Thus |B| = 0.
17. Ch03_17
Theorem 3.4
Let A and B be n n matrices and c be a nonzero scalar.
(a) |cA| = cn|A|.
(b) |AB| = |A||B|.
(c) |At| = |A|.
(d) (assuming A–1 exists)
A
A
1
1
Proof
(a)
A
c
cA
cA
A n
cRn
cR
cR
...,
,
2
,
1
(d)
A
A
I
A
A
A
A
1
1 1
1
1
18. Ch03_18
Example 4
If A is a 2 2 matrix with |A| = 4, use Theorem 3.4 to compute
the following determinants.
(a) |3A| (b) |A2| (c) |5AtA–1|, assuming A–1 exists
Solution
(a) |3A| = (32)|A| = 9 4 = 36.
(b) |A2| = |AA| =|A| |A|= 4 4 = 16.
(c) |5AtA–1| = (52)|AtA–1| = 25|At||A–1| .
25
1
25
A
A
Example 5
Prove that |A–1AtA| = |A|
Solution
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A t
t
t
t
1
)
( 1
1
1
1
19. Ch03_19
Example 6
Prove that if A and B are square matrices of the same size, with A
being singular, then AB is also singular. Is the converse true?
Solution
()
|A| = 0 |AB| = |A||B| = 0
Thus the matrix AB is singular.
()
|AB| = 0 |A||B| = 0 |A| = 0 or |B| = 0
Thus AB being singular implies that either A or B is singular.
The inverse is not true.
20. Ch03_20
Homework
Exercise 3.2 pp. 170-171: 1, 3, 5, 7, 9, 13
Exercise 11
Prove the following identity without evaluating the determinants.
w
v
u
r
q
p
f
d
b
w
v
u
r
q
p
e
c
a
w
v
u
r
q
p
f
e
d
c
b
a
v
u
q
p
f
e
w
u
r
p
d
c
w
v
r
q
b
a
w
v
u
r
q
p
f
e
d
c
b
a
)
(
)
(
)
(
21. Ch03_21
3.3 Numerical Evaluation of a
Determinant
Definition
A square matrix is called an upper triangular matrix if all the
elements below the main diagonal are zero.
It is called a lower triangular matrix if all the elements above the
main diagonal are zero.
triangular
upper
1
0
0
0
9
0
0
0
5
3
2
0
7
0
4
1
,
9
0
0
5
1
0
2
8
3
triangular
lower
1
8
5
4
0
2
0
7
0
0
4
1
0
0
0
8
,
8
9
3
0
1
2
0
0
7
22. Ch03_22
.
find
,
5
0
0
4
3
0
9
1
2
Let A
A
Theorem 3.5
The determinant of a triangular matrix is the product of its
diagonal elements.
Example 1
Proof
nn
nn
n
n
nn
n
n
nn
n
n
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
22
11
4
44
3
34
33
22
11
3
33
2
23
22
11
2
22
1
12
11
0
0
0
0
0
0
0
0
0
.
30
)
5
(
3
2
ol.
A
S
Numerical Evaluation of a Determinant
26. Ch03_26
Example 5
Evaluation the determinant.
1
5
6
6
4
3
2
2
3
2
1
1
2
0
1
1
Solution
11
5
0
0
0
3
0
0
5
2
0
0
2
0
1
1
R1
)
6
(
R4
R1
)
2
(
R3
R1
R2
1
5
6
6
4
3
2
2
3
2
1
1
2
0
1
1
diagonal element is zero and
all elements below this
diagonal element are zero.
0
27. Ch03_27
3.3 Determinants, Matrix Inverse,
and Systems of Linear Equations
Definition
Let A be an n n matrix and Cij be the cofactor of aij.
The matrix whose (i, j)th element is Cij is called the matrix of
cofactors of A.
The transpose of this matrix is called the adjoint of A and is
denoted adj(A).
cofactors
of
matrix
2
1
2
22
21
1
12
11
nn
n
n
n
n
C
C
C
C
C
C
C
C
C
matrix
adjoint
2
1
2
22
21
1
12
11
t
C
C
C
C
C
C
C
C
C
nn
n
n
n
n
28. Ch03_28
Example 1
Give the matrix of cofactors and the adjoint matrix of the
following matrix A.
5
3
1
2
4
1
3
0
2
A
Solution The cofactors of A are as follows.
14
5
3
2
4
11
C 3
5
1
2
1
12
C 1
3
1
4
1
13
C
9
5
3
3
0
21
C 7
5
1
3
2
22
C 6
3
1
0
2
23
C
12
2
4
3
0
31
C 1
2
1
3
2
32
C 8
4
1
0
2
33
C
The matrix of cofactors
of A is
8
6
1
1
7
3
12
9
14
)
(
adj A
8
1
12
6
7
9
1
3
14
The adjoint of A is
29. Ch03_29
Theorem 3.6
Let A be a square matrix with |A| 0. A is invertible with
)
(
adj
1
1
A
A
A
Proof
Consider the matrix product Aadj(A). The (i, j)th element of this
product is
jn
in
j
i
j
i
jn
j
j
in
i
i
C
a
C
a
C
a
C
C
C
a
a
a
A
j
A
i
j
i
2
2
1
1
2
1
2
1
))
adj(
of
(column
)
of
(row
element
th
)
,
(
30. Ch03_30
Therefore
j
i
j
i
A
j
i
if
0
if
element
th
)
.
( A adj(A) = |A|In
Proof of Theorem 3.6
.
2
2
1
1 A
C
a
C
a
C
a jn
in
j
i
j
i
If i = j,
If i j, let .
by
replaced
is
B
A
Ri
Rj
.
0
So 2
2
1
1
B
C
a
C
a
C
a jn
in
j
i
j
i
Since |A| 0, n
I
A
A
A
)
(
adj
1
Similarly, .
n
I
A
A
A
)
(
adj
1
Thus )
(
adj
1
1
A
A
A
row i = row j in B
Matrices A and B have the same cofactors
Cj1, Cj2, …, Cjn.
31. Ch03_31
Theorem 3.7
A square matrix A is invertible if and only if |A| 0.
Proof
() Assume that A is invertible.
AA–1 = In.
|AA–1| = |In|.
|A||A–1| = 1
|A| 0.
() Theorem 3.6 tells us that if |A| 0, then A is invertible.
A–1 exists if and only if |A| 0.
32. Ch03_32
Example 2
Use a determinant to find out which of the following matrices are
invertible.
0
8
2
2
1
1
1
2
1
1
0
1
7
12
4
3
4
2
1
2
2
4
2
3
1
1 D
C
B
A
Solution
|A| = 5 0. A is invertible.
|B| = 0. B is singular. The inverse does not exist.
|C| = 0. C is singular. The inverse does not exist.
|D| = 2 0. D is invertible.
33. Ch03_33
Example 3
Use the formula for the inverse of a matrix to compute the inverse
of the matrix
5
3
1
2
4
1
3
0
2
A
Solution
|A| = 25, so the inverse of A exists.We found adj(A) in Example 1
8
6
1
1
7
3
12
9
14
)
(
adj A
25
8
25
6
25
1
25
1
25
7
25
3
25
12
25
9
25
14
8
6
1
1
7
3
12
9
14
25
1
)
adj(
1
1
A
A
A
34. Ch03_34
Homework
Exercise 3.3 page 178-179: 1, 3, 5, 7.
Exercise
Show that if A = A-1, then |A| = 1.
Show that if At = A-1, then |A| = 1.
35. Ch03_35
Theorem 3.8
Let AX = B be a system of n linear equations in n variables.
(1) If |A| 0, there is a unique solution.
(2) If |A| = 0, there may be many or no solutions.
Proof
(1) If |A| 0
A–1 exists (Thm 3.7)
there is then a unique solution given by X = A–1B (Thm 2.9).
(2) If |A| = 0
since A C implies that if |A|0 then |C|0 (Thm 3.2).
the reduced echelon form of A is not In.
The solution to the system AX = B is not unique.
many or no solutions.
36. Ch03_36
Example 4
Determine whether or not the following system of equations has
an unique solution.
9
4
7
5
3
4
2
2
3
3
3
2
1
3
2
1
3
2
1
x
x
x
x
x
x
x
x
x
Solution
Since
0
1
4
7
3
1
4
2
3
3
Thus the system does not have an unique solution.
37. Ch03_37
Theorem 3.9 Cramer’s Rule
Let AX = B be a system of n linear equations in n variables such
that |A| 0. The system has a unique solution given by
Where Ai is the matrix obtained by replacing column i of A with B.
A
A
x
A
A
x
A
A
x n
n
,
...
,
, 2
2
1
1
Proof
|A| 0 the solution to AX = B is unique and is given by
B
A
A
B
A
X
)
(
adj
1
1
38. Ch03_38
xi, the ith element of X, is given by
)
(
1
1
)]
(
adj
of
row
[
1
2
2
1
1
2
1
2
1
ni
n
i
i
n
ni
i
i
i
C
b
C
b
C
b
A
b
b
b
C
C
C
A
B
A
i
A
x
Thus
A
A
x i
i
Proof of Cramer’s Rule
the cofactor expansion of |Ai|
in terms of the ith column
39. Ch03_39
Example 5
Solving the following system of equations using Cramer’s rule.
6
3
2
5
5
2
2
3
3
2
1
3
2
1
3
2
1
x
x
x
x
x
x
x
x
x
Solution
The matrix of coefficients A and column matrix of constants B are
6
5
2
and
3
2
1
1
5
2
1
3
1
B
A
It is found that |A| = –3 0. Thus Cramer’s rule be applied. We
get
6
2
1
5
5
2
2
3
1
3
6
1
1
5
2
1
2
1
3
2
6
1
5
5
1
3
2
3
2
1 A
A
A
40. Ch03_40
Giving
Cramer’s rule now gives
9
,
6
,
3 3
2
1
A
A
A
3
3
9
,
2
3
6
,
1
3
3 3
3
2
2
1
1
A
A
x
A
A
x
A
A
x
The unique solution is .
3
,
2
,
1 3
2
1
x
x
x
41. Ch03_41
Example 6
Determine values of for which the following system of
equations has nontrivial solutions.Find the solutions for each
value of .
0
)
1
(
2
0
)
4
(
)
2
(
2
1
2
1
x
x
x
x
Solution
homogeneous system
x1 = 0, x2 = 0 is the trivial solution.
nontrivial solutions exist many solutions
= – 3 or = 2.
0
1
2
4
2
0
)
3
)(
2
(
0
6
2
0
)
4
(
2
)
1
)(
2
(
42. Ch03_42
= – 3 results in the system
This system has many solutions, x1 = r, x2 = r.
0
2
2
0
2
1
2
1
x
x
x
x
= 2 results in the system
This system has many solutions, x1 = – 3r/2, x2 = r.
0
3
2
0
6
4
2
1
2
1
x
x
x
x