In encercise 10, Find the inverse of the given matnine (if it euists) using Theon-em 3.8 10.11/a 1/C 1/b 1/2 where neither a, b,e nor dis o
In encercise 10, Find the inverse of the given matnine (if it euists) using Theon-em 3.8 10.11/a 1/C 1/b 1/2 where neither a, b,e nor dis o
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
I have added theorem 3.8 as per the question
![166
Theorem 3.8
Chapter 3 Matrices
If A =
and
b]
then A is invertible if ad-bc0, in which case
A¹ =
If ad bc= 0, then A is not invertible.
Similarly,
€ [a b]₁
The expression ad-bc is called the determinant of A, denoted det A. The formula
for the inverse of
(when it exists) is thus
times the matrix obtained by
det A
interchanging the entries on the main diagonal and changing the signs on the other
two entries. In addition to giving this formula, Theorem 3.8 says that a 2 x 2 matrix
A is invertible if and only if det A # 0. We will see in Chapter 4 that the determinant
can be defined for all square matrices and that this result remains true, although there
is no simple formula for the inverse of larger square matrices.
Proof Suppose that det A = ad-bc # Then
[a b][-d-d]=
ad-bc-c
[ad-bc ab + ba] [ad-bc
=
cd-de -cb + da
0
[-][a] =det A[i]
Since det A # 0, we can multiply both sides of each equation by 1/det A to obtain
[1-6]
-01-01
[Note that we have used property (d) of Theorem 3.3.] Thus, the matrix
1 d
det A-c
d
1
det A-c
satisfies the definition of an inverse, so A is invertible. Since the inverse of A is unique,
by Theorem 3.6, we must have
1
A = de
d -b
In the first case,
A = [a b]-[aca
[ac/a
Conversely, assume that ad
bc = 0. We will consider separately the cases where
a 0 and where a = 0. If a # 0, then d= bc/a, so the matrix can be written as
aw
-c
kaw
d-b
b
bc/a]
where k = c/a. In other words, the second row of A is a multiple of the first. Referring
to Example 3.23(b), we see that if A has an inverse
then
ka kb
and the corresponding system of linear equations
+ by
ky-61
+ kby
kax
0
ad - be] = det A[!]
= 1
+bz 0
= 0
+ kbz 1
has no solution. (Why?)
If a = 0, then ad bc = 0 implies that bc = 0, and therefore either b or c is 0.
Thus, A is of the form
о
[2] or [1]
d]
[oo]
• [[ ]]-[9]+[1]
have an inverse. (Verify this.)
Consequently, if ad-bc= 0, then A is not invertible.
Similarly,
2
cannot](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F46695490-535e-4e6d-be60-a5ac8c464218%2Fd2824ba6-4ee0-4677-8ded-9c8aaa4094a5%2Fn32veqk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:166
Theorem 3.8
Chapter 3 Matrices
If A =
and
b]
then A is invertible if ad-bc0, in which case
A¹ =
If ad bc= 0, then A is not invertible.
Similarly,
€ [a b]₁
The expression ad-bc is called the determinant of A, denoted det A. The formula
for the inverse of
(when it exists) is thus
times the matrix obtained by
det A
interchanging the entries on the main diagonal and changing the signs on the other
two entries. In addition to giving this formula, Theorem 3.8 says that a 2 x 2 matrix
A is invertible if and only if det A # 0. We will see in Chapter 4 that the determinant
can be defined for all square matrices and that this result remains true, although there
is no simple formula for the inverse of larger square matrices.
Proof Suppose that det A = ad-bc # Then
[a b][-d-d]=
ad-bc-c
[ad-bc ab + ba] [ad-bc
=
cd-de -cb + da
0
[-][a] =det A[i]
Since det A # 0, we can multiply both sides of each equation by 1/det A to obtain
[1-6]
-01-01
[Note that we have used property (d) of Theorem 3.3.] Thus, the matrix
1 d
det A-c
d
1
det A-c
satisfies the definition of an inverse, so A is invertible. Since the inverse of A is unique,
by Theorem 3.6, we must have
1
A = de
d -b
In the first case,
A = [a b]-[aca
[ac/a
Conversely, assume that ad
bc = 0. We will consider separately the cases where
a 0 and where a = 0. If a # 0, then d= bc/a, so the matrix can be written as
aw
-c
kaw
d-b
b
bc/a]
where k = c/a. In other words, the second row of A is a multiple of the first. Referring
to Example 3.23(b), we see that if A has an inverse
then
ka kb
and the corresponding system of linear equations
+ by
ky-61
+ kby
kax
0
ad - be] = det A[!]
= 1
+bz 0
= 0
+ kbz 1
has no solution. (Why?)
If a = 0, then ad bc = 0 implies that bc = 0, and therefore either b or c is 0.
Thus, A is of the form
о
[2] or [1]
d]
[oo]
• [[ ]]-[9]+[1]
have an inverse. (Verify this.)
Consequently, if ad-bc= 0, then A is not invertible.
Similarly,
2
cannot
Transcribed Image Text:In encercise 10, find the inverse of the
given matnine (if it euists) using
Theorem
3.8
10.11/a
1/C
1/6
1/2
where neither a, b,e nor
dis o
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