The following matrices are used in Exercises 35-45. 31 A-1 - - - - - - - - [19] [3₂2]· [ B = 12 21 = 02 12 (9) In Exercises 35-45, use Theorem 17 and the matrices in (9) to form Q-¹, where Q is the given matrix. 35. Q = AC 36. Q = CA 37. Q = AT 38. Q = ATC 39. Q = CTAT 41. Q = CB-1 43. Q = 2A 45. Q = (AC)B-¹ 40. Q = B-¹A 42. Q = B-1 44. Q = 10C
The following matrices are used in Exercises 35-45. 31 A-1 - - - - - - - - [19] [3₂2]· [ B = 12 21 = 02 12 (9) In Exercises 35-45, use Theorem 17 and the matrices in (9) to form Q-¹, where Q is the given matrix. 35. Q = AC 36. Q = CA 37. Q = AT 38. Q = ATC 39. Q = CTAT 41. Q = CB-1 43. Q = 2A 45. Q = (AC)B-¹ 40. Q = B-¹A 42. Q = B-1 44. Q = 10C
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

Transcribed Image Text:THEOREM 17
Proof
Let A and B be (n x n) matrices, each of which has an inverse. Then:
A¹ has an inverse, and (A-1)-¹ = A.
1-1
1.
2.
AB has an inverse, and (AB)-¹ = B-¹A-¹.
3.
If k is a nonzero scalar, then kA has an inverse, and (kA)−¹ = (1/k)A¯¹.
4. AT has an inverse, and (AT)-¹ = (A−¹)ª.
1. Since AA-¹ = A-¹A = I, the inverse of A-¹ is A; that is, (A¯¹)−¹ = A.
2. Note that (AB)(B−¹A−¹) = A(BB¯¹)A¯¹ = A(IA¯¹) = AA-¹ = I. Simi-
larly, (B-¹A−¹)(AB) = I, so, by Definition 13, B¯¹A-¹ is the inverse for AB.
Thus (AB)-¹ = B-¹A-¹.
3. The proof of property 3 is similar to the proofs given for properties 1 and 2 and
is left as an exercise.
![The following matrices are used in Exercises 35-45.
31
12
- [8₂] - - [1 ]·~~- - [ + ]
[³
B
-1 1
2
02
(9)
In Exercises 35-45, use Theorem 17 and the matrices in
(9) to form Q-¹, where Q is the given matrix.
35. Q = AC
36. Q = CA
37. Q = AT
38. Q = ATC
39. Q = CTAT
41. Q = CB-1
43. Q = 2A
45. Q = (AC)B-¹
40. Q = B-¹A
42. Q = B-1
44. Q = 10C](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F624dcfab-19f1-4c4b-b3ec-94b9a5b18d08%2Ffabc55b0-aa60-4aa0-9410-5ca243241dfa%2Fs3vo7u4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The following matrices are used in Exercises 35-45.
31
12
- [8₂] - - [1 ]·~~- - [ + ]
[³
B
-1 1
2
02
(9)
In Exercises 35-45, use Theorem 17 and the matrices in
(9) to form Q-¹, where Q is the given matrix.
35. Q = AC
36. Q = CA
37. Q = AT
38. Q = ATC
39. Q = CTAT
41. Q = CB-1
43. Q = 2A
45. Q = (AC)B-¹
40. Q = B-¹A
42. Q = B-1
44. Q = 10C
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