[50 Cki 3 1 L1 4 Therefore, aijaij : бікскi = = Ckk = [5 c. aijajk = 3 1 L1 4 3 11 [61 33 5 615 3 0 1 4 = 33 19 7 0ll6 3 0J 7 17 5 0 6] [5 0 51 [31 24 301 1 13 3 3 oll1 4 61 + 19 + 17 = 97 [5 0 6] d. aj bj = 3 1 3 L1 3 = 21 117 18-8 (18) = 13 rank 1, vector rank 0, scalar. 21 rank 2, matrix 18. e. aij bibj = aij bj bi Let's C₁ = aij bj (16 (16 From Problem 2d, ci = 13. Therefore, aij bi bj = cibi = 13 (18) (18) 13 x 4 + 18 x 1 = 102 rank 0. scalar 4 = 16 x 2 +
[50 Cki 3 1 L1 4 Therefore, aijaij : бікскi = = Ckk = [5 c. aijajk = 3 1 L1 4 3 11 [61 33 5 615 3 0 1 4 = 33 19 7 0ll6 3 0J 7 17 5 0 6] [5 0 51 [31 24 301 1 13 3 3 oll1 4 61 + 19 + 17 = 97 [5 0 6] d. aj bj = 3 1 3 L1 3 = 21 117 18-8 (18) = 13 rank 1, vector rank 0, scalar. 21 rank 2, matrix 18. e. aij bibj = aij bj bi Let's C₁ = aij bj (16 (16 From Problem 2d, ci = 13. Therefore, aij bi bj = cibi = 13 (18) (18) 13 x 4 + 18 x 1 = 102 rank 0. scalar 4 = 16 x 2 +
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
Can you write out the steps for c I don’t get where the values in the final matrix are coming from. (Please show multiplication)

Transcribed Image Text:Therefore, aijaij =
c. aijajk
[5
= 3
L1
15
d. aij bj = 3
[1
Cki
=
[5
3
11
0
1 3
4
0 615
3 11 [61 33 5
1 3 0 1 4 = 33 19 7
4
0ll6 3 0J
5
7 171
Sik Cki = Ckk = 61 + 19 + 17 = 97
O 615 0 1 [31 24 301
1 3 3 1 3 = 21 13
4
0ll1 4
0
[17 4
61
18-8
(18)
= 13 rank 1, vector
rank 0, scalar.
21 rank 2, matrix
18
e. aij bibj = aij bj bi
Let's C₁ = aij bj
(16)
From Problem 2d, c₁ = 13. Therefore, aij bi bj = c₁b₁ = 13
(18)
13 x 4 + 18 x 1 = 102 rank 0. scalar
(16)
-88--
(18)
= 16 x 2 +
![The matrix aj and vector b; are specified by
= {3}
[506]
1
4
alj = 3
1
3
0J
b =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8ea503b5-9839-4a05-b482-50f8a035eb54%2F01601e2a-6a3e-4001-a549-2d154cfd7aa4%2Fooz6vw_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The matrix aj and vector b; are specified by
= {3}
[506]
1
4
alj = 3
1
3
0J
b =
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