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[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
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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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)
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 +
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 =
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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