Vector Spaces Prove property 3 by rearranging the flig Property 3: For every element A, B, C_₂in V=M3x3, (ABB) CAD (BDC) A) Let aij, bij, cij, dij, ei; and the aith entries in A₁B₁C₁D = (ABB) # C, E = A (B+C), respectively 2 Suppose A, B, and C are in V = M₂x3 • from the definition of AB, dij -fai jx bij) x cij and eij aijx (bij x cij) So, (AB) C = A + (PDC) Since aij bij cij are are real #'s dijseij answer for ex FEDCBA M
Vector Spaces Prove property 3 by rearranging the flig Property 3: For every element A, B, C_₂in V=M3x3, (ABB) CAD (BDC) A) Let aij, bij, cij, dij, ei; and the aith entries in A₁B₁C₁D = (ABB) # C, E = A (B+C), respectively 2 Suppose A, B, and C are in V = M₂x3 • from the definition of AB, dij -fai jx bij) x cij and eij aijx (bij x cij) So, (AB) C = A + (PDC) Since aij bij cij are are real #'s dijseij answer for ex FEDCBA M
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
linear alg,
![Vector Spaces
Prove property 3 by rearranging
the flig
Property 3: For every element A, B, C in V=M3x3,
(AB) C= A + (BDC)
A) Let
aij, bij, cij, dij, ei; and the with entries in
A₁B₁C₁D = (ABB) + C, E = A (BOC), respectively
B) Suppose A, B, and C are in V = M3x3
в)
c) from the definition of AB, dij -(aijx bij) x cij and eij =
a i j x (bij x c i j )
420932
D) SO, (AⓇB) ⓇC = A + (PDC)
E
Since aij, bij, cij are real #'s dij=eij
ansue
answer for ex FEDCBA
AXA
wh tast](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F80a45a6e-0b8a-4aa3-8be8-79493f18df43%2F1e77b161-d639-448b-84d5-2218795e8bd6%2Fjjg8zol_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Vector Spaces
Prove property 3 by rearranging
the flig
Property 3: For every element A, B, C in V=M3x3,
(AB) C= A + (BDC)
A) Let
aij, bij, cij, dij, ei; and the with entries in
A₁B₁C₁D = (ABB) + C, E = A (BOC), respectively
B) Suppose A, B, and C are in V = M3x3
в)
c) from the definition of AB, dij -(aijx bij) x cij and eij =
a i j x (bij x c i j )
420932
D) SO, (AⓇB) ⓇC = A + (PDC)
E
Since aij, bij, cij are real #'s dij=eij
ansue
answer for ex FEDCBA
AXA
wh tast
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