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a)�
A=
1 2 3
6 5 4
7 8 -6
TRACE = 1+5-6 = 0 �..ANSWER
b)�.
LET A = [A(I,J)]�.AND�.B=[B(I,J)]
WITH I & J VARYINF FROM 1 TO N
SO �C=[C(I,J)] = [A+B] = [{A(I,J)+B(I,J)}]
THAT IS �
TRACE OF C = [C(1,1)+C(2,2)+�.+C(N,N)] �THAT IS �
TRACE OF C =[{A(1,1+B(1,1)}+{A(2,2)+B(2,2)}+�..+{A(N,N)+B(N,N)}]��������1
TRACE OF A = [A(1,1)+A(2,2)+�.+A(N,N)] ���������..2
TRACE OF B = [B(1,1)+B(2,2)+�.+B(N,N)] ���������.3
ADDING 2 & 3 & COMPARING WITH 1 , WE CAN EASILY SEE THAT
TRACE[A]+TRACE[B]=TRACE[C]=TRACE[A+B] �����PROVED
c)��
LET A = [A(I,J)]��SO �.A' = A TRANSPOSE = [A(J,I)]
WITH I & J VARYINF FROM 1 TO N
IN CASE OF TRANSPOSE , THE DIAGONAL ELEMENTS REMAIN THE SAME
SINCE �.A(I,J) = A(J,I) �.IF I=J �.SO
TRACE OF A = [A(1,1)+A(2,2)+�.+A(N,N)] ���������..2
TRACE OF A' = [A(1,1)+A(2,2)+�.+A(N,N)] ���������.3
COMPARING WITH 2 & 3 , WE CAN EASILY SEE THAT
TRACE[A] = TRACE[ A' ]=TRACE [ A TRANSPOSE ] �����PROVED
d)�
A= B=
1 2 4 6
4 3 5 7
TRACE [A]= 1+3=4 TRACE [B]=4+7=11
A*B =
14 20
31 45
TRACE[A*B]=14+45= 59
[TRACE(A)]*[TRAC(B)] = 4*11 = 44
SO WE CONCLUDE THAT ..
TRACE[A*B] IS NOT EQUAL TO [TRACE(A)]* [ TRACE(B) ]

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Jul 24th, 2015

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