# need perfect work ....

label Calculus
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schedule 1 Day
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The trace of a square matrix A is the sum of the diagonal terms of A and is denoted by tr(A).

(a) Find a 3x3 matrix A with nonzero entries suchthat tr(A) = 0.

(b) If A and B are both n x n matrices, show that tr(A + B) = tr(A) + tr(A).

(c) Show that tr(A) = tr(AT ).

(d) Select two nonzero 2x2 matrices A and B of your choosing, and check if tr(AB) = tr(A)tr(B).

Jul 24th, 2015

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) ]

Jul 24th, 2015

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Jul 24th, 2015
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Jul 24th, 2015
Nov 22nd, 2017
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