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in Linear Algebra by Loyal (5.4k points)
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Please add proper tags and subtopics. This is from Linear Algebra.

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The diagonal elements at Row i, 1<=i<=100 are written down below in the table. 

1 2 3 4 5 6 7 8 ...... 100
1+3 0 2x3+1 4+3 0 2x6+1 7+3 0   100+3

 

Trace is the sum of the diagonal elements of a matrix so no need to bother about other elements. Moreover it's given that it is diagonal matrix so other elements=0.

For all i%3==2 the elements at dii will be 0. So they won't contribute anything to the trace.

For all i%3==1 are i=1,4,7,10...100

dii=i+3

The series will be :

(1+3)+(4+3) +(7+3)......(100+3)

How many such terms are there? 34 [from 1 to 99 there will be exactly 33 such terms whose mod3 will be 1 and then +1 for 100]

So, 34*3 + (1+4+7+10+....100)

Let a=1, d=3 then we can write this series as

34*3 + (a + (a+d) + (a+2d) + (a+3d) ..... + (a+33d) )

=102 + { 34*a + d(1+2+3+......33) }

=102 + {34 + 3*(33)(34)/2 } = 102 + 1717=1819.

 

For all i%3==0, i's are 3,6,9....100

dii=2i+1

There will be 33 such terms

So series is like :

(2*3+1) + (2*6+1)....(2*99+1)

=1*33 + 2(3+6+9....99)

=33 + 2*3(1+2+3...33)

=33 + 6*(33)*(34)/2 =  3399

So total = 1819+3399=5218

 

by Boss (21.9k points)
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we have to find trace of matrix of order i , where 1<=i<=100.

so from the question we can say that ti follows a triplet where.

if ---> imod3=0  : di =2i+1            [email protected]

        imod3=1  : di=i+3                  [email protected]

       i mod3=2 : di=0                    [email protected]

for i = { 3,6,9................................99 } we use  @1   (33 terms)

di={7,13,19.................................199}                      [email protected]

for i= {1,4,7..................................100} we use @2   (34 terms)

di={4,7,10,....................................103}                      [email protected]

for i={2,5,8.................................98} we use @3     

di = {0,0,0,0,0,...............................}                         [email protected]

therefore trace of matrix = summation of di (1<=i<=100)

                                     =using arithmetic progression sum eqn.

                                     = (n/2) (first term +last term)

for eqn [email protected]

di={7,13,19.................................199}       : n=33, a=7 ,d=6, l=199

sum= n/2 (a+l) = 33/2 (199+7) =3399

for eqn @5 :

di={4,7,10,....................................103}     : n=34 a=4 , d=3 ,l=103

sum = n/2 (a+l) = 34/2 (4+103)  =1819

for eqn @6 :

di = {0,0,0,0,0,...............................}  

sum =0

so total sum = 3399+1819+0 = 5218 answer.

 

 

 

by Boss (11.9k points)

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