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The entries of an n × n array of numbers are denoted by xij , with 1 = i, j = n. For all i, j, k, 1 = i, j, k = n the following equality holds xij + xjk +xki = 0. Prove that there exist numbers t1, t2, . . . , tn such that xij = ti -tj for all i, j, 1 = i, j = n.

The entries of an n × n array of numbers are denoted by xij , with
1 = i, j = n. For all i, j, k, 1 = i, j, k = n the following equality holds
xij + xjk +xki = 0.
Prove that there exist numbers t1, t2, . . . , tn such that
xij = ti -tj
for all i, j, 1 = i, j = n.

Grade:11

1 Answers

bharat bajaj IIT Delhi
askIITians Faculty 122 Points
7 years ago
First, set i = j = k, yields 3.Xii = 0, Hence Xii = 0
For j = k, we have
Xij + Xjj + Xji = 0
So, Xij = -Xji
Now, fix i and j, k goes on from 1 to n,
nXij + Sum(Xjk) + Sum(Xki) = 0
nXij + Sum(Xjk) - Sum(Xik) = 0
now define Ti = Sum(Xik)/n
Therefore, Xij = Ti - Tj
Hence proved

Thanks
Bharat bajaj
IIT Delhi
askiitians faculty

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