The entries of an n × n array of numbers are denot

Question

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.

bharat bajaj · Accepted Answer

First, set i = j = k, yields 3.Xii = 0, Hence Xii = 0For j = k, we haveXij + Xjj + Xji = 0So, Xij = -XjiNow, fix i and j, k goes on from 1 to n,nXij + Sum(Xjk) + Sum(Xki) = 0nXij + Sum(Xjk) - Sum(Xik) = 0now define Ti = Sum(Xik)/nTherefore, Xij = Ti - TjHence provedThanksBharat bajajIIT Delhiaskiitians faculty

Algebra> The entries of an n × n array of numbers ...

The entries of an n × n array of numbers are denoted by xij , with\n1 = i, j = n. For all i, j, k, 1 = i, j, k = n the following equality holds\nxij + xjk +xki = 0.\nProve that there exist numbers t1, t2, . . . , tn such that\nxij = ti -tj\nfor all i, j, 1 = i, j = n.

Shrey , 10 Years ago

bharat bajaj

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Algebra> The entries of an n × n array of numbers ...

The entries of an n × n array of numbers are denoted by xij , with\n1 = i, j = n. For all i, j, k, 1 = i, j, k = n the following equality holds\nxij + xjk +xki = 0.\nProve that there exist numbers t1, t2, . . . , tn such that\nxij = ti -tj\nfor all i, j, 1 = i, j = n.

Shrey , 10 Years ago

bharat bajaj

Provide a better Answer & Earn Cool Goodies

LIVE ONLINE CLASSES

Ask a Doubt

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