Lat`A = [a_(ij)]_(3xx 3).` If tr is arithmetic mean of elements of rth row
and `a_(ij )+ a_( jk) + a_(ki)=0` holde for all `1 le i, j, k le 3.`
Matrix A is

Options

A. `t_(1) + t_(2) + t_(3)`
B. zero
C. `(det(A))^(2)`
D. `t_(1) t_(2)t_(3)`

Answer

D

Explanation

`therefore A = [[a_(11) , a_(12),a_(13)],[a_(21),a_(22), a_(23) ],[a_(31), a_(32),a_(33)]]`
`rArr t_(1) = (a_(11) + a_(12)+a_(23))/3 = 0, [because a_(ij) + a_(jk) + a_(ki)=0]`
`t_(2) = (a_(21) + a_(22) + a_(23))/3 = 0`
and `t_(3) = (a_(31) + a_(32) + a_(33))/3 = 0`
`sum _(1lei), sum _(jle 3) a_(ij) =3 (t_(1) + t_(2) + t_(3)) = 0 = t_(1) + t_(2) + t_(3) `
`ne t_(1) t_(2) t_(3) " "[because t_(1) = 0 , t_(2 ) = 0, t_(3) = 0]`
and det ` A = [[a_(11) , a_(12),a_(13)],[a_(21),a_(22), a_(23) ],[a_(31), a_(32),a_(33)]]`
Applying `C_(1)rarr C_(1)+ C_(2) +C_(3),` we get
` = [[0 , a_(12),a_(13)],[0,a_(22), a_(23) ],[0, a_(32),a_(33)]]=0`
`therefore (detA)^(2) = 0`