In a square matrix A of order 3 the elements `a_(ij)` 's are the
sum of the roots of the equation `x^(2) - (a+b) x + ab =0,`
`a_(I,i+1)`'s are the prodeuct of the roots, `a_(I,i-1)` 's are all unity
and the rest of the elements are all zero. The value of the det (A) is equal to

Options

A. 0
B. `(a+b)^(3)`
C. a^(3) - b^(3)`
D. (a^(2)+ b^(2)) (a+b)`

Answer

D

Explanation

`because a_(11) = a_(22) = a_(33) = a+b,`
`a_(12) = a_(23) = ab, a_(21) = a_(32) = 1, a_(13) = a_(31)=0`
`therefore A = [[a+b,ab,0],[1, a+b,ab],[0,1,a+b]]`
`rArr abs(A)= abs((a+b, ab,0),(1, a+b,ab),(0,1,a+b))`
`= (a+b) [(a+b)^(2) - ab]- ab (a+b) = (a+b)(a^(2)+b^(2))`