find the larget value of a third- order determinant whose elements are 0 or 1.

Options

A. 1
B. 0
C. 2
D. 3

Answer

C

Explanation

Let `Delta = |(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)),(a_(3),b_(3),c_(3))|` be a determinant of order 3.
Then,
`Delta = a_(1) b_(2) c_(3) + a_(3) b_(1) c_(2) + a_(2) b_(3) c_(1) - a_(1) b_(3) c_(2) - a_(2) b_(1) c_(3) - a_(3) b_(2) c_(1)`
`Delta = (a_(1) b_(2) c_(3) + a_(3) b_(1) c_(2) + a_(2) b_(3) c_(1)) - (a_(1) b_(3) c_(2) + a_(2) b_(1) c_(3) + a_(3) b_(2) c_(1))`
Since each element of `Delta` is either 1 or 0. Therefore, the value of the determinant cannot exceed 3.
Clearly, the value of `Delta` is maximum when the value of each term in first bracket is 1 and the value of each term in the second bracket is zero. But `a_(1) b_(2) c_(3) = a_(3) b_(1) c_(2) = a_(2) b_(3) c_(1) = 1` implies that every element of the determinant `Delta` is 1 and in that case `Delta = 0`. Thus, we may have
`Delta = |(0,1,1),(1,0,1),(1,1,0)| =2`