If `|(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)),(a_(3),b_(3),c_(3))| =5`, then the value of
`Delta = |(b_(2) c_(3) - b_(3) c_(2),a_(3) c_(2) - a_(2) c_(3),a_(2) b_(3) -a_(3) b_(2)),(b_(3) c_(1) - b_(1) c_(3),a_(1) c_(3) - a_(3) c_(1),a_(3) b_(1) - a_(1) b_(3)),(b_(1) c_(2) - b_(2) c_(1),a_(2) c_(1) - a_(1) c_(2),a_(1) b_(2) - a_(2) b_(1))|` is

Options

A. 5
B. 25
C. 125
D. 0

Answer

B

Explanation

We know that if A is a square matrix of order n and B is the matrix of cofactors of elements of A. Then,
`|B| = |A|^(n-1)`
Here, `Delta` is the determinant of cofactors of elements of matrix A given by
`[(a_(1),b_(1),c_(2)),(a_(2),b_(2),c_(2)),(a_(3),b_(3),c_(3))]`
`:. Delta = |A|^(3 -1) = |A|^(2) = 5^(2) = 25`