Determinant of order 3

Consider the set A of all determinants of order 3 with entries 0 or 1 only.Let B be the subset of A consisting of all determinants with value 1.Let C be the subset of the set of all determinants with value -1. Then

If the determinant |a+p l+x u+fb+q x+y v+gc+r n+z w+h| splits into exactly K determinants of order 3, each element of which contains only one term, then the values of K is

If the determinant |(x+a,p+u,l+f),(y+b,q+v,m+g),(z+c,r+w,n+h)| splits into exactly k determinants of order 3, each element of which contains onlyone term, then the value of k is 8.

Consider the determinants Delta=|{:(2,-1,3),(1,1,1),(1,-1,1):}|,Delta'=|{:(2,0,-2),(-2,-1,1),(-4,1,3):}| STATEMENT-1 Delta'=Delta^2 . STATEMENT-2 : The determinant formed by the cofactors of the elements of a determinant of order 3 is equal in value to the square of the original determinant.

Without expanding a determinant at any stage, show that |{:(x^(2)+x, x+1, x+2),(2x^(2) +3x-1, 3x, 3x-3),(x^(2) +2x+3, 2x-1, 2x-1):}|=xA+B where A and B are determinants of order 3 not involving x.

Definition and Determinant of order 1order 2

If D is determinant of order three of Delta is a determinant formed by the cofactors of determinant D: then

A determinant is chosen at random from the set of all determinants of order two with elements 0 or 1 only. Probability that the determinant chosen has positive value is