If `Delta=[(a-b-c,2a,2a),(2b,b-c-a,2b),(2c,2c,c-a-b)]`Perform `R_1rarrR_1+R_2+R_3`on`Delta`. Take (a+b+c) common from `R_1` & rewrite `Delta`.

Prove that abs[[a-b-c,2a,2a],[2b,b-c-a,2b],[2c,2c,c-a-b]]=(a+b+c)^3

Evaluate Delta=|[1, a, bc],[ 1, b, c a],[ 1, c, a b]|

Given Delta=[(a+b,b+c,c+a),(b+c,c+a,a+b),(c+a,a+b,b+c)] Express given determinant as the sum of 8 determinants.

Using properties of determinants prove the following. abs[[a-b-c,2a,2a],[2b,b-c-a,2b],[2c,2c,c-a-b]]=(a+b+c)^3

Show that abs[[1,a,a^2],[1,b,b^2],[1,c,c^2]]=(a-b)(b-c)(c-a)

Show that Delta=abs[[-a^2,ab,ac],[ba,-b^2,bc],[ac,bc,-c^2]]=4a^2b^2c^2

If a,b, and c are non - zero real numbers, then Delta=|(b^(2)c^(2),bc,b+c),(c^(2)a^(2),ca,c+a),(a^(2)b^(2),ab,a+b)| is equal to a) abc b) a^(2)b^(2)c^(2) c)bc+ca+ab d)None of these

If Delta_(1)|(a_(1)^(2)+b_(1)+c_(1),a_(1)a_(2)+b_(2)+c_(2),a_(1)a_(3)+b_(3)+c_(3)),(b_(1)b_(2)+c_(1),b_(2)^(2)+c_(2),b_(2)b_(3)+c_(3)),(c_(3)c_(1),c_(3)c_(2),c_(3)^(2))|andDelta_(2)=|(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)),(a_(3),b_(3),c_(3))| then Delta_(1) // Delta_(2) is equal to a) a_(1)b_(2)c_(3) b) a_(1)a_(2)a_(3) c) a_(3)b_(2)c_(1) d) a_(1)b_(1)c_(1) +a_(2)b_(2)c_(2)+a_(3)b_(3)c_(3)

If (b-c)^(2) , (c-a)^(2), (a-b)^(2) are in A.P. then prove that (1)/(b-c) , (1)/(c-a), (1)/(a-b) are also in A.P.

Let Delta_(1)=|(ap^(2),2ap,1),(aq^(2),2aq,1),(ar^(2),2ar,1)|andDelta_(2)=|(apq,a(p+q),1),(aqr,a(q+r),1),(arp,a(r+p),1)| then a) Delta_(1)= Delta_(2) b) Delta_(2)=2Delta_(1) c) Delta_(1) = 2Delta_(2) d) Delta_(1) +2Delta_(2) = 0