`A={a, b, c}, B={2, 1, 0}, f={(a, 2), (b, 0), (c, 2)}.` Then f is

If f={(a, 1), (b, -2), (c, 3)}, g={(a, -2), (b, 0), (c, 1)} then f^(2)+g^(2)=

If f={(a, 0), (b, 2), (c, -3)}, g={(a, -1), (b, 1), (c, 2)} then 2f-3g=

If f={(a, 0), (b, -2), (c, 3)}, g={a, -2), (b, 0), (c,1)} then f//g=

If ={(a,1),(b,-2),(c,3)},g={(a,-2),(b,0),(c,1)} then f^(2)+g^(2)=

If a!=b!=c\ a n d\ |{:(a, b, c), (a^2,b^2,c^2), (b+c, c+a, a+b):}|=0 then a+b+c=0 b. a b+b c+c a=0 c. a^2+b^2+c^2=a b+b c+c a d. a b c=0

If a!=b!=c\ and\ |{:(a, b, c), (a^2,b^2,c^2), (b+c, c+a, a+b):}|=0 then A. a+b+c=0 B. a b+b c+c a=0 C. a^2+b^2+c^2=a b+b c+c a D. a b c=0

If a!=b!=c\ a n d\ |{:(a, b, c), (a^2,b^2,c^2), (b+c, c+a, a+b):}|=0 then a+b+c=0 b. a b+b c+c a=0 c. a^2+b^2+c^2=a b+b c+c a d. a b c=0

A=[(a,b),(c,d)] a,b,c,d in {-3,-2,-1,0,1,2,3} , f(A)=det(A) then find the probability that f(A)=15

If a!=1, b!=1, c!=1, f(x)= 1/(1-x) and |(a,1,1),(1,b,1),(1,1,c)|=0 then (A) f(a)+f(b)+f(c)=0 (B) f(a)+f(b)+f(c)=1 (C) f(a)+f(b)+f(c)=-1 (D) f(a)f(b)f(c)=1

Let f(x)=a x^(2)+b x+c , a, b, in R, a neq 0 satisfying f(1)+f(2)=0 . Then the equation f(x)=0 has