If `Delta_1=|(1, 1 ,1),(a, b, c ),(a^2,b^2,c^2)|,Delta_2=|(1,b c, a),(1,c a, b),(1,a b, c)|,t h e n` (a)`Delta_1+Delta_2=0` (b) `Delta_1+2Delta_2=0` (c)`Delta_1=Delta_2` (d) none of these

If Delta_1=|(1,0),(a,b)| and Delta_2=|(1,0),(c,d)| , then Delta_2Delta_1 is equal to :

If Delta_1=|[2a,b,e],[2d,e,f],[4x,2y,2z]|,Delta_2=|[f,2d,e],[2z,4x,2y],[e,2a,b]| , then the value of Delta_1-Delta_2 is

If Delta_1=|[a, b, c],[ x, y, z],[ p, q, r]| and Delta_2=|[q,-b, y],[-p, a,-x],[ r,-c, z]|, without expanding or evaluating Delta_1 and Delta_2 , show that Delta_1+Delta_2=0

Prove that Delta = |{:(1,,1,,1),(a,,b,,c),(bc+a^(2),,ac+b^(2),,ab+c^(2)):}| = 2(a-b)(b-c)(c-a)

If "Delta"_1=|x bb a x b a a x|a n d"Delta"_2=|x b a x| are the given determinants, then "Delta"_1=3("Delta"_2)^2 b. d/(dx)("Delta"_1)=3"Delta"_2 c. d/(dx)("Delta"_1)=3("Delta"_2)^2 d. "Delta"_1=3"Delta"2 3//2

Let Delta_1=|{:(a_1,a_2,a_3),(b_1,b_2,b_3),(c_1,c_2,c_3):}| , Delta_2=|{:(6a_1,2a_2,2a_3),(3b_1,b_2,b_3),(12c_1, 4c_2,4c_3):}| and Delta_3=|{:(3a_1+b_1 , 3a_2+b_2 , 3a_3+b_3),(3b_1,3b_2,3b_3),(3c_1,3c_2,3c_3):}| then Delta_3-Delta_2=kDelta_1 , find k.

If Delta=|(a_1,b_1,c_1),(a_2,b_2,c_2),(a_3,b_3,c_3)| and Delta_1=|(a_1+pb_1,b_1+qc_1,c_1+ra_1),(a_2+pb_2,b_2+qc^2,c^2+ra^2),(a_3+pb_3,b_3+qc_3,c_3+ra_3)| then Delta_1=

Without expanding evaluate the determinant "Delta"=|(1, 1, 1),(a, b, c),( a^2,b^2,c^2)| .

The determinant Delta=|(a^2(a+b),a b,a c),(a b,b^2(a+k),b c),(a c,b c,c^2(1+k))| is divisible by

Let a,b,c , in R not all are equal and Delta_(1)= |{:(a,,b,,c),(b,,c,,a),(c,,a,,b):}| Delta_(2)= |{:(a+2b,,b+3c,,c+4a),(b+2c,,c+3a,,a+4b),(c+2a,,a+3b,,b+4c):}|" then " (Delta_(2))/(Delta_(1)) = "____"