Let `triangle =|[2a_1b_1, a_1b_2+a_2b_1, a_1b_3+a_3b_1], [a_1b_2+a_2b_1, 2a_2b_2, a_2b_3+a_3b_2], [a_1b_3+a_3b_1, a_3b_2+a_2b_3, 2a_3b_3]|` . Expressing `` as the product of two determinants, show that `triangle=0`

Show that if x_1, x_2, x_3!=0 |x_1+a_1b1a_1b_2a_1b_3a_2b_1x_2+a_2b_2a_2b_3a_3b_1a_3b_2x_3+a_3b_3|=x_1x_2x_3(1+(a_1b_1)/(x_1)+(a_2b_2)/(x_2)+(a_3b_3)/(x_3)) .

If the points (a_1, b_1),\ \ (a_2, b_2) and (a_1+a_2,\ b_1+b_2) are collinear, show that a_1b_2=a_2b_1 .

If w is a complex cube root of unity, then value of =|[a_1+b_1w, a_1w^2+b_1,c_1+b_1 w],[ a_2+b_2w, a_2w^2+b_2,c_2+b_2 w],[ a_3+b_3w ,a_3w^2+b_3,c_3+b_3 w] | is a. 0 b. -1 c. 2 d. none of these

Prove that |[a_1alpha_1+b_1beta_1, a_1alpha_2+b_1beta_2, a_1alpha_3+b_1beta_3], [a_2alpha_1+b_2beta_1, a_2alpha_2+b_2beta_2, a_2alpha_3+b_2beta_3], [a_3alpha_1+b_3beta_1, a_3alpha_2+b_3beta_2, a_3alpha_3+b_3beta_3]|=0

If Delta =|(a_1,b_1,c_1),(a_2,b_2,c_2),(a_3,b_3,c_3)| then the value of |(2a_1+3b_1+4c_1,b_1,c_1),(2a2+3b_2+4c_2,b_2,c_2),(2a_3+3b_3+4c_3,b_3,c_3)| is equal to

The value of the determinant Delta = |(1 + a_(1) b_(1),1 + a_(1) b_(2),1 + a_(1) b_(3)),(1 + a_(2) b_(1),1 + a_(2) b_(2),1 + a_(2) b_(3)),(1 + a_(3) b_(1) ,1 + a_(3) b_(2),1 + a_(3) b_(3))| , is

The determinant |(b_(1)+c_(1),c_(1)+a_(1),a_(1)+b_(1)),(b_(2)+c_(2),c_(2)+a_(2),a_(2)+b_(2)),(b_(3)+c_(3),c_(3)+a_(3),a_(3)+b_(3))|

Consider the determinant Delta = |[a_1+b_1x^2,a_1x^2+b_1,c_1],[a_2+b_2x^2,a_2x^2+b_2,c_2],[a_3+b_3x^2,a_3x^2+b_3,c_3]| = 0 , \ w h e r e \ a_i ,b_i , c_i in R \ (i = 1,2,3) \ a n d \ x in R . Statement 1: The value of x satisfying Delta=0 are x=1,-1. Statement 2: If |[a_1,b_1,c_1],[a_2,b_2,c_2],[a_3,b_3,c_3]|=0,t h e n \ Delta=0.

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=

Suppose a_(1),a_(2),a_(3) are in A.P. and b_(1),b_(2),b_(3) are in H.P. and let Delta=|(a_(1)-b_(1),a_(1)-b_(2),a_(1)-b_(3)),(a_(2)-b_(1),a_(2)-b_(2),a_(2)-b_(3)),(a_(3)-b_(1),a_(3)-b_(2),a_(3)-b_(3))| then prove that