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))` .

Show that if x_(1),x_(2),x_(3) ne 0 |{:(x_(1) +a_(1)b_(1),,a_(1)b_(2),,a_(1)b_(3)),(a_(2)b_(1),,x_(2)+a_(2)b_(2),,a_(2)b_(3)),(a_(3)b_(1),,a_(3)b_(2),,x_(3)+a_(3)b_(3)):}| =x_(1)x_(2)x_(3) (1+(a_(1)b_(1))/(x_(1))+(a_(2)b_(2))/(x_(2))+(a_(3)b_(3))/(x_(3)))

|[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]|=

If a_i, b_i in N for i 1,2,3, then coefficient of x in the determinant; |((1+x)^(a_1b_1),(1+x)^(a_1b_2),(1+x)^(a_1b_3)),((1+x)^(a_2b_1),(1+x)^(a_2b_2),(1+x)^(a_2b_3)), ((1+x)^(a_3b_1),(1+x)^(a_3b_2),(1+x)^(a_3b_3))|

Let =|2a_1b_1a_1b_2+a_2b_1a_1b_3+a_3b_1a_1b_2+a_2b_1 2a_2b_2a_2b_3+a_3b_2a_1b_3+a_3b_1a_3b_2+a_2b_3 2a_3b_3| . Expressing as the product of two determinants, show that =0. Hence, show that if a x^2+2h x y+b y^2+2gx+2fy+c=(l x+m y+n)(l^(prime)x+m^(prime)y+n),t h e n|a hgh bfgfc|=0.

If x ,ya n dz are not all zero and connected by the equations a_1x+b_1y+c_1z=0,a_2x+b_2y+c_2z=0,a n d(p_1+lambdaq_1)x+(p_2+lambdaq_2)+(p_3+lambdaq_3z=0) , show that lambda=-|a_1b_1c_1a_2b_2c_2p_1p_2p_3|-:|a_1b_1c_1a_2b_2c_2q_1q_2q_3|

If A=(2x+1)/(2x-1), B=(2x-1)/(2x+1) find (1)/(A-B)-(2B)/(A^(2)-B^(2))

If a x1 2+b y1 2+c z1 2=a x2 2+b y2 2+c z2 2=a x3 2+b y3 2+c z3 2=d ,a x2 3+b y_2y_3+c z_2z_3=a x_3x_1+b y_3y_1+c z_3z_1=a x_1x_2+b y_1y_2+c z_1z_2=f, then prove that |x_1y_1z_1x_2y_2z_2x_3y_3z_3|=(d-f){((d+2f))/(a b c)}^(1//2)

Consider the determinant "Delta"=|a_1+b_1x^2a_1x^2+b_1c_1a_2+b_2x^2a_2x^2+b_2c_2a_3+b_3x^2a_3x^2+b_3c_3|0, w h e r ea_i ,b_i , c_i R(1,2,3)a n dx Rdot Statement 1: The value of x satisfying "Delta"=0 are x=1,-1. Statement 2: If |a_1b_1c_1a_2b_2c_2a_3b_3c_3|=0,t h e n"Delta"=0.

If a_(1) x^(3) + b_(1)x^(2) + c_(1)x + d_(1) = 0 and a_(2)x^(3) + b_(2)x^(2) + c_(2)x + d_(2) = 0 a pair of repeated roots common, then prove that |{:(3a_(1)", "2b_(1) ", "c_(1)),(3a_(2)", " 2b_(2)", "c_(1)),(a_(2)","b_(1)- a_(1)b_(2)", "c_(2)a_(1)-c_(2)a_(1)", "d_(1)a_(2)-d_(2)a_(1)):}|=0