Home
Class 12
MATHS
If A=|(1,1,1),(a,b,c),(a^3,b^3,c^3)|, B=...

If `A=|(1,1,1),(a,b,c),(a^3,b^3,c^3)|, B=|(1,1,1),(a^2,b^2,c^2),(a^3,b^3,c^3)|, C=|(a,b,c),(a^2,b^2,c^2),(a^3,b^3,c^3)|` , then which relation is correct :

A

A=B

B

A=C

C

B=C

D

None of these

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the determinants of the matrices \( A \), \( B \), and \( C \) given by: \[ A = \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{vmatrix}, \quad B = \begin{vmatrix} 1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix}, \quad C = \begin{vmatrix} a & b & c \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix} \] ### Step 1: Calculate the Determinant of Matrix \( A \) Using the properties of determinants, we can expand the determinant \( A \): \[ A = 1 \cdot \begin{vmatrix} b & c \\ b^3 & c^3 \end{vmatrix} - 1 \cdot \begin{vmatrix} a & c \\ a^3 & c^3 \end{vmatrix} + 1 \cdot \begin{vmatrix} a & b \\ a^3 & b^3 \end{vmatrix} \] Calculating these 2x2 determinants: 1. \( \begin{vmatrix} b & c \\ b^3 & c^3 \end{vmatrix} = bc^3 - b^3c = b(c^3 - b^2) \) 2. \( \begin{vmatrix} a & c \\ a^3 & c^3 \end{vmatrix} = ac^3 - a^3c = a(c^3 - a^2) \) 3. \( \begin{vmatrix} a & b \\ a^3 & b^3 \end{vmatrix} = ab^3 - a^3b = a(b^3 - a^2) \) Thus, we have: \[ A = b(c^3 - b^2) - a(c^3 - a^2) + a(b^3 - a^2) \] ### Step 2: Calculate the Determinant of Matrix \( B \) Similarly, we can calculate the determinant \( B \): \[ B = 1 \cdot \begin{vmatrix} b^2 & c^2 \\ b^3 & c^3 \end{vmatrix} - 1 \cdot \begin{vmatrix} a^2 & c^2 \\ a^3 & c^3 \end{vmatrix} + 1 \cdot \begin{vmatrix} a^2 & b^2 \\ a^3 & b^3 \end{vmatrix} \] Calculating these 2x2 determinants: 1. \( \begin{vmatrix} b^2 & c^2 \\ b^3 & c^3 \end{vmatrix} = b^2c^3 - b^3c^2 = b^2(c^3 - bc) \) 2. \( \begin{vmatrix} a^2 & c^2 \\ a^3 & c^3 \end{vmatrix} = a^2c^3 - a^3c^2 = a^2(c^3 - ac) \) 3. \( \begin{vmatrix} a^2 & b^2 \\ a^3 & b^3 \end{vmatrix} = a^2b^3 - a^3b^2 = a^2(b^3 - ab) \) Thus, we have: \[ B = b^2(c^3 - bc) - a^2(c^3 - ac) + a^2(b^3 - ab) \] ### Step 3: Calculate the Determinant of Matrix \( C \) Now, we calculate the determinant \( C \): \[ C = \begin{vmatrix} a & b & c \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix} \] This is a Vandermonde determinant, which can be calculated as: \[ C = (b-a)(c-a)(c-b) \] ### Step 4: Compare the Determinants Now we need to compare the determinants \( A \), \( B \), and \( C \): - \( A \) and \( B \) are not equal because they have different structures and terms. - \( A \) and \( C \) are also not equal since \( C \) is a product of differences. - \( B \) and \( C \) are not equal for the same reasons. ### Conclusion Since none of the matrices \( A \), \( B \), or \( C \) are equal to each other, the correct relation is: **None of these.**
Promotional Banner

Similar Questions

Explore conceptually related problems

1,1,1a,b,ca^(3),b^(3),c^(3)]|=(a-b)(b-c)(c-a)(a+b+c)

Three linear equations a_1x+b_1y+c_1z=0, a_2x+b_2y+c_2z=0,a_3x+b_3y+c_3z=0 are consistent if (A) |(a_1,b_1,c_1),(a_2,b_2,c_2),(a_3,b_3,c_3)|=0 (B) |(a_1,b_1,c_1),(a_2,b_2,c_2),(a_3,b_3,c_3)|=-1 (C) a_1b_1c_1+a_2b_2c_2+a_3b_3c_3=0 (D) none of these

Prove that: (i) |{:(,1,1,1),(,a,b,c),(,a^(3),b^(3),c^(3)):}|=(a-b)(b-c)(c-a)(a+b+c)

|(1,a^(2)+bc,a^(3)),(1,b^(2)+ac,b^(3)),(1,c^(2)+ab,c^(3))|=-(a-b)(b-c)(c-a)(a^(2)+b^(2)+c^(2))

If a!=b!=c such that |[a^3-1,b^3-1,c^3-1] , [a,b,c] , [a^2,b^2,c^2]|=0 then

if A=[{:(1,2,-3),(5,0,2),(1,-1,1):}],B=[{:(3,-1,2),(4,2,5),(2,0,3):}]and c=[{:(4,1,2),(0,3,2),(1,-2,3):}], then compure (A+B) and (B-C), Also , verify that A+(B-C)=(A+B)-C.

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=

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