MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 12
        Let a and b be the roots of equation Ax^2+Bx+C=0.  Evaluate the value of determinant | 3                      1+a+b              1+a^2 +b^2   || 1+a+b.            1+a^2+b^2      1+a^3+b^3    || 1+a^2+b^2.   1+a^3+b^3.     1+a^4+b^4.   |
2 years ago

Answers : (1)

mycroft holmes
272 Points
							
The given determinant can be written as a product:
\begin{vmatrix} 1 & 1 &1 \\ 1 & a& b\\ 1& a^2& b^2 \end{vmatrix} \ \times \begin{vmatrix} 1 & 1 &1 \\ 1 & a& b\\ 1& a^2 & b^2 \end{vmatrix}
 
\begin{vmatrix} 1 & 1 &1 \\ 1 & a& b\\ 1& a^2& b^2 \end{vmatrix} ^2
 
Using the well known expansion for the Vandermonde Determinant, we can evaluate this as (a-b)^2 (a-1)^2(b-1)^2
 
Since a,b are roots of the quadratic Ax2+Bx+C, we have
 
(a-b)^2 = \frac{B^2-4AC}{A^2}
 
and (a-1)^2 (b-1)^2 = [(a-1)(b-1)]^2 = P(1)^2 = (A+B+C)^2
 
Hence given expressions equals
 
\frac{(B^2-4AC)(A+B+C)^2}{A^2}
2 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 101 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details