Question

Determinants -
If f(x) is a polynomial of degree < 3 ; prove that

Answer 1

\[=\frac{ f(x) }{ (x-a)(x-b)(x-c) }\]

Answer 2

one observation is :
for the denominator : a-b, b-c, and c-a are factors

Answer 3

not sure if thats useful, but i think we need to take cases when degree = 0, 1, 2 ?

Answer 4

I found LHS
\[=\frac{ f(a) }{ (x-a)(b-a)(c-a) }+\frac{ f(b) }{ (a-b)(x-b)(c-b) }+\frac{ f(c) }{ (a-c)(b-c)(x-c) }\]

Donno what to do next.. :(

Answer 5

Solution in the book says -
"RHS will also be the same if we split it into partial fractions by the method of suppression by putting x=a,b,c successively."

IDK what does that ^^ means..

Answer 6

thats just fancy way of saying "decompose RHS into partial fractions by using cover up method"

Answer 7

\(\large \frac{ f(x) }{ (x-a)(x-b)(x-c) } = \frac{A}{x-a} + \frac{B}{x-b} + \frac{C}{x-c}\)

Answer 8

since f(x) degree is less < 3, we can do that^

Answer 9

\[
f(x) = k_2x^2+k_1x+k_0
\]plug an chug.

Answer 10

multiply thru by x-a, and cover up A :

you wud get A = \(\frac{f(a)}{(a-b)(a-c)}\)

Answer 11

Yea, I got it now..thanx ^_^

Answer 12

BTW, what does "cover-up" means ?

Answer 13

http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/video-lectures/lecture-29-partial-fractions/

Answer 14

watch this from 5:50 to 10:00

Answer 15

just an easy trick for doing partial fractions...

Answer 16

Thanxx :)

Answer 17

np :)