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Prove that [bc−a2ca−b2ab−c2ca−b2ab−c2bc−a2ab−c2bc−a2ca−b2] is divisible by a+b+c and find the quotient.

0 3 Views | Posted 3 months ago
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    Answered by

    Vishal Baghel | Contributor-Level 10

    3 months ago

    This is a Long Answer Type Questions as classified in NCERT Exemplar

    Sol:

    L e t     Δ = | b c − a 2 c a − b 2 a b − c 2 c a − b 2 a b − c 2 b c − a 2 a b − c 2 b c − a 2 a c − b 2 | C 1 → C 1 + C 2 + C 3 ⇒                                                 [ a b + b c + a c − a 2 − b 2 − c 2 c a − b 2 a b − c 2 a b + b c + a c − a 2 − b 2 − c 2 a b − c 2 b c − a 2 a b + b c + a c − a 2 − b 2 − c 2 b c − a 2 a c − b 2 ] ⇒                         ( a b + b c + a c − a 2 − b 2 − c 2 ) | 1 c a − b 2 a b − c 2 1 a b − c 2 b c − a 2 1 b c − a 2 a c − b 2 |                   ( T a k i n g     a b + b c + a c − a 2 − b 2 − c 2     c o m m o n     f r o m     C 1 ) R 1 → R 1 − R 2     a n d     R 2 → R 2 − R 3 ⇒                               ( a b + b c + a c − a 2 − b 2 − c 2 ) | 0 c a − b 2 − a b + c 2 a b − c 2 − b c + a 2 0 a b − c 2 − b c + a 2 b c − a 2 − a c + b 2 1 b c − a 2 a c − b 2 | ⇒                               ( a b + b c + a c − a 2 − b 2 − c 2 ) | 0 a ( c − b ) + ( c + b ) ( c − b ) b ( a − c ) + ( a + c ) ( a − c ) 0 b ( a − c ) + ( a + c ) ( a − c ) c ( b − a ) + ( b + a ) ( b − a ) 1 b c − a 2 a c − b 2 | ⇒                               ( a b + b c + a c − a 2 − b 2 − c 2 ) | 0 ( c − b ) ( a + b + c ) ( a − c ) ( a + b + c ) 0 ( a − c ) ( a + b + c ) ( b − a ) ( a + b + c ) 1 b c − a 2 a c − b 2 | ⇒                               ( a b + b c + a c − a 2 − b 2 − c 2 ) ( a + b + c ) ( a + b + c ) | 0 c − b a − c 0 a − c b − a 1 b c − a 2 a c − b 2 | ⇒                               ( a + b + c ) 2 ( a b + b c + a c − a 2 − b 2 − c 2 ) | 0 c − b a − c 0 a − c b − a 1 b c − a 2 a c − b 2 | E x p a n d i n g     a l o n g     C 1 ⇒         ( a + b + c ) 2 ( a b + b c + a c − a 2 − b 2 − c 2 ) [ 1 | c − b a − c a − c b − a | ] ⇒         ( a + b + c ) 2 ( a b + b c + a c − a 2 − b 2 − c 2 ) [ ( c − b ) ( b − a ) − ( a − c ) 2 ] ⇒         ( a + b + c ) 2 ( a b + b c + a c − a 2 − b 2 − c 2 ) ( b c − c a − b 2 + a b − a 2 − c 2 + 2 a c ) ⇒         ( a + b + c ) 2 ( a b + b c + a c − a 2 − b 2 −

Similar Questions for you

A = [ 1 0 0 0 α β 0 β α ] then α is (if α, β Î I)
A
alok kumar singh

|2A| = 27

8|A| = 27

Now |A| = α2–β2 = 24

α2 = 16 + β2

α2– β2 = 16

(α–β) (α+β) = 16

->α + β = 8 and

α – β = 2

->α = 5 and β = 3

If A = [ 2 1 − 1 2 ] , B = [ 1 0 1 1 ] , C = ABAT and X = AC2AT then |X| is equal to
A
alok kumar singh

|A| = 3

|B| = 1

->|C| = |ABAT| = |A|B|A7| = |A|2|B|

= 9

->|X| = |A|C|2|AT|

= 3 * 92 * 3 = 9 * 92 = 729

Matrix A of order 3 × 3 is such that |A| = 2 if n = | a d j ( a d j ( a d j . . . . ( a ) ) ) | ︸ 2 0 2 4   t i m e s  then remainder when n is divided by 9 is
A
alok kumar singh

|A| = 2

a d j ( a d j ( a d j . . . . ( a ) ) ) ? 2 0 2 4   t i m e s = | A | ( n − 1 ) 2 0 2 4            

= | A | ( n − 1 ) 2 0 2 4                                             

= 2 2 2 0 2 4                                          

2 2 0 2 4 = ( 2 2 ) 2 2 0 2 2 = 4 ( 8 ) 6 7 4 = 4 ( 9 − 1 ) 6 7 4             

-> 2 2 0 2 4 ≡ 4 ( m o d   9 )

->,  2 2 0 2 4 ≡ 9 m + 4  m ¬ even

2 9 m + 4 ≡ 1 6 ⋅ ( 2 3 ) 3 m ≡ 1 6 ( m o d   9 )

7

If ∫ 0 2 ( 2 x − 2 x − x 2 ) d x = ∫ 0 1 ( 1 − 1 − y 2 − y 2 2 ) d y + ∫ 1 2 ( 2 − y 2 2 ) d y + I     t h e n     I     e q u a l s  
V
Vishal Baghel

∫ 0 2 2 x d x − ∫ 0 2 2 x − x 2 d x = ∫ 0 1 d y − ∫ 0 1 1 − y 2 d y − ∫ 0 2 y 2 2 d y + ∫ 1 2 2 d y + I

⇒ 8 3 − ∫ 0 1 1 − t 2 d t = 1 − 8 6 + 2 + I

I = 1 − ∫ 0 1 1 − t 2 d t

If  ∫ 0 2 ( 2 x − 2 x − x 2 ) d x = ∫ 0 1 ( 1 − 1 − y 2 − y 2 2 ) d y + ∫ 1 2 ( 2 − y 2 2 ) d y + I     t h e n     I     e q u a l s

A
alok kumar singh

∫ 0 2 2 x d x − ∫ 0 2 2 x − x 2 d x = ∫ 0 1 d y − ∫ 0 1 1 − y 2 d y − ∫ 0 2 y 2 2 d y + ∫ 1 2 2 d y + I

⇒ 8 3 − ∫ 0 1 1 − t 2 d t = 1 − 8 6 + 2 + I

I = 1 − ∫ 0 1 1 − t 2 d t

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