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Kindly Consider the following

If a1,a2,a3,...,ar are in G.P., then prove that the determinant [ar+1ar+5ar+9ar+7ar+11ar+15ar+11ar+17ar+21] is

independent of r .

0 7 Views | Posted 3 months ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    3 months ago

    This is a Short Answer Type Questions as classified in NCERT Exemplar
    Sol:

    I f     a 1 , a 2 , a 3 , … a r     b e     t h e     t e r m s     o f     G . P . ,     t h e n                                                                                       a n = A R n − 1 ( w h e r e     A     i s     t h e     f i r s t     t e r m     a n d     R     i s     t h e     c o m m o n     r a t i o     o f     t h e     G . P . ) ∴               a r + 1 = A R r + 1 − 1 = A R r ;     a r + 5 = A R r + 5 − 1 = A R r + 4                     a r + 9 = A R r + 9 − 1 = A R r + 8 ;     a r + 7 = A R r + 7 − 1 = A R r + 6                     a r + 1 1 = A R r + 1 1 − 1 = A R r + 1 0 ;     a r + 1 5 = A R r + 1 5 − 1 = A R r + 1 4                     a r + 1 7 = A R r + 1 7 − 1 = A R r + 1 6 ;     a r + 2 1 = A R r + 2 1 − 1 = A R r + 2 0 ∴  The  determinant  becomes                                                                       | A R r A R r + 4 A R r + 8 A R r + 6 A R r + 1 0 A R r + 1 4 A R r + 1 0 A R r + 1 6 A R r + 2 0 | T a k i n g     A R r ,   A R r + 6     a n d     A R r + 1 0     c o m m o n     f r o m     R 1 ,   R 2 ,       a n d     R 3       r e s p e c t i v e l y .                                                                       A R r . A R r + 6 . A R r + 1 0 | 1 R 4 R 8 1 R 4 R 8 1 R 6 R 1 0 |                                                                     = A R r . A R r + 6 . A R r + 1 0 | 0 |                   [ ?     R 1       a n d     R 2       a r e     i d e n t i c a l     r o w s ]                                                                     = 0 Hence,  the  given  determinant  is  independent  of  r.

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

|2A| = 27

8|A| = 27

Now |A| = α2–β2 = 24

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(α–β) (α+β) = 16

->α + β = 8 and

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->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
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alok kumar singh

|A| = 3

|B| = 1

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

= 9

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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
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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             

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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  
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∫ 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

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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

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