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If the system of equations

x + y + z = 6

2x + 5y + aZ = b

x + 2y + 3z = 14

has infinitely many solutions, then a + b is equal to

Option 1 -

8

Option 2 -

36

Option 3 -

44

Option 4 -

48

0 2 Views | Posted 2 months ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    2 months ago
    Correct Option - 3


    Detailed Solution:

    Δ = 0

    | 1 1 1 2 5 α 1 2 3 | = 0

    15 - 2a + a - 6 – 1 = 0

    a = 8

    For a = 8, equations are

    x + y + 3 = 6

    2x + 5y + 8z = b

    x + 2y + 3z = 14

    ( 2 , 5 , 8 ) = l ( 1 , 1 , 1 ) + m ( 1 , 2 , 3 )

    2 = l + m 5 = l + 2 m ] → 3 = m , l = − 1              

    8 =  l+3m

    β = 6 l + 1 4 m

    = -6 + 42 = 36

    a + b = 8 + 36 = 44

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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
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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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