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78. |xx21+px3yy21+py3zz21+pz3|=(1+pxyz)(x−y)(y−z)(z−x) 

0 2 Views | Posted 4 months ago
Asked by Shiksha User

  • 1 Answer

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    Answered by

    Vishal Baghel | Contributor-Level 10

    4 months ago

    L . H . S . = | x x 2 1 + p x 3 y y 2 1 + p y 3 z z 2 1 + p z 3 | = | x x 2 1 y y 2 1 z z 2 1 | + | x x 2 p x 3 y y 2 p y 3 z z 2 p z 3 | = ? 1 + ? 2 − − − − − ( 1 ) N o w ? 2 = | x x 2 p x 3 y y 2 p y 3 z z 2 p z 3 | = p x y z | 1 x x 2 1 y y 2 1 z z 2 | c 1 ↔ c 3 = − p x y z | x 2 x 1 y 2 y 1 z 2 z 1 | c 1 ↔ c 2 = p x y z | x x 2 1 y y 2 1 z z 2 1 | = p x y z ? 1

    Putting value in (1)

    ⇒?1+pxyz?1=(1+pxyz)?1−−−−−(2)Now?1=|xx21yy21zz21|

    R2→R2−R1 and R3→R3−R1

    =|xx21y−xy2−x20z−xz2−x20|

    Expanding along c3

    ?1=|(y−x)(y−x)(y+x)(z−x)(z−x)(z+x)|=(y−x)(z−x)|1y+x1z+x|=(y−x)(z−x)(z+x−y−x)=(x−y)(y−z)(z−x)

    Putting ?1 in (2)

    L.H.S.=(1+pxyz)(x−y)(y−z)(z−x)=R.H.S

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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