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V
Vishal Baghel

Contributor-Level 10

A = [ 0 2 k 1 ] t h e n A 2 = [ 0 2 k 1 ] [ 0 2 k 1 ] = [ 2 k 2 k 2 k + 1 ]

A 3 = [ 2 k 2 k 2 k + 1 ] [ 0 2 k 1 ] = [ 2 k 4 k + 2 2 k 2 + k 4 k 1 ]              

Now, A(A3 + 3l) = 2l gives A3 + 3l = 2A-1

2 k + 3 = 1 k 2 k 2 3 k + 1 = 0    

  k = 1 2 , 1            

& 4 k + 2 = 2 k o r 2 k + 1 1 k s o o r 2 k 2 + k 1 = 0

=> k = 1 2

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A
alok kumar singh

Contributor-Level 10

x + y + z = 4

3x + 2y + 5z = 3

9 x + 4 y + ( 2 8 + [ λ ] ) z = [ λ ]           

Δ = | 1 1 1 3 2 5 9 4 2 8 + [ λ ] | = 5 6 + 2 [ λ ] 2 0 ( 8 4 + 3 [ λ ] 4 5 ) + ( 6 )

= [ λ ] 9              

I f [ λ ] + 9 0 then unique solution

& if  [ λ ] + 9 = 0 t h e n Δ 1 = Δ 2 = Δ 3 = 0 so infinite solution will exist

Hence λ R  

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A
alok kumar singh

Contributor-Level 10

| A | = | [ x + 1 ] [ x + 2 ] [ x + 3 ] [ x ] [ x + 3 ] [ x + 3 ] [ x ] [ x + 2 ] [ x + 4 ] | = | [ x ] + 1 [ x ] + 2 [ x ] + 3 [ x ] [ x ] + 3 [ x ] + 3 [ x ] [ x ] + 2 [ x ] + 4 |

R 1 R 1 R 3 & R 2 R 2 R 3

| 1 0 1 0 1 1 [ x ] [ x ] + 2 [ x ] + 4 | = 1 9 2

[ x ] = 6 2

x [ 6 2 , 6 3 )              

 

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A
alok kumar singh

Contributor-Level 10

A 2 = ( 1 0 0 0 1 1 1 0 0 ) ( 1 0 0 0 1 1 1 0 0 ) = ( 1 0 0 1 1 1 1 0 0 )

A 3 = ( 1 0 0 1 1 1 1 0 0 ) ( 1 0 0 0 1 1 1 0 0 ) = ( 1 0 0 2 1 1 1 0 0 )

A 2 0 2 5 A 2 0 2 0

= ( 1 0 0 2 0 2 4 1 1 1 0 0 ) ( 1 0 0 2 0 1 9 1 1 1 0 0 ) = ( 1 0 0 5 0 0 1 0 0 ) = A 6 A

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5 months ago

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V
Vishal Baghel

Contributor-Level 10

AAT = ATA = l

A ( A T B A ) 2 0 2 1 A T              

B 2 0 2 1 = ( I + P ) 2 0 2 1 = l + 2 0 2 1 P = [ 1 0 2 0 2 1 i 1 ]

( B 2 0 2 1 ) 1 = [ 1 0 2 0 2 1 i 1 ]

where P = [ 0 0 i 0 ]

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V
Vishal Baghel

Contributor-Level 10

Δ = 0

sin 3 θ=  1 2

but 0 < θ < π 2

θ= 70°

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P
Payal Gupta

Contributor-Level 10

A=[1211] EA=[acbd][1211]

=[a+c2acb+d2bd]

For a = c For a+c=02ac=1]a=1,c=1E=[1101]

d = b + 1, d = 1, b = 0

b+d=12bd=1]b=0,d=1R1R1R2[1001]

For a+c=12ac=1]a=0,c=1

Fora+c=12ac=2]a=1,c=0

b+d=22bd=7]b=5,d=3[1053][1001]

R2 5R1 + 3R2

For Fora+c=12ac=2]a=1,c=1

b+d=12bd=3]b=2,d=1

(A) R1 R1 + R2

(B) R2 R2 + 2R1 [1021][1001]

(C) R2 3R2 + 5R1

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A
alok kumar singh

Contributor-Level 10

Δ = | k 3 1 4 1 5 4 k 4 1 3 |

= ( k ) ( 1 2 k ) + 3 ( 4 k + 4 5 ) 1 4 ( 1 5 + 1 6 )

Δ = 0 k = ± 1 1             

 For k = -11,

->11x + 3y – 14z = 25

-4x + y + 3z = 4

{ 1 1 x + 3 y 1 4 z = 2 5 1 5 x + 4 y 1 1 z = 3 4 x + y + 3 z = 4 } No solution for k = ± 11

               

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5 months ago

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A
alok kumar singh

Contributor-Level 10

Let A = [ a b c d e f 9 h i ]  

Now ATA

trace will be    a 2 + b 2 + c 2 + d 2 + e 2 + f 2 + 9 2 + h 2 = 6

total ways = 

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