Maths Linear Programming

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New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

Δ | 3 2 k 2 4 2 1 2 1 | = | 3 2 k 0 8 0 1 2 1 | , [ R 2 R 1 2 R 3 ]

= 8 (3 + k)

For inconsistent Δ = 0 k = 3

Δ x = | 1 0 2 k 6 4 2 5 m 2 1 | = 3 2 4 0 m 0 m 4 5

New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

Clearly all equations can be λ x - 2 y - 6 z = 0  form where λ = 1 , 4 3 , 6 5  

So for x = 0 y = - 3 z

So x 2 + y 2 + z 2 = 10 z 2  (as  ) x = 0 y = - 3 z

ATQ 10 10 z 2 200

1 z 2 20 z = ± 1 , ± 2 , ± 3 , ± 4

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

( a 1 ) x + 0 y + z = α        ……(1)

x + ( b 1 ) y + 0 z = β                       ……(2)

0 x + y + ( c 1 ) z = γ                       ……(3)

| ( a 1 ) 0 1 1 ( b 1 ) 0 0 1 ( c 1 ) | = 0           For no unique solution D = 0

( a 1 ) ( b 1 ) ( c 1 ) + 1 = 0            

a = 2 ; b = 2 ; c = 0            

Hence, |a + b + c| = 4

 

New answer posted

a month ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

  Δ | 3 2 k 2 4 2 1 2 1 | = | 3 2 k 0 8 0 1 2 1 | , [ R 2 R 1 2 R 3 ]

= -8 (-3 + k)

For inconsistent Δ = 0 k = 3  

Δ x = | 1 0 2 k 6 4 2 5 m 2 1 | = 3 2 4 0 m 0 m 4 5

New answer posted

2 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

New question posted

2 months ago

0 Follower 2 Views

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