in Linear Algebra
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#Matrices
Determine b such that the system of homogeneous equation
2x+y+2z=0
x+y+3z=0
4x+3y+bz=0
has trivial solution
This problem is taken from HK Das Book Page# 72
and solution by book is 
For trivial solution We know that x=0,y=0 and z=0.So b can have any value but i have the point that 
For homogeneous solution,
Rank of A= n(no of unknowns) as per book 
Here no of unknown is three(x,y,z)
and if we solve the matrix

2 1 2
1 1 3
4 3 b

then matrix is like this

1 1 3
0 -1 -4
0 0 b-8

So according to me b is not equal to 8 for unique or trivial solution as per definition Rank of A=n(no of unknowns) for unique solution but book said anyvalue of b
Please tell if anyone know.
Thanks in advance.

in Linear Algebra
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Except B =8 , The system will have trivial solution for any other  value of B

 

OR

B!=8 , The system will have trivial solution .
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except b=8 , system will have  trivial solution for any value of 'b'
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except b=8 , system will have non trivial solution or trivaial solution  for any value of 'b'
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trivial solution
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but when we put x=0,y=0,z=0 in the equation then we put any value of b .Isnt it ?
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no it is not true. it can be shown by ur own calculation .

at last step u calculated final matrix as

1 1 3
0 -1 -4
0 0 b-8

now if we solve it , then we can c that (b-8)Z=0

therefore z= 0/(b-8)

if u put b=8 , then z=0/0 which is undefined hence except b=8 , we will get z=0 for all values of b.
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