+ If a pair of linear equations a, x + b,y + c=0 a,x + b2y + c =0 in two variables have unique solution then correct relation among the following is A) b. ab a B) a . - G = a2 b C2 az b2 a C) 4 a2 b, b D) Q _6179 b a2 C2 10 296
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Step-by-step explanation:
For existence of a non-trivial solution of the
first system
∣
∣
∣
∣
∣
∣
∣
∣
a
b
c
b
c
a
c
a
b
∣
∣
∣
∣
∣
∣
∣
∣
=0
⇒(a+b+c)
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∣
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∣
∣
∣
1
1
1
b
c
a
c
a
b
∣
∣
∣
∣
∣
∣
∣
∣
=0
⇒
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∣
∣
∣
∣
∣
∣
1
1
1
b
c
a
c
a
b
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∣
∣
∣
∣
∣
∣
=0 or (a+b+c)=0
⇒
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1
0
0
b
c−b
a−b
c
a−c
b−c
∣
∣
∣
∣
∣
∣
∣
∣
=0 or (a+b+c)=0
⇒
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∣
∣
∣
∣
c−b
a−b
a−c
b−c
∣
∣
∣
∣
∣
∣
=0 or
(a+b+c)=0 ...(1)
The second system will have a non-trivial solution if we can prove that
△=
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b+c
c+a
a+b
c+a
a+b
b+c
a+b
b+c
c+a
∣
∣
∣
∣
∣
∣
∣
∣
=0
Now, △=2(a+b+c)
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1
1
1
c+a
a+b
b+c
a+b
b+c
c+a
∣
∣
∣
∣
∣
∣
∣
∣
=2(a+b+c)
∣
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∣
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1
0
0
c+a
b−c
b−a
a+b
c−a
c−b
∣
∣
∣
∣
∣
∣
∣
∣
=2(a+b+c)
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∣
b−c
b−a
c−a
c−b
∣
∣
∣
∣
∣
∣
=2(a+b+c)
∣
∣
∣
∣
∣
∣
c−b
a−b
a−c
b−c
∣
∣
∣
∣
∣
∣
=0 (using equation (1))
∴ The second system will have a non-trivial solution.
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