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Question

Investigate for what values of λ,μ the simultaneous equation x+y+z=6;x+2y+3z=10 & x+2y+λz=μ have a unique solution

A
λ3
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B
λ5
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C
λ1
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D
λ2
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Solution

The correct option is B λ3
Given,
x+y+z=6;x+2y+3z=10
x+2y+λz=μ

We have,
a11=1,a12=1,a13=1,b11=1,b12=2,b13=3,c11=1,c12=2,c13=λ, d11=6,d12=10,d13=μ

By cramer's rule we have,
Δ=∣ ∣a11a12a13b11b12b13c11c12c13∣ ∣

Δ1=∣ ∣d1a12a13d2b12b13d3c12c13∣ ∣

We have,
Δ=∣ ∣11112312λ∣ ∣=1(2λ6)1(λ3)+1(22)=2λ6λ+3+0=λ3

Δ1=∣ ∣6111023μ2λ∣ ∣=6(2λ6)1(10λ3μ)+1(202μ)

=12λ3610λ+3μ+202μ=2λ+μ16

For unique solution λ3

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