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Question

Consider the system of equations
kx+(c1)y+z=2
cx+(k+1)y+kz=4
x+cy+z=1

Then correct statement is/are

A
If c=1, then the system will be inconsistent.
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B
If c=1, then the may have infinite solutions.
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C
If k=2cosπ5 then the system will be inconsistent.
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D
The system can never possess infinite solutions.
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Solution

The correct option is C If k=2cosπ5 then the system will be inconsistent.
Δ=∣ ∣kc11ck+1k1c1∣ ∣=k((k+1)ck)(c1)(ck)+(c2(k+1))
=(1c)k2+(c1)k+(c1)
=(1c)(k2k1)
If Δ=0c=1 or k=1±1+42k=1±52k=1+52=2 cosπ5
Δ1=∣ ∣2c114k+1k1c1∣ ∣=2(k+1ck)(c1)(4k)+1(4ck1)
=2k+22ck4c+ck+4k+4ck1=ck+5
For c =1 and infinitely many solutions,
Δ1=0(5k)=0k=5

Δ2=∣ ∣k21c4k111∣ ∣=k(4k)2(ck)+1(c4)
=4kk22c+2k+c4
=(k2+6kc4)
For c =1 and infinitely many solutions,
Δ2=0(k26k+5)=0k=1 or 5
Δ3=∣ ∣kc12ck+141c1∣ ∣=k(k+14c)(c1)(c4)+2(c2k1)=k2+k4ckc2+4c+c4+2c22k2=k2+c24ckk+5c6
For c =1 and infinitely many solutions,
Δ3=0k25k=0k=0 or 5
For c = 1 and k = 5, we have
Δ1, Δ2, Δ3=0
Therefore, the given system of equations can have infinitely many solutions for c = 1 and k =5.
If k=2 cos π5
Δ1, Δ2, Δ3 are not simultaneously zero for any value c, so the system will be inconsistent.

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