Question

# Consider the system of equations kx+(c−1)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 options are B If c=1, then the may have infinite solutions. C If k=2cosπ5 then the system will be inconsistent.Δ=∣∣ ∣∣kc−11ck+1k1c1∣∣ ∣∣=k((k+1)−ck)−(c−1)(c−k)+(c2−(k+1)) =(1−c)k2+(c−1)k+(c−1) =(1−c)(k2−k−1) If Δ=0⇒c=1 or k=1±√1+42⇒k=1±√52k=1+√52=2 cosπ5 Δ1=∣∣ ∣∣2c−114k+1k1c1∣∣ ∣∣=2(k+1−ck)−(c−1)(4−k)+1(4c−k−1) =2k+2−2ck−4c+ck+4−k+4c−k−1=−ck+5 For c =1 and infinitely many solutions, Δ1=0⇒(5−k)=0⇒k=5 Δ2=∣∣ ∣∣k21c4k111∣∣ ∣∣=k(4−k)−2(c−k)+1(c−4) =4k−k2−2c+2k+c−4 =(−k2+6k−c−4) For c =1 and infinitely many solutions, Δ2=0⇒−(k2−6k+5)=0⇒k=1 or 5 Δ3=∣∣ ∣∣kc−12ck+141c1∣∣ ∣∣=k(k+1−4c)−(c−1)(c−4)+2(c2−k−1)=k2+k−4ck−c2+4c+c−4+2c2−2k−2=k2+c2−4ck−k+5c−6 For c =1 and infinitely many solutions, Δ3=0⇒k2−5k=0⇒k=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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