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Question

If the system of equation 2x+3y−z=0,x+ky−2z=0 and 2x−y+z=0 has a non trivial solution (x,y,z) then xy+yz+zx+k is equal to:

A
34
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B
4
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C
12
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D
14
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Solution

The correct option is C 12
Given 2x+3yz=0,x+Ky2z=0,2xy+z=0

for Non-trivial solution

∣ ∣2311K2211∣ ∣=0

By expanding along R2

1[31]+K[2+2](2)[26]=0

K=92

the given equations become

2x+3yz=0......(1)

x+92y2z=0.......(2)

2xy+z=0..........(3)

(1)(3)4y2z=0

2y=z........(4)

yz=12.......(5)

put z from equation (4) into (1)

2x+3y2y=0

2x+y=0

xy=12.....(6)

(6)×(5)xz=14zx=4

xy+yz+zx+k=12124+92

xy+yz+zx+k=12


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