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Q.

If the system of equations 2x+3yz=0,x+ky2z=0and 2xy+z=0has a non trivial solution. Then xy+yz+zx+k=

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a

12

b

32

c

52

d

72

answer is A.

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Detailed Solution

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23-11k-22-11=0   K=92

Given equations are 2x+3yz=0(1)

x+92y2z=0(2),     2xy+z=0(3)

(1)(3)  4y2z=0  zy=2

(1)+(3)  xy=12(5)

(4)×(5)yz×xy=14  zx=4

xy+yz+zx+k=12+124+92=12

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If the system of equations 2x+3y−z=0, x+ky−2z=0and 2x−y+z=0has a non trivial solution. Then xy+yz+zx+k=