Matrices and Determinants 4 Question 13

14. If the system of linear equations

x+ky+3z=0,3x+ky2z=02x+4y3z=0

has a non-zero solution (x,y,z), then xzy2 is equal to

(a) -10

(b) 10

(c) -30

(d) 30

(2018 Main)

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Answer:

Correct Answer: 14. (b)

Solution:

  1. We have,

x+ky+3z=0;3x+ky2z=0;2x+4y3z=0

System of equation has non-zero solution, if

|1k33k2243| =0

(3k+8)k(9+4)+3(122k)=03k+8+9k4k+366k=04k+44=0k=11 Let z=λ, then we get x+11y+3λ=03x+11y2λ=0 and 2x+4y3λ=0

Solving Eqs. (i) and (ii), we get

x=5λ2,y=λ2,z=λxzy2=5λ22×λ2=10



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