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Question

# Let the system of lineat equations 2x+3y−z=0,2x+ky−3z=0 and 2x−y+z=0 have non trivial non trivial solution then xy+yz+zx+k will be

A
2
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B
3
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C
1
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D
4
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Solution

## The correct option is C 3∣∣ ∣∣23−12k−32−11∣∣ ∣∣=0R2→R2−R1 and R3→R3−R1⇒∣∣ ∣∣23−10k−3−20−42∣∣ ∣∣=0⇒k=7Equation 1⇒2xy+3−zy=0 ......(iv)Equation 3⇒2xy−1+zy=0 adding these two equation we get 4xy+2=0⇒xy=−12.....(v)Putting this value in equation (iv) we get zy=2 ....(vi)Equation (v) divided by equation (vi) ⇒xz=−14So, xy+yz+zx+k=−12+12−4+7=3

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