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Question

If the system of linear equations
2x+py+6z=8
x+2y+qz=5
x+y+3z=4
has infinitely many solutions, then a possible pair of p and q is

A
p=3,q=3
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B
p2,q3
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C
p2,q=3
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D
p=2,q3
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Solution

The correct option is D p=2,q3
2x+py+6z=8
x+2y+qz=5
x+y+3z=4

For infinitely many solutions,
Δ=Δx=Δy=Δz=0

Δ=∣ ∣2p612q113∣ ∣=(p2)(q3)

Δx=∣ ∣8p652q413∣ ∣

=308q15p+4pq =(p2)(4q15)

Δy=∣ ∣28615q143∣ ∣=0

Δz=∣ ∣2p8125114∣ ∣=p2

At p=2, Δ=Δx=Δy=Δz=0
So, at p=2 and qR, the system of linear equations has infinitely many solutions.

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