wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
p2,q3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
p2,q=3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
p=2,q3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
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.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon