CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Consider the system of equations:
x+y+z=0
αx+βy+γz=0
α2x+β2y+γ2z=0
Then the system of equations has

A
A unique solution for all values α,β,γ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
Infinite number of solutions if any two of α,β,γ are equal
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
A unique solution if α,β,γ are distinct
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
More than one, but finite number of solutions depending on values of α,β,γ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct options are
B Infinite number of solutions if any two of α,β,γ are equal
C A unique solution if α,β,γ are distinct
x+y+z=0
αx+βy+γz=0
α2x+β2y+γ2z=0
=∣ ∣ ∣111αβγα2β2γ2∣ ∣ ∣
If any of the two values (α,β) or (α,γ) or (β,γ) are equal then =0
Infinite solution
Option B
For all different values of α,β,γ
0
Unique solution
Option C

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