wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The system of linear equations λx+2y+2z=5;2λx+3y+5z=8;4x+λy+6z=10has:


A

no solution when λ=2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

infinitely many solution when λ=2

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

no solution when λ=8

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

a unique solution when λ=8

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A

no solution when λ=2


Explanation for the correct option:

Solving the given set of equations:

The given equations are;

λx+2y+2z=52λx+3y+5z=84x+λy+6z=10

The determinant form of these equations is

D=λ222λ354λ6

Solving the determinant

D=18λ5λ224λ+40+4λ224𝐷=λ26λ+16Now,sinceD=0λ2+6λ16=0λ2+8λ-2λ-16=0λλ+8-2λ+8=0λ+8λ-2=0λ=8or2

For λ=2

D1=2224351026D1=218-10-224-50+28-30D1=28-2-26+2-22D1=16+52-44D1=240

Therefore, the given equations have no solution for λ=2

Therefore, the correct answer is option (A).

Explanation for the incorrect options:

Option (B):

Since λ=2 has no solution.

Therefore, option (B) is the incorrect answer.

Option (C):

For λ=8

D1=82216351086D1=818-40-296-50+230-128D1=-176-92-196D1=-463<0

The equations will have a solution.

Therefore, option (C) is the incorrect option.

Option (D):

Since at λ=8 D1<0 so it will not have a unique solution.


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