CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
102
You visited us 102 times! Enjoying our articles? Unlock Full Access!
Question

For how many values of x in the closed interval [4,1] the matrix 3x1231x+2x+312 is singular.

A
1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 1
A matrix is singular only if its determinant is zero
∣ ∣3x1231x+2x+312∣ ∣=03(2+x+2)+(x+1)(6(x+2)(x+3))+2(3+x+3)=03x+(x+1)(x25x)+2x=0x3+6x2=0x=0,6
0[4,1]
Hence, option A is correct.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Adjoint and Inverse of a Matrix
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon