CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
7
You visited us 7 times! Enjoying our articles? Unlock Full Access!
Question

The value of determinant ∣ ∣ ∣1+a2b22ab2b2ab1a2+b22a2b2a1a2b2∣ ∣ ∣ is equal to

A
(1a2b2)3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(a+b+1)2(ab+b+a)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
(1+a2+b2)3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
(1a2+b2)3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C (1+a2+b2)3
Δ=1ab∣ ∣ ∣b(1+a2b2)2ab22b22a2ba(1a2+b2)2a22b2a1a2b2∣ ∣ ∣by(R1×b,R2×a)∣ ∣ ∣1+a2b22b22b22a21a2+b22a2221a2b2∣ ∣ ∣by(C1b,C2a)∣ ∣ ∣1+a2+b202b21+a2+b21+a2+b22a20(1+a2+b2)1a2b2∣ ∣ ∣(C1C1+C2,C2C2+C3)=(1+a2+b2)3

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