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

Find the value of Δ∣ ∣ ∣2bca2c2b2c22cab2a2b2a22abc2∣ ∣ ∣.

Open in App
Solution

Δ by method of sarrus is a11a22a33+a12a23a31+a31a21a32a31a22a13a32a23a11a33a21a12(2bca2)(2cab2)(2abc2)+c2a2b2+b2c2a2b2(2cab2)b2a2a2(2bca2)(2abc2)c2c28a2b2c24abc44ab4c+2b3c34a4bc+2a3c3+2a3b3a2b2c2+2a2b2c22ab4c+b62a4bc+a62abc4+c6a6+b6+c66abc(a3+b3+c3)+2(a3b3+b3c3+a3c3)+9a2b2c2

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