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CBSE Class 12 Math 2018 Solved Paper

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Question : 13
Total: 29
Using properties of determinants, prove that
|
111+3x
1+3y11
11+3z1
| = 9 (3xyz + xy + yz + zx)
Solution:  
Let Δ = |
111+3x
1+3y11
11+3z1
|
Apply
R2 → R2−R1 , R3 → R3−R1
Δ = |
111+3x
3y0−3x
03z−3x
|
Expanding along R1 , we get
Δ = 9xz + 9xy + 9yz + 27xyz
= 9 (xz + xy + yz + 3xyz)
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