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CBSE Class 12 Math 2018 Solved Paper

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Question : 13 of 29
Marks: +1, -0
Using properties of determinants, prove that
∣111+3x1+3y1111+3z1∣\begin{vmatrix} 1 & 1 & 1+3x \\ 1+3y & 1 & 1 \\ 1 & 1+3z & 1 \end{vmatrix} = 9 (3xyz + xy + yz + zx)
Solution:  
Let Δ = ∣111+3x1+3y1111+3z1∣\begin{vmatrix} 1 & 1 & 1+3x \\ 1+3y & 1 & 1 \\ 1 & 1+3z & 1 \end{vmatrix}
Apply
R2R_2 → R2−R1R_2 - R_1 , R3R_3 → R3−R1R_3 - R_1
Δ = ∣111+3x3y0−3x03z−3x∣\begin{vmatrix} 1 & 1 & 1+3x \\ 3y & 0 & -3x \\ 0 & 3z & -3x \end{vmatrix}
Expanding along R1R_1 , we get
Δ = 9xz + 9xy + 9yz + 27xyz
= 9 (xz + xy + yz + 3xyz)
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