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Question
Using properties of determinants, prove that `|(1,1,1+3x),(1+3y, 1,1),(1,1+3z,1)| = 9(3xyz + xy + yz+ zx)`
Solution
Let `triangle = |(1,1,(1+3z)),(1+3y, 1,1),(1,1+3z,1)| = 9(3xyz + xy + yz + zx)`
Applying R2→R2 − R1, R3→R3 − R1
`=>triangle = |(1,1,1+3x),(3y,0,-3x),(0,3z,-3x)|`
Expanding along R1 ,we get
`triangle = 1(0 + 9xz) - 1(-9xy - 0) + (1+3x)(9yz -0)`
`= 9xz + 9xy + 9yz + 27xyz`
`= 9(3xyz +xy +yz+zx)`
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