Using the properties of determinants in prove that:


Taking LHS,

By applying R1 R1 + R2 + R3, we get




Taking 2 common from the first row, we get



Applying R1 R1 – R2, we get




Applying R3 R3 - R1, we get




Applying R2 R2 – R1, we get




Taking y, z, x common from R1, R2 and R3 respectively, we get



Expanding along C1, we get



= 2xyz [(1){(1) – 0} – (1){0 – 1} + 0}]


= 2xyz [1 + 1]


= 4xyz


= RHS


Hence,


LHS = RHS


Hence Proved


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