Q. 45.0( 1 Vote )

Using the propert

Answer :

Let

By applying C1 C1 + C2 + C3, we have




Taking (x + y + z) common from Column C1, we get



By applying R2 R2 – R1, we get





By applying R3 R3 – R1, we get





Applying C2 C2 – C3, we get





Now, expanding along C1, we get


= (x + y + z) [1×{(3y)(2z + x) – (-3z)(x – y)}]


= (x + y + z) [6yz + 3yx + (3z)(x – y)]


= (x + y + z) [6yz + 3yx + 3zx – 3zy]


= (x + y + z) [3yz + 3zx + 3yx]


= 3(x + y + z)(yz + zx + yx)


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