Q. 45.0( 1 Vote )

# Using the propert

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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