Answer :

Let Δ =

Applying Elementary Row Transformations


R2 R2 – R1 and R3 R3 – R1


Δ =


Taking (y – x) and (z – x) common from R2 and R3 respectively


Δ = (y – x) (z – x)


Applying R3 R3 – R2


Δ = (y – x) (z – x)


Taking (z – y) common from R3


Δ = (y – x) (z – x) (z – y)


Expanding along R3, we have


Δ = (x – y) (y – z) (z – x) [0 – 1 {x × p(y2 + x2 + xy) – 1 × (1 + px3)} + p (x + y + z) {x × (y + x) – 1 × x2}


Δ = (x – y) (y – z) (z – x) (-px3 – pxy2 – px2y + 1 + px3 + px2y + pxy2 + pxyz)


Δ = (x – y) (y – z) (z – x) (1 + pxyz)


Hence, the given result is proved.


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