Ex 4.2, 9 - Show that |x x2 yz y y2 zx z z2 xy| = (x-y) (y-z) - Solving by simplifying det.

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  1. Chapter 4 Class 12 Determinants
  2. Serial order wise
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Ex 4.2, 9 By using properties of determinants, show that: x﷮x2﷮yz﷮y﷮y2﷮zx﷮z﷮z2﷮xy﷯﷯ = (x – y) (y – z) (z – x)(xy + yz + zx) Taking L.H.S x﷮x2﷮yz﷮y﷮y2﷮zx﷮z﷮z2﷮xy﷯﷯ Applying R1→ R1 – R2 = x−𝑦﷮x2−y2﷮yz−yz﷮y﷮y2﷮zx﷮z﷮z2﷮xy﷯﷯ = 𝐱−𝒚﷮ 𝐱−𝒚﷯(𝑥+𝑦)﷮−z (𝐱−𝐲)﷮y﷮y2﷮zx﷮z﷮z2﷮xy﷯﷯ Taking out (x – y) common from R1 = (x – y) 1﷮𝑥+𝑦﷮−z﷮y﷮y2﷮zx﷮z﷮z2﷮xy﷯﷯ Applying R2→ R2 – R3 = (x – y) 1﷮𝑥+𝑦﷮−z﷮y−z﷮y2−z2﷮zx−xy﷮z﷮z2﷮xy﷯﷯ = (x – y) 1﷮𝑥+𝑦﷮−z﷮𝐲−𝐳﷮(𝐲−𝐳)(𝑦+𝑧)﷮−x(𝒚−𝒛)﷮z﷮z2﷮xy﷯﷯ Taking out (y – z) common from R2 = (x – y) (y – z) 1﷮𝑥+𝑦﷮−z﷮1﷮𝑦+𝑧﷮−x﷮z﷮z2﷮xy﷯﷯ Applying R1→ R1 – R2 = (x – y) (y – z) 1−𝟏﷮𝑥+𝑦−(𝑦+𝑧)﷮−z−(−x)﷮1﷮𝑦+𝑧﷮−x﷮z﷮z2﷮xy﷯﷯ = (x – y) (y – z) 𝟎﷮𝑥+𝑦−𝑦−𝑧﷮−z+x﷮1﷮𝑦+𝑧﷮−x﷮z﷮z2﷮xy﷯﷯ = (x – y) (y – z) 0﷮𝑥−𝑧﷮x−z﷮1﷮𝑦+𝑧﷮−x﷮z﷮z2﷮xy﷯﷯ = (x – y) (y – z) 0﷮−(𝒛−𝒙)﷮−(𝐳−𝐱)﷮1﷮𝑦+𝑧﷮−x﷮z﷮z2﷮xy﷯﷯ Taking out (z – x) from R1 = (x – y) (y – z) (z – x) 0﷮−1﷮−1﷮1﷮𝑦+𝑧﷮−x﷮z﷮z2﷮xy﷯﷯ Applying C2→ C2 – C3 = (x – y) (y – z) (z – x) 0﷮−1−(−1)﷮−1﷮1﷮𝑦+𝑧−(−𝑥)﷮−x﷮z﷮z2−xy﷮xy﷯﷯ = (x – y) (y – z) (z – x) 0﷮−1+1﷮−1﷮1﷮𝑦+𝑧+𝑥﷮−x﷮z﷮z2−xy﷮xy﷯﷯ = (x – y) (y – z) (z – x) 0﷮𝟎﷮−1﷮1﷮𝑥+𝑦+𝑧﷮−x﷮z﷮z2−xy﷮xy﷯﷯ Expanding Determinant along R1 = (x – y) (y – z) (z – x) 0 𝑥+𝑦+𝑧﷮−𝑥﷮𝑧2−𝑥𝑦﷮𝑥𝑦﷯﷯−0 1﷮−𝑥﷮𝑧﷮𝑥𝑦﷯﷯+ −1﷯ 1﷮𝑥+𝑦+𝑧﷮𝑧﷮𝑧2−𝑥𝑦﷯﷯﷯ = (x – y) (y – z) (z – x) 0−0+ −1﷯ 1﷮𝑥+𝑦+𝑧﷮𝑧﷮𝑧2−𝑥𝑦﷯﷯﷯ = (x – y) (y – z) (z – x) (0 – 0 – 1((z2 – xy) – z (x + y + z))) = (x – y) (y – z) (z – x) ( –(z2 – xy) + z (x + y + z)) = (x – y) (y – z) (z – x) ( – z2 + xy + + zx + zy + z2) = (x – y) (y – z) (z – x) ( xy + zx + zy + z2 – z2) = (x – y) (y – z) (z – x) ( xy + yz + zx) = R.H.S Hence proved

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