Example 32 - Show that Determinant = 2xyz (x +y+z)3 - Solving by simplifying det.

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Example 32 Show that Δ = 𝑦+𝑧﷯2﷮𝑥𝑦﷮𝑧𝑥﷮𝑥𝑦﷮ 𝑥+𝑧﷯2﷮𝑦𝑧﷮𝑥𝑧﷮𝑦𝑧﷮ 𝑥+𝑦﷯2﷯﷯ = 2xyz (x + y + z)3 Taking L.H.S Δ = 𝑦+𝑧﷯3﷮𝑥𝑦﷮𝑧𝑥﷮𝑥𝑦﷮ 𝑥+𝑧﷯2﷮𝑦𝑧﷮𝑥𝑧﷮𝑦𝑧﷮ 𝑥+𝑦﷯2﷯﷯ Divide & Multiply by xyz = 𝑥𝑦𝑧﷮𝑥𝑦𝑧﷯ 𝑦+𝑧﷯3﷮𝑥𝑦﷮𝑧𝑥﷮𝑥𝑦﷮ 𝑥+𝑧﷯2﷮𝑦𝑧﷮𝑥𝑧﷮𝑦𝑧﷮ 𝑥+𝑦﷯2﷯﷯ = 1﷮𝑥𝑦𝑧﷯ x. y. z 𝑦+𝑧﷯3﷮𝑥𝑦﷮𝑧𝑥﷮𝑥𝑦﷮ 𝑥+𝑧﷯2﷮𝑦𝑧﷮𝑥𝑧﷮𝑦𝑧﷮ 𝑥+𝑦﷯2﷯﷯ = 1﷮𝑥𝑦𝑧﷯ x. y. z 𝑦+𝑧﷯3﷮𝑥𝑦﷮𝑧𝑥﷮𝑥𝑦﷮ 𝑥+𝑧﷯2﷮𝑦𝑧﷮𝑥𝑧﷮𝑦𝑧﷮ 𝑥+𝑦﷯2﷯﷯ Multiplying R1 by x , R2 by y & R3 by z = 1﷮𝑥𝑦𝑧﷯ 𝒙 𝑦+𝑧﷯2﷮𝒙(𝑥𝑦)﷮𝒙(𝑧𝑥)﷮𝒚(𝑥𝑦)﷮𝒚 𝑥+𝑧﷯2﷮𝒚(𝑦𝑧)﷮𝒛(𝑥𝑧)﷮𝑦𝒛2﷮𝒛 𝑥+𝑦﷯2﷯﷯ Taking out x common from C1, y common from C2 & z common from C3 = 𝑥𝑦𝑧﷮𝑥𝑦𝑧﷯ y+z﷯2﷮x2﷮x2﷮y2﷮ x+z﷯2﷮y2﷮𝑧2﷮z2﷮ x+y﷯2﷯﷯ Applying C2→ C2 – C1 = y+z﷯2﷮x2− 𝑦+𝑧﷯2﷮x2﷮y2﷮ x+z﷯2−y2﷮y2﷮𝑧2﷮z2−z2﷮ x+y﷯2﷯﷯ = y+z﷯2﷮(x− 𝑦+𝑧﷯)(𝑥+ 𝑦+𝑧﷯)﷮x2﷮y2﷮ (x+z)−𝑦﷯(𝑥+𝑧+𝑦)﷮y2﷮𝑧2﷮0﷮ x+y﷯2﷯﷯ = y+z﷯2﷮(𝑥−𝑦−𝑧)(𝒙+𝒚+𝒛)﷮x2﷮y2﷮ x+z−y﷯(𝒙+𝒚+𝒛)﷮y2﷮𝑧2﷮0﷮ x+y﷯2﷯﷯ Taking out (𝑥+𝑦+𝑧) common from C2 = (𝑥+𝑦+𝑧) y+z﷯2﷮𝑥−𝑦−𝑧﷮x2﷮y2﷮𝑥+𝑧−𝑦﷮y2﷮𝑧2﷮0﷮ x+y﷯2﷯﷯ Applying C3 → C3 – C1 = (𝑥+𝑦+𝑧) y+z﷯2﷮𝑥−𝑦−𝑧﷮𝑥2 − 𝑦+𝑧﷯2﷮y2﷮𝑥+𝑧−𝑦﷮𝑦2−𝑦2﷮𝑧2﷮0﷮ 𝑥+𝑦﷯2−𝑧2﷯﷯ = (𝑥+𝑦+𝑧) y+z﷯2﷮𝑥−𝑦−𝑧﷮(𝒙+𝒚+𝒛)(𝑥−(𝑦+𝑧))﷮y2﷮𝑥+𝑧−𝑦﷮0﷮𝑧2﷮0﷮(𝒙+𝒚+𝒛)((𝑥+𝑦)−𝑧)﷯﷯ Taking out (𝑥+𝑦+𝑧) Common from C3 = (𝑥+𝑦+𝑧)(𝑥+𝑦+𝑧) y+z﷯2﷮𝑥−𝑦−𝑧﷮𝑥−𝑦−𝑧﷮y2﷮𝑥+𝑧−𝑦﷮0﷮𝑧2﷮0﷮𝑥+𝑦−𝑧﷯﷯ = 𝑥+𝑦+𝑧﷯2 y+z﷯2﷮𝑥−𝑦−𝑧﷮𝑥−𝑦−𝑧﷮y2﷮𝑥+𝑧−𝑦﷮0﷮𝑧2﷮0﷮𝑥+𝑦−𝑧﷯﷯ Applying R1→ R1 – R2 – R3 = 𝑥+𝑦+𝑧﷯2 y+z﷯2−𝑦2−𝑧2﷮ 𝑥−𝑦−𝑧﷯− 𝑥+𝑧−𝑦﷯−0﷮ 𝑥−𝑦−𝑧﷯−0−(𝑥+𝑦−𝑧)﷮y2﷮𝑥+𝑧−𝑦﷮0﷮𝑧2﷮0﷮𝑥+𝑦−𝑧﷯﷯ = 𝑥+𝑦+𝑧﷯2 𝑦2+𝑧2+2𝑦𝑧−𝑦2−𝑧2﷮𝑥−𝑥−𝑦+𝑦−𝑧−𝑧﷮𝑥−𝑥−𝑦−𝑦−𝑧+𝑧﷮y2﷮𝑥+𝑧−𝑦﷮0﷮𝑧2﷮0﷮𝑥+𝑦−𝑧﷯﷯ = 𝑥+𝑦+𝑧﷯2 2𝑦𝑧﷮−2𝑧﷮−2𝑦﷮y2﷮𝑥+𝑧−𝑦﷮0﷮𝑧2﷮0﷮𝑥+𝑦−𝑧﷯﷯ Applying C2→ C2 + 1﷮𝑦﷯ C1 = 𝑥+𝑧+𝑦﷯2 2𝑦𝑧﷮−2𝑧+ 𝟏﷮𝒚﷯(𝟐𝒚𝒛)﷮2𝑦﷮y2﷮x−𝑦+𝑧+ 𝟏﷮𝒚﷯ (𝒚𝟐)﷮0﷮𝑧2﷮0+ 𝟏﷮𝒚﷯(𝒛𝟐)﷮𝑥+𝑦−𝑧﷯﷯ = 𝑥+𝑧+𝑦﷯2 2𝑦𝑧﷮0﷮2𝑦﷮y2﷮x+𝑧﷮0﷮𝑧2﷮ 𝑧﷮2﷯﷮𝑦﷯﷮𝑥+𝑦−𝑧﷯﷯ Applying C3→ C3 + 1﷮𝑧﷯ C1 = 𝑥+𝑦+𝑧﷯2 2𝑦𝑧﷮0﷮−2𝑦+ 𝟏﷮𝒛﷯(𝟐𝒚𝒛)﷮y2﷮𝑥+𝑧﷮0+ 𝟏﷮𝒛﷯ (𝒚𝟐)﷮𝑧2﷮ 𝑧﷮2﷯﷮𝑦﷯﷮ 𝑥+𝑦−𝑧﷯+ 𝟏﷮𝒛﷯ (𝒛𝟐)﷯﷯ = 𝑥+𝑦+𝑧﷯2 2𝑦𝑧﷮0﷮0﷮y2﷮𝑥+𝑧﷮ 𝑦﷮2﷯﷮𝑧﷯ ﷮𝑧2﷮ 𝑧﷮2﷯﷮𝑦﷯﷮𝑥+𝑦﷯﷯ Expanding Determinant along R1 = 𝑥+𝑦+𝑧﷯2 2𝑦𝑧 𝑥+𝑧﷮ 𝑧﷮2﷯﷮𝑦﷯﷮ 𝑧﷮2﷯﷮𝑦﷯﷮𝑥+𝑦﷯﷯−0 𝑦2﷮ 𝑦﷮2﷯﷮𝑧﷯﷮ 𝑧﷮2﷯﷮𝑥+𝑦﷯﷯+0 𝑦2﷮𝑥+𝑦﷮ 𝑧﷮2﷯﷮ 𝑧﷮2﷯﷮𝑦﷯﷯﷯﷯ = 𝑥+𝑦+𝑧﷯2 2𝑦𝑧 𝑥+𝑧﷮ 𝑧﷮2﷯﷮𝑦﷯﷮ 𝑧﷮2﷯﷮𝑦﷯﷮𝑥+𝑦﷯﷯−0+0﷯ = 𝑥+𝑦+𝑧﷯2 2yz (x + z) (x + y) – 𝑧2﷮𝑦﷯ 𝑦2﷮𝑧﷯﷯﷯ – 0 + 0﷯ = 𝑥+𝑦+𝑧﷯2 (2yz ((x + z) (x + y) – zy ) = 𝑥+𝑦+𝑧﷯2 (2yz) ((x + z) (x + y) – zy ) = 𝑥+𝑦+𝑧﷯2 (2yz) (x2 + xy + zx + zy – zy) = 𝑥+𝑦+𝑧﷯2 (2yz) (x2 + xy + xz) = 𝑥+𝑦+𝑧﷯2 (2yz) . x (x + y + z) = 𝑥+𝑦+𝑧﷯2 (2xyz) = (2xyz) 𝑥+𝑦+𝑧﷯2 = R.H.S Hence Proved

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