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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Misc 5 (Method 1) Evaluate |■8(𝑥&𝑦&𝑥+𝑦@𝑦&𝑥+𝑦&𝑥@𝑥+𝑦&𝑥&𝑦)| Let ∆ = |■8(𝑥&𝑦&𝑥+𝑦@𝑦&𝑥+𝑦&𝑥@𝑥+𝑦&𝑥&𝑦)| = 𝑥[(𝑥+𝑦)𝑦−𝑥^2 ]−𝑦[𝑦^2−𝑥(𝑥+𝑦)]+(𝑥+𝑦)[𝑥𝑦−(𝑥+𝑦)^2 ] = 𝑥[𝑥𝑦+𝑦^2−𝑥^2 ]−𝑦[𝑦^2−𝑥^2−𝑥𝑦]+(𝑥+𝑦)[𝑥𝑦−𝑥^2−𝑦^2−2𝑥𝑦] = 𝒙[𝒙𝒚+𝒚^𝟐−𝒙^𝟐 ]−𝒚[𝒚^𝟐−𝒙^𝟐−𝒙𝒚]+(𝒙+𝒚)[−𝒙^𝟐−𝒚^𝟐−𝒙𝒚] = 𝑥^2 𝑦+𝑥𝑦^2−𝑥^3−𝑦^3+𝑥^2 𝑦+𝑥𝑦^2−𝑥^3−𝑥𝑦^2−𝑥^2 𝑦−𝑥^2 𝑦−𝑦[𝑦^2−𝑥^2−𝑥𝑦]+(𝑥+𝑦)[−𝑥^2−𝑦^2−𝑥𝑦] = 𝒙[𝒙𝒚+𝒚^𝟐−𝒙^𝟐 ]−𝒚[𝒚^𝟐−𝒙^𝟐−𝒙𝒚]+(𝒙+𝒚)[−𝒙^𝟐−𝒚^𝟐−𝒙𝒚] = 𝑥^2 𝑦+𝑥𝑦^2−𝑥^3−𝑦^3+𝑥^2 𝑦+𝑥𝑦^2−𝑥^3−𝑥𝑦^2−𝑥^2 𝑦−𝑥^2 𝑦−𝑦^3−𝑥𝑦^2 = 𝑥^2 𝑦+𝑥𝑦^2−𝑥^3−𝑦^3+𝑥^2 𝑦+𝑥𝑦^2−𝑥^3−𝑥𝑦^2−𝑥^2 𝑦−𝑥^2 𝑦−𝑦^3−𝑥𝑦^2 = −2𝑥^3−2𝑦^3 = − 2(x3+y3) Hence , ∆ = – 2(𝐱𝟑+𝐲𝟑) Misc 5 (Method 2) Evaluate |■8(𝑥&𝑦&𝑥+𝑦@𝑦&𝑥+𝑦&𝑥@𝑥+𝑦&𝑥&𝑦)| Let ∆ = |■8(𝑥&𝑦&𝑥+𝑦@𝑦&𝑥+𝑦&𝑥@𝑥+𝑦&𝑥&𝑦)| Applying R1→ R1 + R2 + R3 = |■8(𝑥+𝑦+𝑥+𝑦&𝑦+𝑥+𝑦+𝑥&𝑥+𝑦+𝑥+𝑦@𝑦&𝑥+𝑦&𝑥@𝑥+𝑦&𝑥&𝑦)| = |■8(2x+2y&2x+2y&2x+2y@y&x+y&x@x+y&x&y)| = |■8(𝟐(𝐱+𝐲)&𝟐(𝐱+𝐲)&𝟐(𝐱+𝐲)@y&x+y&x@x+y&x&y)| Taking common 2(x + y), from R1 = 𝟐(𝐱+𝐲) |■8(1&1&1@y&x+y&x@x+y&x&y)| Applying C2→ C2 – C1 = 2(x+y) |■8(1&𝟏−𝟏&1@y&x+y−𝑦&x@x+y&x−x−y&y)| = 2(x+y) |■8(1&𝟎&1@y&x&x@x+y&−y&y)| Applying C3 →C3 – C1 = 2(x+y) |■8(1&0&𝟏−𝟏@y&x&x−y@x+y&−y&y−(x+y))| = 2(x+y) |■8(1&0&𝟎@y&x&x−y@x+y&−y&−x)| Expanding determinant along R1 = 2(x+y) (1|■8(𝑥&𝑥−𝑦@−𝑦&−𝑥)|−0|■8(𝑦&𝑥−𝑦@𝑥+𝑦&−𝑥)|+0|■8(𝑦&𝑥@𝑥+𝑦&−𝑦)|) = 2(x+y) (1|■8(𝑥&𝑥−𝑦@−𝑦&−𝑥)|−0+0) = 2(x+y) (1( – x2 – ( –y) (x – y)) ) = 2(x+y) ( – x2 + y (x – y)) = 2(x+y) ( – x2 + xy – y2) = – 2(x+y) ( x2 + y2 – xy) = − 2(x3+y3) Hence , ∆ = – 2(𝐱𝟑+𝐲𝟑)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.