Misc 9 - Chapter 4 Class 12 Determinants (Term 1)

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Misc 9 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&[email protected]&x+y&[email protected]+y&x&y)| = |■8(𝟐(𝐱+𝐲)&𝟐(𝐱+𝐲)&𝟐(𝐱+𝐲)@y&x+y&[email protected]+y&x&y)| Taking common 2(x + y), from R1 = 𝟐(𝐱+𝐲) |■8(1&1&[email protected]&x+y&[email protected]+y&x&y)| Applying C2→ C2 – C1 = 2(x+y) |■8(1&𝟏−𝟏&[email protected]&x+y−𝑦&[email protected]+y&x−x−y&y)| = 2(x+y) |■8(1&𝟎&[email protected]&x&[email protected]+y&−y&y)| Applying C3 →C3 – C1 = 2(x+y) |■8(1&0&𝟏−𝟏@y&x&x−[email protected]+y&−y&y−(x+y))| = 2(x+y) |■8(1&0&𝟎@y&x&x−[email protected]+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) (Using a3 + b3 = (a + b) (a2 + b2 – ab))