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

Ex 4.2, 9 - Chapter 4 Class 12 Determinants - Part 2
Ex 4.2, 9 - Chapter 4 Class 12 Determinants - Part 3 Ex 4.2, 9 - Chapter 4 Class 12 Determinants - Part 4 Ex 4.2, 9 - Chapter 4 Class 12 Determinants - Part 5 Ex 4.2, 9 - Chapter 4 Class 12 Determinants - Part 6

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Transcript

Question 9 By using properties of determinants, show that: |■8(x&x2&yz@y&y2&zx@z&z2&xy)| = (x – y) (y – z) (z – x)(xy + yz + zx) Solving L.H.S |■8(𝑥&𝑥^2&𝑦𝑧@𝑦&𝑦^2&𝑧𝑥@𝑧&𝑧^2&𝑥𝑦)| Applying R1→ R1 – R2 = |■8(𝑥−𝑦&𝑥^2−𝑦^2&𝑦𝑧−𝑥𝑧@𝑦&𝑦^2&𝑧𝑥@𝑧&𝑧^2&𝑥𝑦)| Question 9 By using properties of determinants, show that: |■8(x&x2&yz@y&y2&zx@z&z2&xy)| = (x – y) (y – z) (z – x)(xy + yz + zx) Solving L.H.S |■8(𝑥&𝑥^2&𝑦𝑧@𝑦&𝑦^2&𝑧𝑥@𝑧&𝑧^2&𝑥𝑦)| Applying R1→ R1 – R2 = |■8(𝑥−𝑦&𝑥^2−𝑦^2&𝑦𝑧−𝑥𝑧@𝑦&𝑦^2&𝑧𝑥@𝑧&𝑧^2&𝑥𝑦)| = |■8(𝒙−𝒚&(𝒙−𝒚)(𝑥+𝑦)&−𝑧 (𝒙−𝒚)@𝑦&𝑦^2&𝑧𝑥@𝑧&𝑧^2&𝑥𝑦)| Taking out (x – y) common from R1 = (x – y) |■8(1&𝑥+𝑦&−𝑧@𝑦&𝑦^2&𝑧𝑥@𝑧&𝑧^2&𝑥𝑦)| Applying R2→ R2 – R3 = (x – y) |■8(1&𝑥+𝑦&−𝑧@𝑦−𝑧&𝑦^2−𝑧^2&𝑧𝑥−𝑥𝑦@𝑧&𝑧^2&𝑥𝑦)| = (x – y) |■8(1&𝑥+𝑦&−𝑧@𝒚−𝒛&(𝒚−𝒛)(𝑦+𝑧)&−𝑥(𝒚−𝒛)@𝑧&𝑧^2&𝑥𝑦)| Taking out (y – z) common from R2 = (x – y) (y – z)|■8(1&𝑥+𝑦&−𝑧@1&𝑦+𝑧&−𝑥@𝑧&𝑧2&𝑥𝑦)| Applying R1→ R1 – R2 = (x – y) (y – z)|■8(1−𝟏&𝑥+𝑦−(𝑦+𝑧)&−𝑧−(−𝑥)@1&𝑦+𝑧&−𝑥@𝑧&𝑧2&𝑥𝑦)| = (x – y) (y – z)|■8(𝟎&𝑥+𝑦−𝑦−𝑧&−𝑧+𝑥@1&𝑦+𝑧&−𝑥@𝑧&𝑧2&𝑥𝑦)| = (x – y) (y – z)|■8(0&𝑥−𝑧&𝑥−𝑧@1&𝑦+𝑧&−𝑥@𝑧&𝑧2&𝑥𝑦)| = (x – y) (y – z)|■8(0&−(𝒛−𝒙)&−(𝒛−𝒙)@1&𝑦+𝑧&−𝑥@𝑧&𝑧2&𝑥𝑦)| Taking out (z – x) from R1 = (x – y) (y – z) (z – x) |■8(0&−1&−1@1&𝑦+𝑧&−𝑥@𝑧&𝑧2&𝑥𝑦)| Applying C2→ C2 – C3 = (x – y) (y – z) (z – x) |■8(0&−1−(−1)&−1@1&𝑦+𝑧−(−𝑥)&−𝑥@𝑧&𝑧2−𝑥𝑦&𝑥𝑦)| = (x – y) (y – z) (z – x) |■8(0&−1+1&−1@1&𝑦+𝑧+𝑥&−𝑥@𝑧&𝑧2−𝑥𝑦&𝑥𝑦)| = (x – y) (y – z) (z – x) |■8(0&𝟎&−1@1&𝑥+𝑦+𝑧&−x@z&z2−xy&xy)| Expanding Determinant along R1 = (x – y) (y – z) (z – x) (0|■8(𝑥+𝑦+𝑧&−𝑥@𝑧2−𝑥𝑦&𝑥𝑦)|−0|■8(1&−𝑥@𝑧&𝑥𝑦)|+(−1)|■8(1&𝑥+𝑦+𝑧@𝑧&𝑧2−𝑥𝑦)|) = (x – y) (y – z) (z – x) (0−0+(−1)|■8(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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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.