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Ex 4.2
Ex 4.2, 2 Important
Ex 4.2, 3
Ex 4.2, 4
Ex 4.2, 5 Important
Ex 4.2, 6 Important
Ex 4.2, 7 Important
Ex 4.2, 8 (i) Important
Ex 4.2, 8 (ii)
Ex 4.2, 9 Important You are here
Ex 4.2, 10 (i)
Ex 4.2, 10 (ii) Important
Ex 4.2, 11 (i)
Ex 4.2, 11 (ii) Important
Ex 4.2, 12 Important
Ex 4.2, 13 Important
Ex 4.2, 14 Important
Ex 4.2, 15 (MCQ) Important
Ex 4.2, 16 (MCQ)
Last updated at Jan. 22, 2020 by Teachoo
Ex 4.2, 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)| Ex 4.2, 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