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Ex 4.2
Ex 4.2, 2 Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 3 Deleted for CBSE Board 2023 Exams
Ex 4.2, 4 Deleted for CBSE Board 2023 Exams
Ex 4.2, 5 Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 6 Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 7 Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 8 (i) Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 8 (ii) Deleted for CBSE Board 2023 Exams
Ex 4.2, 9 Important Deleted for CBSE Board 2023 Exams You are here
Ex 4.2, 10 (i) Deleted for CBSE Board 2023 Exams
Ex 4.2, 10 (ii) Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 11 (i) Deleted for CBSE Board 2023 Exams
Ex 4.2, 11 (ii) Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 12 Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 13 Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 14 Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 15 (MCQ) Important Deleted for CBSE Board 2023 Exams
Ex 4.2, 16 (MCQ) Deleted for CBSE Board 2023 Exams
Last updated at March 30, 2023 by Teachoo
Ex 4.2, 9 By using properties of determinants, show that: |■8(x&x2&[email protected]&y2&[email protected]&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&[email protected]&y2&[email protected]&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&−[email protected]&𝑦+𝑧&−𝑥@𝑧&𝑧2&𝑥𝑦)| Applying C2→ C2 – C3 = (x – y) (y – z) (z – x) |■8(0&−1−(−1)&−[email protected]&𝑦+𝑧−(−𝑥)&−𝑥@𝑧&𝑧2−𝑥𝑦&𝑥𝑦)| = (x – y) (y – z) (z – x) |■8(0&−1+1&−[email protected]&𝑦+𝑧+𝑥&−𝑥@𝑧&𝑧2−𝑥𝑦&𝑥𝑦)| = (x – y) (y – z) (z – x) |■8(0&𝟎&−[email protected]&𝑥+𝑦+𝑧&−[email protected]&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