Check sibling questions

Example 15  - If x, y, z are different, show 1 + xyz = 0 - Class 12

Example 15 - Chapter 4 Class 12 Determinants - Part 2
Example 15 - Chapter 4 Class 12 Determinants - Part 3 Example 15 - Chapter 4 Class 12 Determinants - Part 4 Example 15 - Chapter 4 Class 12 Determinants - Part 5

Get live Maths 1-on-1 Classs - Class 6 to 12


Transcript

Example 15 If x, y, z are different and Δ = |■8(x&x2&[email protected]&y2&[email protected]&z2&1+z3)| = 0 , then show that 1 + xyz = 0 Solving ∆ = |■8(x&x2&[email protected]&y2&[email protected]&z2&1+z3)| Here, expanding elements of C3 into two determinants = |■8(x&x2&[email protected]&y2&[email protected]&z2&1)| + |■8(x&x2&[email protected]&y2&[email protected]&z2&z3)| = |■8(x&x2&[email protected]&y2&[email protected]&z2&1)| + |■8(x&x2&[email protected]&y2&[email protected]&z2&z3)| = |■8(x&x2&[email protected]&y2&[email protected]&z2&1)|+ xyz |■8(1&x&[email protected]&y&[email protected]&z&z2)| = (−1) |■8(x&1&[email protected]&1&[email protected]&1&z2)| + xyz |■8(1&x&[email protected]&y&[email protected]&z&z2)| = (−1)(−1)|■8(1&x&[email protected]&y&[email protected]&z&z2)| + xyz |■8(1&x&[email protected]&y&[email protected]&z&z2)| Taking x , y , z common from R1, R2, R3 respectively Replacing C3↔ C2 Replacing C1↔ C2 If any two columns of a determinant are interchanged , then sign of determinant changes = |■8(1&x&[email protected]&y&[email protected]&z&z2)| + xyz |■8(1&x&[email protected]&y&[email protected]&z&z2)| = |■8(1&x&[email protected]&y&[email protected]&z&z2)| (1 + xyz) Using R2 → R2 – R1 and R3 → R3 – R1 = |■8(1&x&[email protected]𝟏 −𝟏&y−x&y2 −[email protected]𝟏−𝟏&z−x&z2 −x2)| (1+ xyz) = |■8(1&x&[email protected]𝟎&(y−x)&(y −x)(y+x)@𝟎&(z−x)&(z −x)(z+x))| (1+ xyz) Taking common factor (y – x) from R2 & (z – x) from R3 = (1 + xyz) (y – x) (z – x) |■8(1&x&[email protected]&1&[email protected]&1&z+x)| Expanding determinant = (1 + xyz) (y – x) (z – x) (z – y) (1|■8(1&𝑦+𝑥@1&𝑧+𝑥)|" – 0 " |■8(𝑥&𝑥[email protected]&𝑧+𝑥)|" + 0" |■8(𝑥&𝑥[email protected]&𝑦+𝑥)|) = (1 + xyz) (y – x) (z – x) (1 (y + x) – (y + x) + 0 + 0) = (1 + xyz) (y – x) (z – x) (z + y – y – x) = (1 + xyz) (y – x) (z – x) (z – y) ∴ ∆ = (1 + xyz) (y – x) (z – x) (z – y) Since ∆ = 0 given (1 + xyz) (y – x) (z – x) (z – y) = 0 Since it is given that x, y, z all are different, i.e., y – x ≠ 0, z – x ≠ 0, z – y ≠ 0, So, only Possibility is (1 + xyz) = 0 Hence Proved

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

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.