Solve all your doubts with Teachoo Black (new monthly pack available now!)

Are you in **school**? Do you **love Teachoo?**

We would love to talk to you! Please fill this form so that we can contact you

Examples

Example 1

Example 2

Example 3

Example 4

Example 5 Important

Example 6 Deleted for CBSE Board 2023 Exams

Example 7 Deleted for CBSE Board 2023 Exams

Example 8 Deleted for CBSE Board 2023 Exams

Example 9 Important Deleted for CBSE Board 2023 Exams

Example 10 Important Deleted for CBSE Board 2023 Exams

Example 11 Deleted for CBSE Board 2023 Exams

Example 12 Deleted for CBSE Board 2023 Exams

Example 13 Deleted for CBSE Board 2023 Exams

Example 14 Important Deleted for CBSE Board 2023 Exams

Example 15 Important Deleted for CBSE Board 2023 Exams You are here

Example 16 Important Deleted for CBSE Board 2023 Exams

Example 17

Example 18 Important

Example 19

Example 20

Example 21

Example 22 Important

Example 23

Example 24 Important

Example 25

Example 26 Important

Example 27

Example 28 Important

Example 29

Example 30

Example 31 Important Deleted for CBSE Board 2023 Exams

Example 32 Important Deleted for CBSE Board 2023 Exams

Example 33 Important

Example 34 Important Deleted for CBSE Board 2023 Exams

Chapter 4 Class 12 Determinants

Serial order wise

Last updated at Jan. 23, 2020 by Teachoo

Example 15 If x, y, z are different and Δ = |■8(x&x2&1+x3@y&y2&1+y3@z&z2&1+z3)| = 0 , then show that 1 + xyz = 0 Solving ∆ = |■8(x&x2&1+x3@y&y2&1+y3@z&z2&1+z3)| Here, expanding elements of C3 into two determinants = |■8(x&x2&1@y&y2&1@z&z2&1)| + |■8(x&x2&x3@y&y2&y3@z&z2&z3)| = |■8(x&x2&1@y&y2&1@z&z2&1)| + |■8(x&x2&x3@y&y2&y3@z&z2&z3)| = |■8(x&x2&1@y&y2&1@z&z2&1)|+ xyz |■8(1&x&x2@1&y&y2@1&z&z2)| = (−1) |■8(x&1&x2@y&1&y2@z&1&z2)| + xyz |■8(1&x&x2@1&y&y2@1&z&z2)| = (−1)(−1)|■8(1&x&x2@1&y&y2@1&z&z2)| + xyz |■8(1&x&x2@1&y&y2@1&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&x2@1&y&y2@1&z&z2)| + xyz |■8(1&x&x2@1&y&y2@1&z&z2)| = |■8(1&x&x2@1&y&y2@1&z&z2)| (1 + xyz) Using R2 → R2 – R1 and R3 → R3 – R1 = |■8(1&x&x2@𝟏 −𝟏&y−x&y2 −x2@𝟏−𝟏&z−x&z2 −x2)| (1+ xyz) = |■8(1&x&x2@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&x2@0&1&y+x@0&1&z+x)| Expanding determinant = (1 + xyz) (y – x) (z – x) (z – y) (1|■8(1&𝑦+𝑥@1&𝑧+𝑥)|" – 0 " |■8(𝑥&𝑥2@1&𝑧+𝑥)|" + 0" |■8(𝑥&𝑥2@1&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