# Misc 12 - Chapter 4 Class 12 Determinants (Term 1)

Last updated at Jan. 23, 2020 by Teachoo

Last updated at Jan. 23, 2020 by Teachoo

Transcript

Misc 12 Using properties of determinants, prove that: |■8(x&x2&1+px3@y&y2&1+py3@z&z2&1+pz3)| = (1 + pxyz) (x – y) (y – z) (z – x) Solving L.H.S |■8(x&x2&1+px3@y&y2&1+py3@z&z2&a+pz3)| Expressing elements of 3rd column as sum of two elements = |■8(x&x2&1@y&y2&1@z&z2&1)| + |■8(x&x2&px3@y&y2&py3@z&z2&pz3)| Using Property : If some or all elements of a row or column of a determinant are expressed as sum of two (or more) terms ,then the determinant is expressed as a sum of two (or more) determinants. = |■8(x&x2&1@y&y2&1@z&z2&1)| + |■8(x&x2&px3@y&y2&py3@z&z2&pz3)| = |■8(x&x2&1@y&y2&1@z&z2&1)| + p |■8(x&x2&x3@y&y2&x3@z&z2&x3)| = |■8(x&x2&1@y&y2&1@z&z2&1)| + pxyz |■8(1&x&x2@1&y&x2@1&z&x2)| = |■8(x&x2&1@y&y2&1@z&z2&1)| – pxyz |■8(x&1&x2@y&1&y2@z&1&z2)| Taking p common from C3 Taking x common from R1 , y common from R2 , z common from R3 If any two rows ( or columns) of a determinant are interchanged , then sign of determinant changes. Again Replace C2 ↔ C3 = |■8(x&x2&1@y&y2&1@z&z2&1)| – ( – 1)pxyz |■8(x&x2&1@y&y2&1@z&z2&1)| = |■8(x&x2&1@y&y2&1@z&z2&1)| + pxyz |■8(x&x2&1@y&y2&1@z&z2&1)| Taking Common |■8(𝒙&𝒙𝟐&𝟏@𝒚&𝒚𝟐&𝟏@𝒛&𝒛𝟐&𝟏)| = (1 + pxyz) |■8(x&x2&1@y&y2&1@z&z2&1)| Applying R1 → R1 − R2 = (1 + pxyz) |■8(x−y&x2−y2&𝟏−𝟏@y&y2&1@z&z2&1)| = (1 + pxyz) |■8(x−y&(x−y)(x−y)&𝟎@y&y2&1@z&z2&1)| Taking Common (x – y) From R1 = (1 + pxyz) (x – y) |■8(1& x+y&0@y&y2&1@z&z2&1)| Applying R2 → R2 − R3 = (1 + pxyz) (x – y) |■8(1& x+y&0@y−z&y2−z2&𝟏−𝟏@z&z2&1)| = (1 + pxyz) (x – y) |■8(1& x+y&0@y−z&(y−z)(y+z)&𝟎@z&z2&1)| Taking Common (y – z) from R2 = (1 + pxyz) (x – y) (y – z) |■8(1& x+y&0@1&𝑦+𝑧&0@z&z2&1)| Applying R1 → R1 − R2 = (1 + pxyz) (x – y) (y – z) |■8(𝟏−𝟏& x+y−𝑦−𝑧&0−0@1&𝑦+𝑧&0@z&z2&1)| = (1 + pxyz) (x – y) (y – z) |■8(𝟎& x−𝑧&0@1&𝑦+𝑧&0@z&z2&1)| Expanding determinant along C3 = (1 + pxyz) (x – y) (y – z) (0|■8(1&𝑦+𝑧@𝑧&𝑧2)|−0|■8(0&𝑥−𝑧@𝑧&𝑧2)|+1|■8(0&𝑥−𝑧@1&𝑦+𝑧)|) = (1 + pxyz) (x – y) (y – z) (0 – 0 + 1 (0 – (x – z)) = (1 + pxyz) (x – y) (x – z) (0 – x + z) = (1 + pxyz) (x – y) (y – z) (z – x) = R.H.S Hence Proved

Miscellaneous

Misc 1

Misc. 2 Important Deleted for CBSE Board 2022 Exams

Misc 3

Misc 4 Deleted for CBSE Board 2022 Exams

Misc 5

Misc 6 Important Deleted for CBSE Board 2022 Exams

Misc 7 Important

Misc 8

Misc 9

Misc 10

Misc 11 Important Deleted for CBSE Board 2022 Exams

Misc 12 Important Deleted for CBSE Board 2022 Exams You are here

Misc. 13 Deleted for CBSE Board 2022 Exams

Misc 14 Deleted for CBSE Board 2022 Exams

Misc. 15 Important Deleted for CBSE Board 2022 Exams

Misc. 16 Important

Misc 17 Important Deleted for CBSE Board 2022 Exams

Misc 18

Misc 19 Important

Matrices and Determinants - Formula Sheet and Summary

Chapter 4 Class 12 Determinants (Term 1)

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.