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Misc 12 - Using properties of determinants, prove (1+pxyz) (x-z)

Misc 12 - Chapter 4 Class 12 Determinants - Part 2
Misc 12 - Chapter 4 Class 12 Determinants - Part 3
Misc 12 - Chapter 4 Class 12 Determinants - Part 4
Misc 12 - Chapter 4 Class 12 Determinants - Part 5
Misc 12 - Chapter 4 Class 12 Determinants - Part 6

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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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.