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

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Misc 12 Using properties of determinants, prove that: |■8(x&x2&[email protected]&y2&[email protected]&z2&1+pz3)| = (1 + pxyz) (x – y) (y – z) (z – x) Solving L.H.S |■8(x&x2&[email protected]&y2&[email protected]&z2&a+pz3)| Expressing elements of 3rd column as sum of two elements = |■8(x&x2&[email protected]&y2&[email protected]&z2&1)| + |■8(x&x2&[email protected]&y2&[email protected]&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&[email protected]&y2&[email protected]&z2&1)| + |■8(x&x2&[email protected]&y2&[email protected]&z2&pz3)| = |■8(x&x2&[email protected]&y2&[email protected]&z2&1)| + p |■8(x&x2&[email protected]&y2&[email protected]&z2&x3)| = |■8(x&x2&[email protected]&y2&[email protected]&z2&1)| + pxyz |■8(1&x&[email protected]&y&[email protected]&z&x2)| = |■8(x&x2&[email protected]&y2&[email protected]&z2&1)| – pxyz |■8(x&1&[email protected]&1&[email protected]&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&[email protected]&y2&[email protected]&z2&1)| – ( – 1)pxyz |■8(x&x2&[email protected]&y2&[email protected]&z2&1)| = |■8(x&x2&[email protected]&y2&[email protected]&z2&1)| + pxyz |■8(x&x2&[email protected]&y2&[email protected]&z2&1)| Taking Common |■8(𝒙&𝒙𝟐&𝟏@𝒚&𝒚𝟐&𝟏@𝒛&𝒛𝟐&𝟏)| = (1 + pxyz) |■8(x&x2&[email protected]&y2&[email protected]&z2&1)| Applying R1 → R1 − R2 = (1 + pxyz) |■8(x−y&x2−y2&𝟏−𝟏@y&y2&[email protected]&z2&1)| = (1 + pxyz) |■8(x−y&(x−y)(x−y)&𝟎@y&y2&[email protected]&z2&1)| Taking Common (x – y) From R1 = (1 + pxyz) (x – y) |■8(1& x+y&[email protected]&y2&[email protected]&z2&1)| Applying R2 → R2 − R3 = (1 + pxyz) (x – y) |■8(1& x+y&[email protected]−z&y2−z2&𝟏−𝟏@z&z2&1)| = (1 + pxyz) (x – y) |■8(1& x+y&[email protected]−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&[email protected]&𝑦+𝑧&[email protected]&z2&1)| Applying R1 → R1 − R2 = (1 + pxyz) (x – y) (y – z) |■8(𝟏−𝟏& x+y−𝑦−𝑧&0−[email protected]&𝑦+𝑧&[email protected]&z2&1)| = (1 + pxyz) (x – y) (y – z) |■8(𝟎& x−𝑧&[email protected]&𝑦+𝑧&[email protected]&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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.