# Misc 12 - Chapter 4 Class 12 Determinants

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Misc 12 Using properties of determinants, prove that: xx21+px3yy21+py3zz21+pz3 = (1 + pxyz) (x – y) (y – z) (z – x) Taking L.H.S xx21+px3yy21+py3zz2a+pz3 Expressing elements of 3rd column as sum of two elements = xx21yy21zz21 + xx2px3yy2py3zz2pz3 = xx21yy21zz21 + xx2px3yy2py3zz2pz3 = xx21yy21zz21 + p xx2x3yy2x3zz2x3 = xx21yy21zz21 + pxyz 1xx21yx21zx2 = xx21yy21zz21 – pxyz x1x2y1y2z1z2 Again Replace C2 ↔ C3 = xx21yy21zz21 – ( – 1)pxyz xx21yy21zz21 = xx21yy21zz21 + pxyz xx21yy21zz21 Taking Common 𝑥𝑥21𝑦𝑦21𝑧𝑧21 = (1 + pxyz) xx21yy21zz21 Applying R1 → R1 − R2 = (1 + pxyz) x−yx2−y2𝟏−𝟏yy21zz21 = (1 + pxyz) x−y x−y(x−y)𝟎yy21zz21 Taking Common (x – y) From R1 = (1 + pxyz) (x – y) 1 x+y0yy21zz21 Applying R2 → R2 − R3 = (1 + pxyz) (x – y) 1 x+y0y−zy2−z2𝟏−𝟏zz21 = (1 + pxyz) (x – y) 1 x+y0y−z(y−z)(y+z)𝟎zz21 Taking Common (y – z) from R2 = (1 + pxyz) (x – y) (y – z) 1 x+y01𝑦+𝑧0zz21 Applying R1 → R1 − R2 = (1 + pxyz) (x – y) (y – z) 𝟏−𝟏 x+y−𝑦−𝑧0−01𝑦+𝑧0zz21 = (1 + pxyz) (x – y) (y – z) 𝟎 x−𝑧01𝑦+𝑧0zz21 Expanding determinant along C3 = (1 + pxyz) (x – y) (y – z) 0 1𝑦+𝑧𝑧𝑧2−0 0𝑥−𝑧𝑧𝑧2+1 0𝑥−𝑧1𝑦+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

Chapter 4 Class 12 Determinants

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.