Misc 12 - Class 12

Transcript

Misc 12 Using properties of determinants, prove that: x﷮x2﷮1+px3﷮y﷮y2﷮1+py3﷮z﷮z2﷮1+pz3﷯﷯ = (1 + pxyz) (x – y) (y – z) (z – x) Taking L.H.S x﷮x2﷮1+px3﷮y﷮y2﷮1+py3﷮z﷮z2﷮a+pz3﷯﷯ Expressing elements of 3rd column as sum of two elements = x﷮x2﷮1﷮y﷮y2﷮1﷮z﷮z2﷮1﷯﷯ + x﷮x2﷮px3﷮y﷮y2﷮py3﷮z﷮z2﷮pz3﷯﷯ = x﷮x2﷮1﷮y﷮y2﷮1﷮z﷮z2﷮1﷯﷯ + x﷮x2﷮px3﷮y﷮y2﷮py3﷮z﷮z2﷮pz3﷯﷯ = x﷮x2﷮1﷮y﷮y2﷮1﷮z﷮z2﷮1﷯﷯ + p x﷮x2﷮x3﷮y﷮y2﷮x3﷮z﷮z2﷮x3﷯﷯ = x﷮x2﷮1﷮y﷮y2﷮1﷮z﷮z2﷮1﷯﷯ + pxyz 1﷮x﷮x2﷮1﷮y﷮x2﷮1﷮z﷮x2﷯﷯ = x﷮x2﷮1﷮y﷮y2﷮1﷮z﷮z2﷮1﷯﷯ – pxyz x﷮1﷮x2﷮y﷮1﷮y2﷮z﷮1﷮z2﷯﷯ Again Replace C2 ↔ C3 = x﷮x2﷮1﷮y﷮y2﷮1﷮z﷮z2﷮1﷯﷯ – ( – 1)pxyz x﷮x2﷮1﷮y﷮y2﷮1﷮z﷮z2﷮1﷯﷯ = x﷮x2﷮1﷮y﷮y2﷮1﷮z﷮z2﷮1﷯﷯ + pxyz x﷮x2﷮1﷮y﷮y2﷮1﷮z﷮z2﷮1﷯﷯ Taking Common 𝑥﷮𝑥2﷮1﷮𝑦﷮𝑦2﷮1﷮𝑧﷮𝑧2﷮1﷯﷯ = (1 + pxyz) x﷮x2﷮1﷮y﷮y2﷮1﷮z﷮z2﷮1﷯﷯ Applying R1 → R1 − R2 = (1 + pxyz) x−y﷮x2−y2﷮𝟏−𝟏﷮y﷮y2﷮1﷮z﷮z2﷮1﷯﷯ = (1 + pxyz) x−y﷮ x−y﷯(x−y)﷮𝟎﷮y﷮y2﷮1﷮z﷮z2﷮1﷯﷯ Taking Common (x – y) From R1 = (1 + pxyz) (x – y) 1﷮ x+y﷮0﷮y﷮y2﷮1﷮z﷮z2﷮1﷯﷯ Applying R2 → R2 − R3 = (1 + pxyz) (x – y) 1﷮ x+y﷮0﷮y−z﷮y2−z2﷮𝟏−𝟏﷮z﷮z2﷮1﷯﷯ = (1 + pxyz) (x – y) 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) 1﷮ x+y﷮0﷮1﷮𝑦+𝑧﷮0﷮z﷮z2﷮1﷯﷯ Applying R1 → R1 − R2 = (1 + pxyz) (x – y) (y – z) 𝟏−𝟏﷮ x+y−𝑦−𝑧﷮0−0﷮1﷮𝑦+𝑧﷮0﷮z﷮z2﷮1﷯﷯ = (1 + pxyz) (x – y) (y – z) 𝟎﷮ x−𝑧﷮0﷮1﷮𝑦+𝑧﷮0﷮z﷮z2﷮1﷯﷯ Expanding determinant along C3 = (1 + pxyz) (x – y) (y – z) 0 1﷮𝑦+𝑧﷮𝑧﷮𝑧2﷯﷯−0 0﷮𝑥−𝑧﷮𝑧﷮𝑧2﷯﷯+1 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