Ex 4.2, 14 - Chapter 4 Class 12 Determinants (Term 1)
Last updated at Jan. 22, 2020 by Teachoo
Last updated at Jan. 22, 2020 by Teachoo
Transcript
Ex 4.2, 14 By using properties of determinants, show that: |โ 8(a2+1&ab&ac@ab&b2+1&bc@ca&cb&c2+1)| = 1 + a2 + b2 + c2 Solving L.H.S |โ 8(a2+1&ab&ac@ab&b2+1&bc@ca&cb&c2+1)| Multiplying & Dividing by abc = ๐๐๐/๐๐๐ |โ 8(a2+1&ab&ac@ab&b2+1&bc@ca&cb&c2+1)| Multiplying 1st row by a, 2nd row by b & 3rd row by c ( R1 โ aR1 , R2 โ bR3 , R3 โ bR3 ) = 1/๐๐๐ |โ 8(๐(a2+1)&๐(ab)&๐(ac)@๐(ab)&๐(b2+1)&๐(bc)@๐(ca)&๐(cb)&๐(c2+1))| = 1/๐๐๐ |โ 8(a3+a&๐2b&๐2c@ab2&b3+b&๐2c@c2a&๐2b&c3+c)| Applying R1 โ R1 + R2 + R3 = 1/๐๐๐ |โ 8(a3+a+๐๐2+๐2๐&๐2b+b3+b+c2b&๐2c+b2c+c3+c@ab2&b3+b&๐2c@c2a&๐2b&c3+c)| = 1/๐๐๐ |โ 8(a(๐๐+๐+๐๐+๐๐)&๐(๐๐+๐๐+๐+๐๐)&๐(๐๐+๐๐+๐๐"+1" )@ab2&b3+b&๐2c@c2a&๐2b&c3+c)| Taking (1+๐2+๐2+๐2) common from 1st Row = ((๐ + ๐๐ + ๐๐ + ๐๐))/๐๐๐ |โ 8(a&๐&๐@ab2&b3+b&๐2c@c2a&๐2b&c(c3+1))| Taking a common from C1 ,b from C2 & c from C3 . = ๐๐๐ ( (1 + ๐2 + ๐2 + ๐2))/๐๐๐ |โ 8(1&1&1@b2&b3+1&๐2@c2&๐2&c2+1)| Applying C1 โ C1 โ C2 = (1+๐2+๐2+๐2) |โ 8(๐โ๐&1&1@b2โ๐2โ1&b2+1&๐2@c2โc2&๐2&c2+1)| = (1+๐2+๐2+๐2) |โ 8(๐&1&1@โ1&b2+1&๐2@0&๐2&c2+1)| Applying C2 โ C2 โ C3 = (1+๐2+๐2+๐2) |โ 8(0&๐โ๐&1@โ1&b2+1โ๐2&๐2@0&๐2โ๐2โ1&c2+1)| = (1+๐2+๐2+๐2) |โ 8(0&๐&1@โ1&1&๐2@0&โ1&c2+1)| Expanding along R1 = (1+๐2+๐2+๐2)(1|โ 8(1&๐2@โ1&๐2+1)|" โ 0" |โ 8(โ1&๐ถ2@0&๐2+1)|" + 0" |โ 8(1&1@0&โ1)|) = (1 + a2 + b1 + c2) (0 โ 0 + 1) = (1 + a2 + b1 + c2) (1) = (1 + a2 + b1 + c2) = R.H.S Hence proved
Using Property 5 (Determinant as sum of two or more determinants)
Using Property 5 (Determinant as sum of two or more determinants)
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