Ex 4.2, 14 - Chapter 4 Class 12 Determinants (Term 1)
Last updated at Jan. 22, 2020 by
Last updated at Jan. 22, 2020 by
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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