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Using Property 5 (Determinant as sum of two or more determinants)
Using Property 5 (Determinant as sum of two or more determinants)
Last updated at May 29, 2023 by Teachoo
Question 14 By using properties of determinants, show that: |β 8(a2+1&ab&[email protected]&b2+1&[email protected]&cb&c2+1)| = 1 + a2 + b2 + c2 Solving L.H.S |β 8(a2+1&ab&[email protected]&b2+1&[email protected]&cb&c2+1)| Multiplying & Dividing by abc = πππ/πππ |β 8(a2+1&ab&[email protected]&b2+1&[email protected]&cb&c2+1)| Multiplying 1st row by a, 2nd row by b & 3rd row by c ( R1 β aR1 , R2 β bR3 , R3 β bR3 ) = 1/πππ |β 8(π(a2+1)&π(ab)&π(ac)@π(ab)&π(b2+1)&π(bc)@π(ca)&π(cb)&π(c2+1))| = 1/πππ |β 8(a3+a&π2b&π[email protected]&b3+b&π[email protected]&π2b&c3+c)| Applying R1 β R1 + R2 + R3 = 1/πππ |β 8(a3+a+ππ2+π2π&π2b+b3+b+c2b&π[email protected]&b3+b&π[email protected]&π2b&c3+c)| = 1/πππ |β 8(a(ππ+π+ππ+ππ)&π(ππ+ππ+π+ππ)&π(ππ+ππ+ππ"+1" )@ab2&b3+b&π[email protected]&π2b&c3+c)| Taking (1+π2+π2+π2) common from 1st Row = ((π + ππ + ππ + ππ))/πππ |β 8(a&π&π@ab2&b3+b&π[email protected]&π2b&c(c3+1))| Taking a common from C1 ,b from C2 & c from C3 . = πππ ( (1 + π2 + π2 + π2))/πππ |β 8(1&1&[email protected]&b3+1&π[email protected]&π2&c2+1)| Applying C1 β C1 β C2 = (1+π2+π2+π2) |β 8(πβπ&1&[email protected]βπ2β1&b2+1&π[email protected]βc2&π2&c2+1)| = (1+π2+π2+π2) |β 8(π&1&1@β1&b2+1&π[email protected]&π2&c2+1)| Applying C2 β C2 β C3 = (1+π2+π2+π2) |β 8(0&πβπ&1@β1&b2+1βπ2&π[email protected]&π2βπ2β1&c2+1)| = (1+π2+π2+π2) |β 8(0&π&1@β1&1&π[email protected]&β1&c2+1)| Expanding along R1 = (1+π2+π2+π2)(1|β 8(1&π2@β1&π2+1)|" β 0" |β 8(β1&πΆ[email protected]&π2+1)|" + 0" |β 8(1&[email protected]&β1)|) = (1 + a2 + b1 + c2) (0 β 0 + 1) = (1 + a2 + b1 + c2) (1) = (1 + a2 + b1 + c2) = R.H.S Hence proved