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Question

Using the properties of determinants, show that:

(i)∣ ∣abc2a2a2bbca2b2c2ccab∣ ∣=(a+b+c)2
(ii)∣ ∣x+y+2zxyzy+z+2xyzxz+x+2y∣ ∣=2(x+y+z)3

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Solution

(i) ∣ ∣abc2a2a2bbca2b2c2ccab∣ ∣

C1C1C2,C2C2C3

=∣ ∣ ∣(a+b+c)02a(a+b+c)(a+b+c)2b0(a+b+c)cab∣ ∣ ∣

Taking a+b+c common from C1,C2
=(a+b+c)2∣ ∣102a112b01cab∣ ∣

R2R2+R1
=(a+b+c)2∣ ∣102a012a+2b01cab∣ ∣

Expanding along first column,
=(a+b+c)2[1(c+a+b2a2b)]
=(a+b+c)3


(ii) ∣ ∣x+y+2zxyzy+z+2xyzxz+x+2y∣ ∣

R1R1R2,R2R2R3

=∣ ∣ ∣x+y+z(x+y+z)00x+y+z(x+y+z)zxz+x+2y∣ ∣ ∣

Taking (x+y+z) common from R1,R2
=(x+y+z)2∣ ∣110011zxz+x+2y∣ ∣

C2C2+C3
=(x+y+z)2∣ ∣100011zx+zz+x+2y∣ ∣

Expanding along first row,
=(x+y+z)211x+zx+z+2y
=(x+y+z)2(x+z+2y+x+z)
=2(x+y+z)3


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