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Question

Evaluate the following determinant:

(i) 1+a1111+a1111+a=a3+3a2

(ii) a2+2a2a+112a+1a+21331=a-13

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Solution

(ii) To Prove: a2+2a2a+112a+1a+21331=a-13


LHS=a2+2a2a+112a+1a+21331Applying R1R1-R2 =a2+2a-2a-12a+1-a-21-12a+1a+21331 =a2-1a-102a+1a+21331Taking a-1 common from R1 =a-1a+1102a+1a+21331Applying R2R2-R3 =a-1a+1102a+1-3a+2-31-1331 =a-1a+1102a-2a-10331Taking a-1 common from R2 =a-12a+110210331Expanding through C3 =a-1211a+1-2 =a-121a+1-2 =a-12a-1 =a-13=RHS


Hence, a2+2a2a+112a+1a+21331=a-13.

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