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Miscellaneous Exercise 2(A) · Q72

Q.Show with usual notations that for any matrix A=[aij]3×3A = [a_{ij}]_{3\times3}: a11A11+a12A12+a13A13=∣A∣a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13} = |A|

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Step 1: For A=[aij]3×3A=[a_{ij}]_{3\times3}, the determinant ∣A∣|A| can be evaluated by expanding along any row or column; expanding along row 1 means writing ∣A∣|A| as a sum of each row-1 entry times its own cofactor.

Step 2: By definition, A11=(−1)1+1M11=M11A_{11}=(-1)^{1+1}M_{11}=M_{11} is the cofactor of a11a_{11} (the determinant left after deleting row 1 and column 1); similarly A12=−M12A_{12}=-M_{12} and A13=M13A_{13}=M_{13} are the cofactors of a12a_{12} and a13a_{13}.

Step 3: The row-1 cofactor expansion of a 3×33\times3 determinant is, by definition, ∣A∣=a11M11−a12M12+a13M13=a11A11+a12A12+a13A13|A|=a_{11}M_{11}-a_{12}M_{12}+a_{13}M_{13}=a_{11}A_{11}+a_{12}A_{12}+a_{13}A_{13} (the alternating signs are already built into the cofactors A1jA_{1j}, so writing it with cofactors removes the need to alternate signs by hand). …

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