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My professor gave us this definition for determinants for a $n \times n$ matrix $A$:

$$\det(A) = a_{11}C_{11} + a_{12}C_{12} ... + a_{1n}C_{1n} $$

where $C_{1j}$ is the cofactor of $A$ on $a_{ij}$. He also said that this can be generalized to:

$$\det(A) = a_{q1}C_{q1} + a_{q2}C_{q2} ... + a_{qn}C_{qn} $$

for any row or column $q$. He didn't give us a proof of the above statement, because he said it was too complicated.

Out of curiosity, could someone give me a proof of the above statement? I would appreciate a proof instead of a tip, because of me being completely new to this and the proof being way too advanced for my skills.

What I know so far: Row Reduction, Basis (not change of basis however), Linear Independence, Span, Transformations, Inverses, Subspaces and Dimensions. Please try to keep proofs within my what I know.

Thank you.

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  • $\begingroup$ Any function $D$ (on rows or columns of a $n \times n$ matrix) with three properties: $n$-linear (when other arguments are fixed, $D$ is a linear function on an argument), alternating (changing any two arguments negates the output value, and $D(A) = 0$ whether two rows/columns of $A$ are same), and $D(I) = 1$, is unique and is $\det$. The proof is not very complex, but is quite long... $\endgroup$ – JiminP Nov 9 '13 at 4:01
  • $\begingroup$ It suffices to use the Leibniz formula as an intermediary methinks. $\endgroup$ – anon Nov 9 '13 at 4:03
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    $\begingroup$ It is easier to begin with the third definition $\det(A)=\sum_\sigma\operatorname{sgn}(\sigma)\prod_{i=1}^na_{i,\sigma(i)}$ and prove that both definitions in your question are equivalent to the third one. $\endgroup$ – user1551 Nov 9 '13 at 4:28
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This article answers your question exactly. It is 9 pages long and, even then, the actual proof of the fact that the determinant can be calculated along any row or column is sketched.

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