Why Does Sum of Matrix Elements x Cofactor in Different Rows Equal Zero?

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The discussion explains that the sum of the elements of a matrix in one row multiplied by the cofactors from a different row equals zero due to properties of determinants. When a row in a square matrix is replaced with another, the determinant remains unchanged, leading to the conclusion that the expansion using elements from one row and cofactors from another results in a determinant that cancels out. This is because the cofactors are defined in relation to their respective rows, and mixing them leads to a zero result. The discussion emphasizes the consistency of determinant properties across row operations. Ultimately, the mathematical reasoning confirms that the determinant's value is invariant under such transformations.
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In a matrix the sum of element of a matrix in a row times it's co factor of that elemt gives the determinant value, but why does the sum of element of a matrix times cofactor of different row is always zero?
 
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In a square matrix A, replace row j by row i, obtaining a new matrix B with row i and row j being equal. Expanding using elements from row i in A and cofactors corresponding to row j in A will be the same as expanding det B along row j (using cofactors corresponding to row j in B also) since the elements and cofactors are the same in both cases. So the result is det B, which is...?
 
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Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

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