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Why adding a row with another multiplied row in square matrix $A$ doesn't change the $\det(A)$ value?

Relure
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user221685
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  • Do you know that the determinant is a multilinear function of the rows of a matrix? And that interchanging two rows changes the sign of the determinant? – littleO Mar 07 '15 at 10:12
  • http://math.stackexchange.com/questions/598219/effect-of-elementary-row-operations-on-determinant, http://math.stackexchange.com/questions/211013/why-is-the-determinant-invariant-under-row-and-column-operations/ – Hans Lundmark Mar 07 '15 at 10:27

2 Answers2

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Because of:

  1. the linearity of the determinant according to the lines
  2. the anti-symmetry of the determinant.
mookid
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To develop the answer of @mookid note that: multilinearity means that that the detrminat is a linear function of his (vector) rows $\mathbf{a}_h$, so you have: $$ \det (\mathbf{a}_1,\cdots,\mathbf{a}_i+k\mathbf{a}_{j},\cdots, \mathbf{a}_{j},\cdots \mathbf{a}_n)^T=\det(\mathbf{a}_1,\cdots,\mathbf{a}_i,\cdots \mathbf{a}_{j},\cdots \mathbf{a}_n)^T +k\det(\mathbf{a}_1,\mathbf{a}_{j},\cdots \mathbf{a}_{j},\cdots \mathbf{a}_n)^T $$ Now, the determinat is also antisymmetric and this means that if two rows are the same the determinat is null, so: $\det(\mathbf{a}_1,\mathbf{a}_{j},\cdots \mathbf{a}_{j},\cdots \mathbf{a}_n)^T=0$.

Emilio Novati
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