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$\begingroup$

$A=(a_{ij})_{nxn}$

$ a_{ij},u_{i} \ \in F $

Prove:

$\det\begin{pmatrix}u_{1}a_{11}& u_{2}a_{12}&...& u_{n}a_{1n}\\ a_{21}& a_{22}& ...& a_{2n}\\ ...&...&...&...\\ a_{n1}& a_{n2}& ...& a_{nn}\end{pmatrix} +\det\begin{pmatrix}a_{11}& a_{12}&...& a_{1n}\\ u_{1}a_{21}& u_{2}a_{22}& ...& u_{n}a_{2n}\\ ...&...&...&...\\ a_{n1}& a_{n2}& ...& a_{nn}\end{pmatrix}+...+\det\begin{pmatrix}a_{11}& a_{12}&...& a_{1n}\\ a_{21}& a_{22}& ...& a_{2n}\\ ...&...&...&...\\ u_{1}a_{n1}& u_{2}a_{n2}& ...& u_{n}a_{nn}\end{pmatrix} =(u_{1}+...+u_{n})\det(A) $

I tried to evaluate each determinant by the row with the scalars, but coudln't get anywhere with that.

Any ideas?

2 Answers 2

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Your idea is good. Use cofactor expansion of the determinant along the row containing $u_j$ and exchange the summation indices:

$$\sum_{i=1}^n\sum_{j=1}^n u_ja_{ij}C_{ij}=\sum_{j=1}^n\sum_{i=1}^nu_ja_{ij}C_{ij}=\sum_{j=1}^nu_j\sum_{i=1}^na_{ij}C_{ij}=\sum_{j=1}^nu_j\det(A)=\det(A)\sum_{j=1}^nu_j$$

where $C_{ij}$ is the cofactor.

  • 0
    Should the index of the second sum should be j ? – 2017-01-26
  • 0
    yes, corrected. it came from cut and paste :) – 2017-01-26
  • 0
    Got it, thanks! – 2017-01-26
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We have by definition: $\det\begin{pmatrix}u_{1}a_{11}& u_{2}a_{12}&...& u_{n}a_{1n}\\ a_{21}& a_{22}& ...& a_{2n}\\ ...&...&...&...\\ a_{n1}& a_{n2}& ...& a_{nn}\end{pmatrix}=\sum_{\sigma\in S_n}sign(\sigma)u_{\sigma(1)}a_{1\sigma(1)}a_{2\sigma(2)}...a_{n\sigma(n)}$

So for: $1\leq i \leq n$ ,we have : $\det\begin{pmatrix}a_{11}& a_{12}&...& a_{1n}\\ ...&...&...&...\\ u_1a_{i1}& u_2a_{i2}& ...& u_na_{in}\\ ...&...&...&...\\ a_{n1}& a_{n2}& ...& a_{nn}\end{pmatrix}=\sum_{\sigma\in S_n}sign(\sigma)u_{\sigma(i)}a_{1\sigma(1)}a_{2\sigma(2)}...a_{i\sigma(i)}...a_{n\sigma(n)}$

Sum all for $1\leq i\leq n$ , we have that : $$\sum_{i=1}^n\sum_{\sigma\in S_n}sign(\sigma)u_{\sigma(i)}a_{1\sigma(1)}a_{2\sigma(2)}...a_{n\sigma(n)}$$ $$=\sum_{\sigma\in S_n}sign(\sigma)(\sum_{i=1}^nu_{\sigma(i)})a_{1\sigma(1)}a_{2\sigma(2)}...a_{n\sigma(n)}$$ $$=\sum_{\sigma\in S_n}sign(\sigma)(\sum_{i=1}^nu_i)a_{1\sigma(1)}a_{2\sigma(2)}...a_{n\sigma(n)}$$ $$=(\sum_{i=1}^nu_i)det(A)$$

Last equal because $\sigma $ is bijective. And we have concluded.