Determinant as Sum of Determinants

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Theorem

Let $\begin{vmatrix} a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{r1} & \cdots & a_{rs} & \cdots & a_{rn} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{ns} & \cdots & a_{nn} \end{vmatrix}$ be a determinant.

Then $\begin{vmatrix} a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{r1} + a'_{r1} & \cdots & a_{rs} + a'_{rs} & \cdots & a_{rn} + a'_{rn} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{ns} & \cdots & a_{nn} \end{vmatrix} = \begin{vmatrix} a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{r1} & \cdots & a_{rs} & \cdots & a_{rn} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{ns} & \cdots & a_{nn} \end{vmatrix} + \begin{vmatrix} a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a'_{r1} & \cdots & a'_{rs} & \cdots & a'_{rn} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{ns} & \cdots & a_{nn} \end{vmatrix}$.


Similarly:

Then $\begin{vmatrix} a_{11} & \cdots & a_{1s} + a'_{1s} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{r1} & \cdots & a_{rs} + a'_{rs} & \cdots & a_{rn} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{ns} + a'_{ns} & \cdots & a_{nn} \end{vmatrix} = \begin{vmatrix} a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{r1} & \cdots & a_{rs} & \cdots & a_{rn} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{ns} & \cdots & a_{nn} \end{vmatrix} + \begin{vmatrix} a_{11} & \cdots & a'_{1s} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{r1} & \cdots & a'_{rs} & \cdots & a_{rn} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a'_{ns} & \cdots & a_{nn} \end{vmatrix}$.


Proof

Let:

$B = \begin{vmatrix} a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{r1} + a'_{r1} & \cdots & a_{rs} + a'_{rs} & \cdots & a_{rn} + a'_{rn} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{ns} & \cdots & a_{nn} \end{vmatrix} = \begin{vmatrix} b_{11} & \cdots & b_{1s} & \cdots & b_{1n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ b_{r1} & \cdots & b_{rs} & \cdots & b_{rn} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ b_{n1} & \cdots & b_{ns} & \cdots & b_{nn} \end{vmatrix}$
$A_1 = \begin{vmatrix} a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{r1} & \cdots & a_{rs} & \cdots & a_{rn} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{ns} & \cdots & a_{nn} \end{vmatrix}$
$A_2 = \begin{vmatrix} a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a'_{r1} & \cdots & a'_{rs} & \cdots & a'_{rn} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{ns} & \cdots & a_{nn} \end{vmatrix}$


Then:

\(\ds B\) \(=\) \(\ds \sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n b_{k \map \lambda k} }\)
\(\ds \) \(=\) \(\ds \sum_\lambda \map \sgn \lambda a_{1 \map \lambda 1} \cdots \paren {a_{r \map \lambda r} + a'_{r \map \lambda r} } \cdots a_{n \map \lambda n}\)
\(\ds \) \(=\) \(\ds \sum_\lambda \map \sgn \lambda a_{1 \map \lambda 1} \cdots a_{r \map \lambda r} \cdots a_{n \map \lambda n} + \sum_\lambda \map \sgn \lambda a_{1 \map \lambda 1} \cdots a'_{r \map \lambda r} \cdots a_{n \map \lambda n}\)
\(\ds \) \(=\) \(\ds A_1 + A_2\)

$\blacksquare$


The result for columns follows directly from Determinant of Transpose.

$\blacksquare$