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How can I prove the following?

If $A$ is an $n \times n$ matrix such that
$$ \sum\limits_{j=1}^n a_{ij} = 0 $$ for $1 \leq i \leq n$ then $\det A = 0$.

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    $\begingroup$ By interpreting the sum as the multiplication of the matrix by a very particular non-zero vector. What have you tried, and where are you having trouble? What facts do you know about determinants? Where did you encounter this problme? $\endgroup$
    – user296602
    Mar 17, 2016 at 22:10
  • $\begingroup$ The expressions you require to be zero are often called row sums of $A$ for short. $\endgroup$
    – hardmath
    Mar 19, 2016 at 0:26

2 Answers 2

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If $\sum_{j=1}^n a_{ij}=0$ for all $1\leq i\leq n$, then in particular $$A\cdot \begin{bmatrix}1\\ 1\\ \vdots \\ 1\end{bmatrix}=\begin{bmatrix}a_{11}\cdot 1 + \cdots + a_{1n}\cdot 1\\ a_{21}\cdot 1 + \cdots + a_{2n}\cdot 1\\ \vdots \\ a_{n1}\cdot 1 + \cdots + a_{nn}\cdot 1 \end{bmatrix}=\begin{bmatrix}0\\0\\\vdots \\ 0\end{bmatrix}$$

So, since the system $A\overrightarrow{x}=\overrightarrow{0}$ does not have a unique solution $\overrightarrow{x}=\overrightarrow{0}$, we can conclude that $A$ is not invertible and thus $\det(A)=0$.

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Hint: Compare that with a linear combination of the column vectors $a_i$. What does it tell you about the set of $a_i$ and their parallelepiped?

Answer:

The expression is a linear combination of the column vectors of $A$ with non-zero coefficients. Thus the column vectors are linear dependent and the parallelepiped they span has zero volume, thus $\DeclareMathOperator{det}{det}0 = V=\det(A)$.

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