How can I prove that A is non-invertible nxn matrix if sum of all elements in rows of A equal to zero?
$$a_{11} + a_{12} + \cdots + a_{1n} = 0$$
$$a_{21} + a_{22} + \cdots + a_{2n} = 0$$
$$\vdots$$
$$a_{n1} + a_{n2} + \cdots + a_{nn} = 0$$
[Math] non-invertible matrix if sum of all elements in rows equals to zero
matrices
Best Answer
Let $A=\left(C_{1},\ldots,C_{n}\right)$ where $C_{j}$ is the $j$ -th column of $A$ . Then your condition is $C_{1}+\ldots+C_{n}=0$ that is $C_{n}=-\left(C_{1}+\ldots+C_{n-1}\right)$ i.e. the columns of $A$ are linearly dependant. Thus the determinant is zero.