Let $A=(a_{ij})\in M_n(\mathbb F)$ such that:
$(a)$ sum of elements in each row equals $0$. Prove $\det A=0$
$(b)$ sum of elements in each row equals $1$. Prove $\det (A-I)=0$
$(c)$ what if the sum of elements in each row equals $5$?
Element $a_{ij}$ in $\text{the j-th column}$ can be written as a linear combination of the rest of the elements in the $i-th\;\text{row}$ depending on the condition in the subtasks.
$(a)$ $$a_{ik}=-\sum_{j=1,\\j\ne k}^{n} a_{ij}\;\forall i\in\{1,\ldots,n\}$$ Therefore we can add all of the columns to the $\text{k-th column}$ to get a zero-column$$a_{ik}=-\sum_{j=1,\\j\ne i}^{n} a_{ij}+\sum_{j=1,\\j\ne i}^{n} a_{ij}=0\;\forall i\in\{1,\ldots,n\}$$ $$\implies r(A)<n\implies \det(A)=0.$$
$(b)$ $$a_{ii}=1-\sum_{j=1,\\j\ne i}^{n} a_{ij}\;\forall i\in\{1,\ldots,n\}$$ Element $\displaystyle(A-I)_{ii}=1-\sum_{j=1,\\j\ne i}^{n} a_{ij}-1=-\sum_{j=1,\\j\ne i}^{n}a_{ij}\;\forall i\in\{1,\ldots,n\}$
Therefore we can add all of the columns to an arbitrary one to get a zero-column again.$$a_{ik}=-\sum_{j=1,\\j\ne i}^{n} a_{ij}+\sum_{j=1,\\j\ne i}^{n} a_{ij}=0\;\forall i\in\{1,\ldots,n\}$$ $$\implies r(A)<n\implies \det(A)=0.$$ $(c)$ My conclusion in general: if the sum of elements in each column equals $n$ $$\det(A-n\cdot I)=0$$ And so it would be for $n=5$: $$\det(A-5\cdot I)=0$$ Is this correct?