Let $A$ be a $n\times n$ matrix over the reals such that the diagonal entries are all positive, the off-diagonal entries are all negative, and the row sums are all positive. Show that $\det A \neq 0$.
To show that $\det A\neq 0$, it would be sufficient to show that the system $AX = 0$ does not have a nontrivial solution. Then I suppose that to show that, one could assume that it has a nontrivial solution and then obtain a contradiction. But I'm stuck on actually doing that part.
Any suggestions? I'm also interested in where I can find questions similar to this one to practice.
Answer
Suppose there exists a vector $x \in \mathbb{R}^n \setminus \{\vec{0}\}$ such that $Ax = \vec{0}$.
Pick an index $k$ such that $|x_k| = \displaystyle\max_{j = 1,\ldots,n}|x_j|$. So we have $|x_j| \le |x_k|$ for all $j = 1,\ldots,n$.
If $x_k = 0$, then we would have $x_j = 0$ for all $j = 1,\ldots,n$, i.e. $x = \vec{0}$, which isn't possible.
If $x_k > 0$, then $x_j \le |x_j| \le |x_k| = x_k$ for all $j$. Then, since $A_{k,k} > 0$, $A_{k,j} < 0$ for all $j \neq k$, and $\displaystyle\sum_{j = 1}^{n}A_{k,j} > 0$, we have $$0 = (Ax)_k = \sum_{j = 1}^{n}A_{k,j}x_j = A_{k,k}x_k + \sum_{j \neq k}A_{k,j}x_j \ge A_{k,k}x_k + \sum_{j \neq k}A_{k,j}x_k = x_k\sum_{j = 1}^{n}A_{k,j} > 0,$$ a contradiction.
Similarly, if $x_k < 0$, then $x_j \ge -|x_j| \ge -|x_k| = x_k$ for all $j$. Then, since $A_{k,k} > 0$, $A_{k,j} < 0$ for all $j \neq k$, and $\displaystyle\sum_{j = 1}^{n}A_{k,j} > 0$, we have $$0 = (Ax)_k = \sum_{j = 1}^{n}A_{k,j}x_j = A_{k,k}x_k + \sum_{j \neq k}A_{k,j}x_j \le A_{k,k}x_k + \sum_{j \neq k}A_{k,j}x_k = x_k\sum_{j = 1}^{n}A_{k,j} < 0,$$ a contradiction.
Therefore, there is no nonzero vector $x$ such that $Ax = \vec{0}$. Hence, $A$ is invertible, and thus, $\det A \neq 0$.
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