Prove A is invertible if its diagonals are non-zero The following answer was from this question:
Prove that an upper triangular matrix is invertible if and only if every diagonal entry is non-zero.
Can someone explain the purpose of the linear combination that is in the solution and how it proves a non-trivial solution? What are they using the k-1 row?
Proof. ($\Rightarrow$)
Let $A=[a_{i,j}]_{n\times n}$ be an invertible upper triangular matrix. Suppose that $A$ has a diagonal entry that is zero, i.e., $a_{k,k}=0$, where $k\in\mathbb{N}$, $1\leq k<n$ (note that if $k=n$, then A would have a zero row, thus making A singular, which should not be the case). Then, the homogenous equation $Ax=0$, i.e.,
$$\left[
\begin{array}{ccc}
a_{1,1} &&\cdots &&a_{1,n}\\
0 &\ddots\\
\vdots & &a_{k,k} &&\vdots\\
&&&\ddots\\
0&&\cdots&0&a_{n,n}
\end{array}
\right]\cdot
\left[
\begin{array}{c}
x_1\\
\vdots\\
x_k\\
\vdots\\
x_n
\end{array}\right]=0,$$
where the $k^{th}$ row is the last row whose entry on the main diagonal is zero, will have a nontrivial solution since 
$$\underbrace{\overbrace{a_{k-1,k-1}x_{k-1}}^{\text{if $k\neq1$}}+a_{k-1,k}x_k}_{(i)}+\underbrace{a_{k-1,k+1}x_{k+1}+\cdots+a_{k-1,n}x_n}_{equals\text{ }to\text{ }0}=0,$$
wherein (i) implies that we have, indeed, a nontrivial solution. Thus, a contradiction exists since if $A$ is invertible, then the homogenous equation $Ax=0$ must only have the trivial solution.
Therefore, all of the diagonal entries of $A$ must be nonzero.  
 A: If you do not want to appeal to determinants, or ranks of linear transformations, you can brute force it as follows: Suppose $U$ is invertible. Then, there is a matrix $A=[a_1,\cdots, a_n]$ of column vectors such that $AU=I.$ Writing this as a system of $vector$ equations, you get 
$a_1u_{11}=e_1$
$a_1u_{12} + a_2u_{22} = e_2$
$\cdots$
$\cdots$
$a_1u_{1n} + a_2u{2n} + \cdots + a_nu_{nn} = e_n$
Start from the top and work down, solving for the $a_j$. You get $a_j = (e_j −\sum^{j−1}_{k=1} a_ku_{kj} )/u_{jj};\  j = 1, \cdots, n.$ So, if $U$ is invertible, the inverse, which is unique, $must$ be given by the above formula. On the other hand, if any of the diagonal elements is zero, then this formula cannot hold and so $U$ does not have an inverse.
A: It seems like a mistake, besides their answer wasn't accepted, they were talking about the k-th row as the row in which the value of $$a_{k,k}=0$$
yet they were pointing out to the k-1 th row.
I would recommend looking at the characteristic polynomial of A, $$P_{A}(\lambda)=det(\lambda I-A)$$
Since the matrix is triangular then computing the determinant is as easy as multiplying the diagonal values:
$$det(\lambda I-A)=(\lambda - a_{11})*(\lambda - a_{22})*...*(\lambda - a_{nn})$$
Suppose that A has a diagonal entry that is zero, without the loss of generality, let's say $$a_{11}=0$$
then,
$$det(\lambda I-A)=\lambda*(\lambda - a_{22})*...*(\lambda - a_{nn})$$
thus we can conclude that 0 is an eigenvalue of A, or in other words, A is singular.

Alternative way to prove it without computing the characteristic polynomial of A would be to perform elementary operations on A, since we suppose that A has a diagonal entry that is zero, then by performing Gaussian elimination we'll eventually find that $$rank(A)=n-1$$ therefore $$dimP(A)=n-rank(A)=1$$
that means that there exists $$v\in F^n$$ such that $$Av=0$$
Or another way to finish the proof would be:
A is invertible if and only if $$rank(A)=n$$
since we found that $$rank(A)=n-1$$ 
A is not invertible in contradiction to our base assumption, which A is invertible.

In order to do it without ranks or determinants, you can take a look at the other answer or try to write the equations and deduct what you need about $$x_{k}$$ assuming that A has a diagonal entry that is zero at the kth entry.
