I'm working through lecture notes for revision and I've come across this theorem and its proof.
Theorem. An $n\times n$ matrix A is invertible if and only if $\det A \ne 0$.
In the proof of $\det A \ne 0$ implies $A$ invertible, we define $D$ by the formula for the inverse, then aim to prove $AD = DA = I_n$. We show for $k = 1, \ldots, n$, $(AD)_{kk} = 1$.
Then, we need to show if $k \ne l$ then $(AD)_{kl} = 0$. We have the line $$ (AD)_{kl} = \frac 1 {\det A}\sum_{i=1}^n A_{ki} (-1)^{i+l}\det \tilde A_{li}. $$ Then, my lecturer claims that the above summation is equal to the determinant of the matrix, $$\begin{pmatrix}A_{11} &\cdots & A_{1n}\\ \vdots &\ddots&\vdots \\ A_{k1} &\cdots & A_{kn}\\ \vdots &\ddots&\vdots \\ A_{k1} &\cdots & A_{kn}\\ \vdots &\ddots&\vdots \\ A_{n1} &\cdots & A_{nn}\\ \end{pmatrix}$$
This is the step I don't fully understand. My lecturer said, and it is apparent, that the summation is almost the definition of a determinant. However, why is it the determinant of this matrix above?
I'm not sure of it being a definition, but I guess you have seen the cofactor expansion of a determinant: By expanding $\det M$ along row $l$ we have $$\det M = \sum_{i=1}^n M_{li} (-1)^{i+l} \det(\tilde M_{li}).$$
Now let $M$ be the matrix obtained from $A$ by replacing row $l$ with row $k$ (which I guess is what is meant by the matrix you show). Then on one hand, $\det M = 0$ because it has two identical rows. On the other hand, $M_{li} = A_{ki}$, and $\tilde M_{li} = \tilde A_{li}$ because $M$ and $A$ become the same after deleting row $l$, so the above sum is the same as the sum in the expression for $(AD)_{kl}$.