Let $A$ be an $n \times n$ invertible matrix such that $A=L U$, where $L$ is a unit lower triangular matrix and $U$ is an upper triangular matrix. For $1 \leq k \leq n$, let $A_k$ denote the principal leading submatrix of $A$ formed by the first $k$ rows and the first $k$ columns of $A$. Show that $\det\left(A_{k}\right) \neq 0$, $k=1,2, \ldots, n$.
I just verified it by taking a matrix $A$ with real constant entities inside it and then finding $A_k$ using $A$ and I proved that determinant is not equal to $0$
But how can we do that in a more generalized way ? Just a little bit of steps showing may also work
Let $A$ be an invertible matrix such that it has an $LU$ factorization : $\,\,A=LU.$
Let $k \in \{1,2,\ldots,n-1\}$, and
$$A=\begin{pmatrix} A_k & * \\ * & * \end{pmatrix} = \begin{pmatrix} L_k & 0 \\ * & * \end{pmatrix} \cdot \begin{pmatrix} U_k & * \\ 0 & * \end{pmatrix}, $$
where $(*)$ indicates a block matrix that we don't need to keep track of.
By doing multiplication block by block, we obtain $A_{k}=L_{k}\cdot U_{k}, \; \; $ and thus : $$\det(A_k)=\det(L_k)\det(U_k)=\det(U_k).$$
Since $\det(U)\neq 0$, then $\det(U_k)\neq 0$ (No coefficient in the diagonal of $U_k$ is zero).
Finally : $\boxed{\forall k \in \{1,\ldots,n\} : \det(A_k)\neq0.}$