Show that $B$ $\in$ $GL_{n-1}($K$)$ $\Rightarrow$ $A$ $\in$ $GL_n$($K$) Let $K$ be a field and A = $ 
 \begin{bmatrix}
-1 & A_{12} \\
 0 & B \\
 \end{bmatrix}
$  $\in$ $K^{n,n}$ and $B$ $\in$  $K^{n-1,n-1}$. Show that $A$ $\in$ $GL_n$($K$) applies if $B$ $\in$ $GL_{n-1}($K$)$. 
My thoughts and questions: $B$ is obviously  a (n - 1) $\times$ (n - 1) matrix in the field $\mathbb{K}$. For example: For n = 4 is $B$ a $\underbrace{3}_{4-1}$ $\times$ $\underbrace{3}_{4-1}$ matrix in the field $\mathbb{K}$. 
But generally I don't understand how $A$ $\in$ $GL_n$($K$) applies if $B$ $\in$ $GL_{n-1}($K$)$. 
I thought about using this lemma: Let $A$ $\in$ $K^{n,m}$ with $A$ = $
\begin{bmatrix}
A_{1} \\
A_{2} \\
\end{bmatrix}$ which means that $A_{1}$ $\in$ $K^{l,m}$, $A_{2}$ $\in$ $K^{n-l,m}$. 
Therefore we have that Rank(A) $\le$ Rank($A_{1}$) + Rank($A_{2}$). But it is made for  $A$ $\in$ $K^{n,m}$ and not for $A$ $\in$ $K^{n,n}$.
How do I beginn this proof? and how does  $A$ $\in$ $GL_n$($K$) applies if $B$ $\in$ $GL_{n-1}($K$)$? (I really don't why this correct). Any hints guiding me to the right direction I much appreciate.
 A: HINT:
$$
\det A=-\det B\neq0
$$
A: Hint:
The matrix $A$ is block-triangular (with square blocks). Hence $\det A$ is the product of the determinants of the blocks.
A: We don't really need determinants to solve this one; we can work directly from the definition of $GL_m(K)$.
We need to show that 
$B \in GL_{n - 1}(K) \Longrightarrow A = \begin{bmatrix} -1 & A_{12} \\ 0 & B \end{bmatrix} \in GL_n(K); \tag 1$
that is, $A$ is invertible if $B$ is.  We recall that invertible matrices are characterized by vanishing kernel; therefore we need to show
$\ker B = \{0\} \Longrightarrow \ker A = \{0\}; \tag 2$
so let
$\vec a \in K^n; \tag 3$
we may write
$\vec a = \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix}; \tag 4$
if
$\vec a \in \ker A, \tag 5$
then we have
$A \vec a = 0, \tag 6$
or
$\begin{bmatrix} -1 & A_{12} \\ 0 & B \end{bmatrix} \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} = 0; \tag 7$
we may also set
$\vec b = \begin{pmatrix} a_2 \\ a_3 \\ \vdots \\ a_n \end{pmatrix} \in K^{n - 1}; \tag 8$
then (7) may be written
$\begin{bmatrix} -1 & A_{12} \\ 0 & B \end{bmatrix} \begin{pmatrix} a_1 \\ \vec b \end{pmatrix} = 0; \tag 9$
thus, performing the matrix-vector multiplication indicated in this equation,
$\begin{pmatrix} -a_1 + A_{12} \vec b \\ B\vec b \end{pmatrix} = 0, \tag{10}$
whence
$-a_1 + A_{12} \vec b = 0, \tag{11}$
and
$B \vec b = 0; \tag{12}$
since we assume $B \in GL_{n - 1}(K)$ we see that (12) implies
$\vec b = 0, \tag{13}$
from which (11) yields
$-a_1 = -a_1 + A_{12} \vec b = 0; \tag{14}$
we thus see from (4), (8), (11), (12), (13) and (14) that
$\vec a = 0 \Longrightarrow \ker A = 0 \Longrightarrow A \in GL_n(K). \tag{15}$
It is worth noting that the above demonstration may be "walked backwards" to show that
$A \in GL_n(K) \Longrightarrow B \in GL_{n - 1}(K) \tag{16}$
as well; we need show that
$\ker A = \{0\} \Longrightarrow \ker B = \{0\}; \tag{17}$
now if
$\vec b \in \ker B, \tag{18}$
that is, 
$B \vec b = 0, \tag{19}$
then with
$a_1 = A_{12} \vec b \tag{20}$
we have 
$\vec a = \begin{pmatrix} a_1 \\ \vec b \end{pmatrix} \tag{21}$
satisfying
$A \vec a = 0; \tag{22}$
therefore
$\vec a \in \ker A \Longrightarrow \vec a = 0 \Longrightarrow \vec b = 0 \Longrightarrow \ker B = \{0\}, \tag{23}$
whence
$B \in GL_{n - 1}(K). \tag{24}$
A: You can find an inverse of $A$, which directly proves that $A$ is invertible:
The matrix $A$ is in upper triangular block form, so the inverse ought to be of the form
$$
  \begin{bmatrix}
    -1 & ?      \\
     0 & B^{-1} 
  \end{bmatrix}.
$$
By considering the product
$$
    \begin{bmatrix}
      -1  & A_{12}  \\
       0  & B
    \end{bmatrix}
    \begin{bmatrix}
      -1  & ?      \\
       0  & B^{-1}
    \end{bmatrix}
  = \begin{bmatrix}
      1 & -? + A_{12} B^{-1} \\
      0 & I_{n-1}
    \end{bmatrix},
$$
so we see that the unknown entry $?$ should be $A_{12} B^{-1}$.
We have guessed the right matrix because
$$
    \begin{bmatrix}
      -1  & A_{12}  \\
       0  & B
    \end{bmatrix}
    \begin{bmatrix}
      -1  & A_{12} B^{-1} \\
       0  & B^{-1}
    \end{bmatrix}
  = \begin{bmatrix}
      1 & 0 \\
      0 & I_{n-1}
    \end{bmatrix}
  = I_n
$$
and
$$
    \begin{bmatrix}
      -1  & A_{12} B^{-1} \\
       0  & B^{-1}
    \end{bmatrix}
    \begin{bmatrix}
      -1  & A_{12}  \\
       0  & B
    \end{bmatrix}
  = \begin{bmatrix}
      1 & 0 \\
      0 & I_{n-1}
    \end{bmatrix}
  = I_n.
$$
This shows that $A$ is invertible with
$$
    \begin{bmatrix}
      -1  & A_{12}  \\
       0  & B
    \end{bmatrix}^{-1}
  = \begin{bmatrix}
      -1  & A_{12} B^{-1} \\
       0  & B^{-1}
    \end{bmatrix}.
$$
A: It is enough to prove that $A$ is surjective as a linear map.
Let $y \in K^n$. We need to find $x \in K^n$ such that $Ax^T=y^T$.
Since $B$ is surjective, there is $(x_2,\dots,x_n) \in K^{n-1}$ such that $B(x_2,\dots,x_n)^T=(y_2,\dots,y_n)^T$. Let $x_1 = A_{12} \cdot (x_2,\dots,x_n)-y_1$. Then $x=(x_1,x_2,\dots,x_n)$ satisfies $Ax^T=y^T$.
A: Using Leibniz formula :
$$\det(A)=\sum_{\sigma \in \mathfrak  S_n}\varepsilon(\sigma )\prod_{i=1}^n a_{i\sigma (i)}=-\sum_{\sigma \in \mathfrak S_n: \sigma (1)=1}\varepsilon(\sigma )\prod_{i=2}^na_{i\sigma (i)}=-\det(B).$$
