Answer
A is invertible
Work Step by Step
$A=\left[\begin{array}{cc}A_{11} & A_{12} \\ 0 & A_{22}\end{array}\right]$
First suppose $A$ is invertible. Then example 5 shows us that $A_{11}$ and $A_{22}$ are invertible.
Now, suppose $A_{11}$ and $A_{22}$ are invertible.
Formula in example 5 provides candidate for inverse of $A$
$\left[\begin{array}{cc}A_{11} & A_{12} \\ 0 & A_{22}\end{array}\right] \cdot\left[\begin{array}{cc}A_{11}^{-1} & -A 11^{-1} A_{12} A_{22}^{-1} \\ 0 & A_{22}^{-1}\end{array}\right]=$
$=\left[\begin{array}{cc}A_{11} A_{11}^{-1}+A_{12} \cdot 0 & A_{11} A_{11}^{-1} A_{12} A_{22}^{-1}+A_{12} A_{22}^{-1} \\ 0 \cdot A_{11}^{-1}+A_{22} \cdot 0 & 0 \cdot\left(-A_{11}^{-1} A_{12} A_{22}^{-1}+A_{22} A_{22}^{-1}\right)\end{array}\right]=$
$=\left[\begin{array}{cc}I & -\left(A_{11} A_{11}^{-1}\right) A_{12} A_{22}^{-1}+A_{12} A_{22}^{-1} \\ 0 & I\end{array}\right]=$
$=\left[\begin{array}{cc}I & -A_{12} A_{22}^{-1}+A_{12} A_{22}^{-1} \\ 0 & I\end{array}\right]=$
$=\left[\begin{array}{ll}I & 0 \\ 0 & I\end{array}\right]=I$
This means that $A$ is invertible.
