linear algebra - Inverse of a lower triangular matrix

I got the following question to solve:

Given the lower triangular matrix

\begin{bmatrix}
A_{11} & 0 \\
A_{21} & A_{22}
\end{bmatrix}

of size $n \times n$ (n is a power of 2) where $A_{11}$, $A_{21}$ and $A_{22}$ are matrices of size $(n/2) \times (n/2)$, show that the inverse is,

\begin{bmatrix}
A_{11}^{-1} & 0 \\
-A_{22}^{-1}A_{21}A_{11} & A_{22}^{-1}
\end{bmatrix}

how do I go about to solve this problem?

Edit: the matrix is invertible.

Edit: the second matrix should be changed to:
\begin{bmatrix}
A_{11}^{-1} & 0 \\
-A_{22}^{-1}A_{21}A_{11}^{\color{red}{-1}} & A_{22}^{-1}
\end{bmatrix}
The inverse was missing.

Answer

Clearly, you are having some trouble evaluating, and I think this is because of a typo! Evaluating with normal matrix multiplication I got
$$
\begin{bmatrix}
A_{11} & 0 \\

A_{21} & A_{22}
\end{bmatrix}\begin{bmatrix}
A_{11}^{-1} & 0 \\
-A_{22}^{-1}A_{21}A_{11} & A_{22}^{-1}
\end{bmatrix} = \begin{bmatrix}
A_{11}A_{11}^{-1} & 0 \\
A_{21} A_{11}^{-1} -A_{22}A_{22}^{-1}A_{21}A_{11} & A_{22}A_{22}^{-1}
\end{bmatrix}
$$
Everything evaluates trivially except for this term

$$A_{21} A_{11}^{-1} - A_{21}A_{11}$$
Which clearly does not equal $0$ all the time.
Thus I believe there was a typo made here and that the $A_{11}$ at the end should be an $A_{11}^{-1}$, as then the above expression reduces to the identity matrix,
$$
\begin{bmatrix}
I & 0 \\
0 & I
\end{bmatrix}
$$