Determine the determinant and the inverse to the matrix A. I am currently working on determining the determinant and the inverse to the matrix A.
$A =\begin{bmatrix} 1 & 0 & 0 & 0 & 0\\0 & 1 & 0 & 0 & 0\\0 & 0 & 1 & 0 & 0\\0 & 0 & 0 & a & b\\0 & 0 & 0 & c & d\\\end{bmatrix}$
The determinant is $ad-bc$. But how do I find out the inverse? I guess there should be some kind of way to see what the inverse would be...
Any help would be much appreciated, as I am a beginner in this area a good explanation is always helpful :)! 

Since the OP asked in several comments whether this can be done by Gauss elimination/row operations, I wil also provide this computation. I will only do this for the right lower block.
$$
\left(\begin{array}{cc|cc}
a & b & 1 & 0\\
c & d & 0 & 1
\end{array}\right)\sim
\left(\begin{array}{cc|cc}
ac & bc & c & 0\\
ac & ad & 0 & a
\end{array}\right)\sim
\left(\begin{array}{cc|cc}
ac & bc & c & 0\\
0 & ad-bc &-c & a
\end{array}\right)\sim
\left(\begin{array}{cc|cc}
ac & bc & c & 0\\
0 & 1 &-\frac{c}{ad-bc} & \frac{a}{ad-bc}
\end{array}\right)\sim
\left(\begin{array}{cc|cc}
ac & 0 & \frac{acd}{ad-bc} & -\frac{abc}{ad-bc}\\
0 & 1 &-\frac{c}{ad-bc} & \frac{a}{ad-bc}
\end{array}\right)\sim
\left(\begin{array}{cc|cc}
1 & 0 & \frac{d}{ad-bc} & -\frac{b}{ad-bc}\\
0 & 1 &-\frac{c}{ad-bc} & \frac{a}{ad-bc}
\end{array}\right)
$$
The above calculation shows that
$$\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}=\frac1{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$$
Note that the operations we did are only valid if $ad-bc\ne0$ (since we have divided by this expression).
