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 :)! 
 A: Hint:see  block matrix inversion $$A =\begin{bmatrix} 1 & 0 & 0 & 0 & 0\\0 & 1 & 0 & 0 & 0\\0 & 0 & 1 & 0 & 0\\0 & 0 & 0 & a & b\\0 & 0 & 0 & c & d\\\end{bmatrix}=\begin{bmatrix} I & 0\\0 & C \end{bmatrix}\to A^{-1}=\begin{bmatrix} I & 0\\0 & C^{-1} \end{bmatrix}$$ such that $C=\begin{bmatrix} a & b\\ c & d\end{bmatrix}$
A: Hint: what is the inverse of $A=\bigl(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\bigr)$? Does putting the inverse $A^{-1}$ in the corner of the matrix in place of what's already there create the desired inverse matrix? Check.
For a theoretical way to see why one might guess this, consider direct sums. If we let the matrix be a transformation of ${\bf R}^3\oplus{\bf R}^2$, it can be written in the form ${\rm Id}\oplus A$, which has inverse ${\rm Id}\oplus A^{-1}$.
A: First off the inverse only exists if $_____________$?
Assuming the above hypothesis, start out by finding the inverse of $A:=\begin{bmatrix} a & b\\ c & d\end{bmatrix}$.
I'm assuming this was given earlier or appeared in a previous exercise, therefore you should know that $\begin{bmatrix}a & b\\ c & d\end{bmatrix}^{-1}=\dfrac{1}{ad-bc}\begin{bmatrix}  d & -b\\ -c & a\end{bmatrix}=(\det (A))^{-1}\begin{bmatrix}  d & -b\\ -c & a\end{bmatrix}$.
Now consider the matrix $\begin{bmatrix}1 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 &d\det (A))^{-1} & -b\det (A))^{-1}\\
0 & 0 & 0 &-c\det (A))^{-1} & a\det (A))^{-1} \end{bmatrix}\color{grey}{=\begin{bmatrix} I_3 & 0\\ 0 & A^{-1}\end{bmatrix}_{5\times 5}}.$
A: To compute the determinant use Laplace expansion. Then, make an inversion on each block.
A: 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).
