Inverse Matrix Question Determine whether there is an inverse and then determine what the inverse is if it exists:
\begin{bmatrix}a&b\\c&d\end{bmatrix}
Where $ad-bc\not= 0$.
I know how to do this for a normal matrix, but I don't understand how to do this for one with the variables.
 A: Try to multiply your matrix by
$$\frac 1{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$$
which is legit since $ad-bc\ne 0$.
What can you conclude?
One way of finding this particular matrix is using Gauss' reduction.
A: If a matrix has an inverse, we can find this inverse using the following method:(in this case, because $ad-bc \neq 0$, we certainly can find an inverse matrix, as you will notice throughout the work)
$(A|I_n) \equiv (I_n|A^{-1}) $
where $\equiv$ means that those 2 matrices are row equivalent. Thus, look for the row reduced echelon form of:
\begin{bmatrix}a&b&1&0\\c&d&0&1\end{bmatrix}.
What appears in the two last column will be the inverse you are looking for.
A: You need to solve this equation:
$$\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}a'&b'\\c'&d'\end{bmatrix}
=\begin{bmatrix}1&0\\0&1\end{bmatrix}.$$
After expansion, you get the system
$$\begin{cases}aa'+bc'=1,\\ca'+dc'=0,\\ab'+bd'=0,\\cb'+dd'=1.\end{cases}$$ which splits in two $2\times2$ subsystems.
Then by Cramer
$$\begin{cases}
a=\frac{\left|\begin{matrix}1&b\\0&d\end{matrix}\right|}{\left|\begin{matrix}a&b\\c&d\end{matrix}\right|},
b=\frac{\left|\begin{matrix}0&b\\1&d\end{matrix}\right|}{\left|\begin{matrix}a&b\\c&d\end{matrix}\right|},\\
c=\frac{\left|\begin{matrix}a&1\\c&0\end{matrix}\right|}{\left|\begin{matrix}a&b\\c&d\end{matrix}\right|},
d=\frac{\left|\begin{matrix}a&0\\c&1\end{matrix}\right|}{\left|\begin{matrix}a&b\\c&d\end{matrix}\right|},
\end{cases}$$
giving
$$\begin{bmatrix}\frac d{ad-bc}&-\frac b{ad-bc}\\-\frac c{ad-bc}&\frac a{ad-bc}\end{bmatrix}.$$
A: There are other answers. But I wanted to include this answer wherein we reduce the matrix to the identity and ensures the existence of the inverse.
Denote the given matrix by $A.$
1) $c=0$
If $c=0,$ then $a$ and $d$ both has to be non-zero, otherwise we would have $ad-bc=0.$
Row-reducing $\begin{bmatrix}a&b\\0&d\end{bmatrix}\to \begin{bmatrix}1&b/a\\0&1\end{bmatrix}\to \begin{bmatrix}1&0\\0&1\end{bmatrix}$
Thus $A$ is invertible.
2) $a=0$
If $a=0$ then $b$ and $c$ both has to be non-zero, otherwise we would have $ad-bc=0.$
Row-reducing $\begin{bmatrix}0&b\\c&d\end{bmatrix}\to\begin{bmatrix}0&1\\1&d/c\end{bmatrix}\to\begin{bmatrix}0&1\\1&0\end{bmatrix}\to\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$A$ is invertible in this case too.
3) $a\ne 0, c\ne 0$ 
Row-reducing $\begin{bmatrix}a&b\\c&d\end{bmatrix}\to \begin{bmatrix}ac&bc\\ac&ad\end{bmatrix}\to\begin{bmatrix}ac&bc\\0&ad-bc\end{bmatrix}\to\begin{bmatrix}ac&bc\\0&1\end{bmatrix}\to\begin{bmatrix}ac&0\\0&1\end{bmatrix}\to\begin{bmatrix}1&0\\0&1\end{bmatrix}.$
Thus $A$ is invertible in all the cases and the inverse exists.
The explicit form of the inverse is given in one of the other answers to the question.
