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I know that a $GL(2, \mathbb{R})$ is a group if and only if, certain conditions hold. One of them is to have an inverse. It is widely known that a matrix has an inverse if its determinant is not equal to $0$.

So how do I show (like a proof) that a 2 X 2 matrix $A$ is in $GL(2, \mathbb{R})$ iff the determinant of $A$ is not $0$?

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    Is $R$ here an arbitrary ring or the real numbers? Use `\mathbb{R}` in the latter case, please. – 2017-01-28
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    $\DeclareMathOperator{\GL}{GL}$$\newcommand{\R}{\mathbb{R}}$Your statement is somewhat messy. You haven't specified what $R$ is, but let us assume you mean the real numbers $\R$. Then $\GL(2, \R)$ is *by definition* the set of $2 \times 2$ invertible matrices, and this is a group under multiplication, period. Then one can show that a real square matrix is invertible iff its determinant is non-zero. – 2017-01-28

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$$\left(\begin{matrix} a & b \\ c & d \end{matrix}\right)\left(\begin{matrix} d & -b \\ -c & a \end{matrix}\right)=\left(\begin{matrix} ad-bc & 0 \\ 0 & ad-bc \end{matrix}\right)$$ Thus if the determinant $ad-bc$ of the first matrix is an invertible (with respect to multiplication) element of the ring $R$, the inverse of the matrix is just the second matrix, divided by the determinant. However, if $ad-bc$ is not invertible, then we can use the fact that determinant of product is product of determinants to show that the determinant of any multiple of this matrix is not 1, thus we cannot obtain an identity matrix by multiplication.

So the matrix is invertible iff its determinant is invertible. For any field (e.g. real numbers) it is equivalent to the fact that the determinant is non-zero.