Proving a subgroup of $\operatorname{GL}(2, \mathbb R)$

Question

Suppose that $\operatorname{GL}(2, \mathbb R)$ is the set of all $2\times2$, invertible matrices with entries from $ \mathbb R$. If
$ U= \left[ {\begin{array}{cc} a & b \\ 0 & d \\ \end{array} } \right] ,$ s.t. $a,b,d \in \mathbb R, a,d \neq 0$ Prove that U is a subgroup of $\operatorname{GL}(2, \mathbb R)$

So, I know that $\operatorname{GL}(2, \mathbb R)$ is the set of all linear transformations from $\mathbb R ^2 $ to $\mathbb R ^2$ and that therefore they're bijections. Also, I know that $\operatorname{GL}(2, \mathbb R)$ has only 2x2 matrices that have non zero determinants and real number entries, as a matrix could only be in $\operatorname{GL}(2, \mathbb R)$ i.f.f. its determinant is nonzero. Because there is a zero in the c position here, we only really have to show that $ad \neq 0$, not sure about notation there. Anyhow my thought was as follows:

Firstly, note that U is upper triangular. Consider another matrix, $ x= \left[ {\begin{array}{cc} d & e \\ 0 & f \\ \end{array} } \right] $, then by matrix multiplication $ U \cdot x$ =

$ Ux= \left[ {\begin{array}{cc} {ad} & {ae+bf} \\ 0 & {cf} \\ \end{array} } \right] $

So, $ac \neq 0$ and $df \neq 0$, which implies that $adcf \neq 0$, and therefore U is closed under matrix multiplication. As for the inverse of U, I believe it to be as follows:

$ {U^{-1}}= \left[ {\begin{array}{cc} {C^{-1}} & -{b(ac)^{-1}} \\ 0 & {a^{-1}} \\ \end{array} } \right] $ But I am not sure if this is right or what conclusions to draw from this matrix.

Answer 1

Hint:

If $\mathbf{A}=\begin{pmatrix}a&b\\c&d\end{pmatrix}$, then $\mathbf{A}^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$

Answer 2

$G = \{ A| A \text{ is upper triangular }, \det A \neq 0 \}$. It should be clear that $G \subset \operatorname{GL}(n, \mathbb R)$.

To show that $G$ is a group is really just a matter of grinding through the definition:

You need to show closure, that is, if $A_1, A_2 \in G$ then $A_1 A_2 \in G$. Invertibility follows from $\det (A_1 A_2) = \det A_1 \det A_2$, you need to confirm that $A_1 A_2$ is also upper triangular.

Associativity follows from associativity of matrix multiplication.

The identity $I $ is a member of $G$.

You need to show invertibility, that if $A \in G$, then $A^{-1} \in G$. The only tricky part here is to show that $A^{-1}$ is upper triangular.