group theory - Invertability of Singular 2x2 Matrix with all same real values.

Question:

Let set G = { matrix [{a a},{a a}] such that a is real but not 0 } represent the set of 2x2 matrices with same elements of the reals excluding a = 0, show that G is a group under matrix multiplication. Explain why each element of G has an inverse even though the matrices have 0 determinants.

Thoughts:

I can show closure, the existence of a unique identity and associativity; but the inverse has me lost. How can a 2x2 singular matrix with identical elements from the reals (excluding zero) be a group under matrix multiplication when no inverse exist that I can see (though the book asserts otherwise)?

Problem is from Contemporary Abstract Algebra (Gallian 8th Ed) Ch2 Problem 52.

Answer

Hint: for arbitrary $a \in \mathbb{R} \setminus \{0\}$, we're after a solution to the equation

$$\begin{bmatrix}
a & a \\
a & a \\
\end{bmatrix}
\begin{bmatrix}
b & b \\
b & b \\
\end{bmatrix}
=

\mathrm{id}_G
=
\begin{bmatrix}
1/2 & 1/2 \\
1/2 & 1/2 \\
\end{bmatrix}$$