# Inverse of any element in a group of matrices [duplicate]

I came across the following question in an under graduate course examination:

Let $$G$$ be a group of all matrices of the form $$\pmatrix{x&x\\x&x}$$, where $$0\neq x\in \Bbb R$$ under matrix multiplication. Find the inverse of any element in $$G$$.

It appears to me that the question is wrongly framed because any matrix of the given form doesn't have inverse due to the fact that its determinant is zero.

• this is correct. Dec 11, 2019 at 9:40
• @Ahmad Bazzi Do you mean that the given question under consideration is correct?
– gete
Dec 11, 2019 at 9:47
• Yes the question is correct as stated. The identity element of the group is not what you expect. Dec 11, 2019 at 9:56
• It is a group but with a different identity element than you first thought. hence the determinant argument does not disprove anything. Dec 11, 2019 at 10:01

The trick is that the identity element is not what you'd expect.

First, note that the group operation is clearly commutative. We can compute $$\pmatrix{x&x\\x&x}\pmatrix{y&y\\y&y} = \pmatrix{2xy&2xy\\2xy&2xy}$$

which tells us that if $$y=\frac{1}{2}$$ (recall $$x \neq 0$$)

$$\pmatrix{x&x\\x&x}\pmatrix{\frac{1}{2}&\frac{1}{2}\\\frac{1}{2}&\frac{1}{2}} = \pmatrix{x&x\\x&x}$$

So the identity element of this group is the matrix with $$x=\frac{1}{2}$$.

To compute the inverse, just look at the first expression again and write $$2xy = \frac{1}{2}$$, which gives $$y = \frac{1}{4x}$$. Indeed, we can verify

$$\pmatrix{x&x\\x&x}\pmatrix{\frac{1}{4x}&\frac{1}{4x}\\\frac{1}{4x}&\frac{1}{4x}} = \pmatrix{\frac{1}{2}&\frac{1}{2}\\\frac{1}{2}&\frac{1}{2}}$$