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.

  • 1
    $\begingroup$ this is correct. $\endgroup$ Dec 11, 2019 at 9:40
  • $\begingroup$ @Ahmad Bazzi Do you mean that the given question under consideration is correct? $\endgroup$
    – gete
    Dec 11, 2019 at 9:47
  • 2
    $\begingroup$ Yes the question is correct as stated. The identity element of the group is not what you expect. $\endgroup$
    – Derek Holt
    Dec 11, 2019 at 9:56
  • $\begingroup$ It is a group but with a different identity element than you first thought. hence the determinant argument does not disprove anything. $\endgroup$ Dec 11, 2019 at 10:01

1 Answer 1


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

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


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