Find some matrix $B\in GL_2 (\mathbb{R})$ such that $B^{-1} = B^T$ and $B \neq I$

What I tried: I tried to create a simultaneous equation i.e. if B = $\begin{bmatrix} a&b\\c & d\end{bmatrix}$, then I equated the elements of the transpose to $\dfrac{1}{ad-bc}$$\begin{bmatrix} d&-b\\-c & a\end{bmatrix}$ but ended up with $ad-bc$ which doesn't make sense. What did I do wrong?

  • $\begingroup$ $\begin{bmatrix}-1 & 0 \\ 0 & 1\end{bmatrix}$ $\endgroup$
    – Decaf-Math
    Commented Feb 14, 2016 at 19:08

2 Answers 2


These kind of matrices are called orthogonal matrices . There are many examples of them, notably the following rotation matrix:

\begin{bmatrix} cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta) \\ \end{bmatrix} Note that this makes use that the determinant $cos^2(\theta) + sin^2(\theta)$ will always be one. Furthermore, the adjoint of this matrix will equal the transpose. Another example would be the identity matrix:

\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}

I would like to note that these kind of matrices have many interesting properties. Notably, the set of all the rows or the set of all the columns will be a set of orthonormal vectors.

A 2x2 matrix will be orthogonal if the following criteria are met:

\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}

$$ 1 = a^2 + x^2 $$ $$ 1 = b^2 + c^2 $$ $$ 0 = ac + bd $$

There are indeed strategies of finding orthogonal matrices of higher dimensions.

  • 1
    $\begingroup$ Really nice solution . . . Thanks! $\endgroup$
    – Raton
    Commented Feb 14, 2016 at 19:19
  • 1
    $\begingroup$ Besides the rotations (which include the identity matrix), the other 2x2 orthogonal matrices are the reflections, obtained (for example) by negating the second column in the arbitrary rotation matrix above. $\endgroup$
    – Ned
    Commented Feb 14, 2016 at 20:16

This is a better approach to solve the problem because it can easily generalized to a matrix in $\mathcal M_n(\Bbb R)$.

Let $B=(c_1\; c_2)$ the desired matrix where $c_i$ its $i$-column then we get

$$B^TB=(c_1\; c_2)^T(c_1\; c_2)=(c_1^T\; c_2^T)(c_1\; c_2)=\begin{pmatrix}c_1^Tc_1&c_1^Tc_2\\c_2^Tc_1&c_2^Tc_2\end{pmatrix}=I_2$$

which means that the columns of $B$ are orthonormal.


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