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}$$\begin{bmatrix} d&-b\\-c & a\end{bmatrix}$ but ended up with $ad-bc$ which doesn't make sense. What did I do wrong?
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$\begingroup$These kind of matrices are called orthogonal matrices . There are many examples of them, notably the following rotation matrix:
\begin{bmatrix} cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta) \\ \end{bmatrix} Note that this makes use that the determinant $cos^2(\theta) + sin^2(\theta)$ will always be one. Furthermore, the adjoint of this matrix will equal the transpose. Another example would be the identity matrix:
\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}
I would like to note that these kind of matrices have many interesting properties. Notably, the set of all the rows or the set of all the columns will be a set of orthonormal vectors.
A 2x2 matrix will be orthogonal if the following criteria are met:
\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}
$$ 1 = a^2 + x^2 $$ $$ 1 = b^2 + c^2 $$ $$ 0 = ac + bd $$
There are indeed strategies of finding orthogonal matrices of higher dimensions.
$\endgroup$ 2 $\begingroup$This is a better approach to solve the problem because it can easily generalized to a matrix in $\mathcal M_n(\Bbb R)$.
Let $B=(c_1\; c_2)$ the desired matrix where $c_i$ its $i$-column then we get
$$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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