Let $\mathbb{R}^n,\; n\geq 2$ be equipped with standard inner product. Let $\{v_1,v_2,......,v_n\}$ be $n$ column vectors forming an orthonormal basis of $\mathbb{R}^n$. Let $A$ be the $n\times n$ matrix formed by the column vectors $v_1,v_2,v_3,......,v_n$. Then
$1.$ $A = A^{-1}$
$2.$ $A = A^T$
$3.$ $A^{-1} = A^T$
$4.$ $\det(A) = 1.$
I have tried it as, since the given basis is orthonormal so the inner product $$\langle v_i,v_j\rangle = \begin{cases}0,\;& i\not=j\\ 1,&i=j \end{cases} ,$$
so we obtain the identity matrix i.e. $\det(A) = 1.$ But option $4$ is not correct, please someone help me to find the matrix.
As a huge hint, observe that if we write $A= (a_{ij})\;,\;\;i,j=1,...,n\;$ , then $\;A^t=(b_{ij})\;$ , with $\;b_{ij}=a_{ji}\;$ , and then
$$AA^t=(a_{ij})(b_{ij})=\left(\sum_{k=1}^na_{ik}b_{kj}\right)=\left(\sum_{k=1}^na_{ik}a_{jk}\right)=\langle v_i,\,v_j\rangle$$
where the last expression on the right denotes the usual, Euclidean inner product in $\;\Bbb R^n\;$ .
Work out this, using the fact that $\;\{v_1,...,v_n\}\;$ is an orthonormal basis, and get then almost at once (1)-(4) ...and in (4) you seem to have forgotten the absolute value: $\;(4)\;|\det A|=1\;$