let V be a vectorspace with $v_1 = (3,2,1), v_2 = (2,2,1), v_3 = (1,1,1)$. Do the two basis $A = (v_1, v_2, v_3)$ and $B = (v_2, v_3, v_1)$ have the same orientation?
Since this is a new thematic for me, I tried out different ways. I hope I did it correctly. Maybe there are some other (easier, faster) options?
(1) $A$ and $B$ have same orientation, written $A \sim B$ if and only if $\det(T_B^A) > 0$.
(1.1) "From $A$ to $B$" I have to present basis-vectors of A as linear combination of basis-vectors from $B$. This easily means $v_1 = 0v_2 + 0v_3 + 1v_1$, and $v_2 = 1v_2+ 0v_3 + 0v_1$ and $v_3 = 0v_2 + 1v_3 + 0v_1$. Writing them down in a Matrix I get
$\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 1 & 0\end{pmatrix}$ (I have to write them downwards, right?
So this matrix is the transformation matrix. Since its determinant is $1$ I have shown: $A$ and $B$ have same orientation
(2) Second way: $A \sim B$ if and only if $\det M(A)\cdot \det M(B) > 0$.
In this case $M(A)$ is matrix with vectors of basis $A$. So I have
$M(A) := \begin{pmatrix} 3 & 2 & 1 \\ 2 & 2 & 1 \\ 1 & 1 & 1 \end{pmatrix}$.
$M(B) := \begin{pmatrix} 2 & 1 & 3 \\ 2 & 1 & 2 \\ 1 & 1 & 1 \end{pmatrix}$.
Since $\det (M(A)) = 1 = \det (M(B))$ I get $\det (M(A))\cdot \det (M(B)) = 1 > 0$.