Compute the cross products of given vectors

Consider the vectors

$A = -\mathbf{i} + 2 \mathbf{k}, \qquad B = 2 \mathbf{i} + \mathbf{j} - \mathbf{k}, \qquad C = \mathbf{i} + 2 \mathbf{j} + 2 \mathbf{k}.$

Compute each of the following.

$A \times B$ ;
$B \times C$ ;
$C \times A$ ;
$A \times (C \times A)$ ;
$(A \times B) \times C$ ;
$A \times (B \times C)$ ;
$(A \times C) \times B$ ;
$(A + B) \times (A - C)$ ;
$(A \times B) \times (A \times C)$ .

We compute
$\begin{align*} A \times B &= \left( \begin{vmatrix*}[r] a_2 & a_3 \\ b_2 & b_3 \end{vmatrix*}, \begin{vmatrix*}[r] a_3 & a_1 \\ b_3 & b_1 \end{vmatrix*}, \begin{vmatrix*}[r] a_1 & a_2 \\ b_1 & b_2 \end{vmatrix*} \right) \\[9pt] &= \left( \begin{vmatrix*}[r] 0 & 2 \\ 1 & -1 \end{vmatrix*}, \begin{vmatrix*}[r] 2 & -1 \\ -1 & 2 \end{vmatrix*}, \begin{vmatrix*}[r] -1 & 0 \\ 2 & 1 \end{vmatrix*} \right) \\[9pt] &= (0-2, 4 + (-1), -1-0) \\[9pt] &= (-2,3,-1). \end{align*}$
We compute
$\begin{align*} B \times C &= \left( \begin{vmatrix*}[r] b_2 & b_3 \\ c_2 & c_3 \end{vmatrix*}, \begin{vmatrix*}[r] b_3 & b_1 \\ c_3 & c_1 \end{vmatrix*}, \begin{vmatrix*}[r] b_1 & b_2 \\ c_1 & c_2 \end{vmatrix*} \right) \\[9pt] &= \left( \begin{vmatrix*}[r] 1 & -1 \\ 2 & 2 \end{vmatrix*}, \begin{vmatrix*}[r] -1 & 2 \\ 2 & 1 \end{vmatrix*}, \begin{vmatrix*}[r] 2 & 1 \\ 1 & 2 \end{vmatrix*} \right) \\[9pt] &= (2+2,-1-4,4-1) \\[9pt] &= (4,-5,3). \end{align*}$
We compute
$\begin{align*} C \times A &= \left( \begin{vmatrix*}[r] c_2 & c_3 \\ a_2 & a_3 \end{vmatrix*}, \begin{vmatrix*}[r] c_3 & c_1 \\ a_3 & a_1 \end{vmatrix*}, \begin{vmatrix*}[r] c_1 & c_2 \\ a_1 & a_2 \end{vmatrix*} \right) \\[9pt] &= \left( \begin{vmatrix*}[r] 2 & 2 \\ 0 & 2 \end{vmatrix*}, \begin{vmatrix*}[r] 2 & 1 \\ 2 & -1 \end{vmatrix*}, \begin{vmatrix*}[r] 2 & 2 \\ -1 & 0 \end{vmatrix*} \right) \\[9pt] &= (4,-2-2,0+2) \\[9pt] &= (4,-4,2). \end{align*}$
Using part (c) we compute
$\begin{align*} A \times (C \times A) &= A \times (4,-4,2) \\[9pt] &= \left( \begin{vmatrix*} 0 & 2 \\ -4 & 2 \end{vmatrix*}, \begin{vmatrix*}[r] -1 & -2 \\ 2 &1 \end{vmatrix*}, \begin{vmatrix*}[r] -1 & 0 \\ 4 & -4 \end{vmatrix*} \right) \\[9pt] &= (8,10,4). \end{align*}$
Using part (a) we compute
$\begin{align*} (A \times B) \times C &= (-2,3,-1) \times C \\[9pt] &= \left( \begin{vmatrix*} 3 & -1 \\ 2 & 2 \end{vmatrix*}, \begin{vmatrix*}[r] -1 & -2 \\ 2 &1 \end{vmatrix*}, \begin{vmatrix*}[r] -2 & 3 \\ 1 & 2 \end{vmatrix*} \right) \\[9pt] &= (8,3,-7). \end{align*}$
Using part (b) we compute
$\begin{align*} A \times (B \times C) &= A \times (4,-5,3) \\[9pt] &= \left( \begin{vmatrix*} 0 & 2 \\ -5 & 3 \end{vmatrix*}, \begin{vmatrix*}[r] 2 & -1 \\ 3 & 4 \end{vmatrix*}, \begin{vmatrix*}[r] -1 & 0 \\ 4 & -5 \end{vmatrix*} \right) \\[9pt] &= (10,11,5). \end{align*}$
Using part (c) we compute
$\begin{align*} (A \times C) \times B &= (-4,4,-2) \times B \\[9pt] &= \left( \begin{vmatrix*} 4 & -2 \\ 1 & -1 \end{vmatrix*}, \begin{vmatrix*}[r] -2 & -4 \\ -1 & 2 \end{vmatrix*}, \begin{vmatrix*}[r] -4 & 4 \\ 2 & 1 \end{vmatrix*} \right) \\[9pt] &= (-2,-8,-12). \end{align*}$
We compute
$\begin{align*} (A + B) \times (A - C) &= (1,1,1) \times (-2,-2,0) \\[9pt] &= \left( \begin{vmatrix*} 1 & 1 \\ -2 & 0 \end{vmatrix*}, \begin{vmatrix*}[r] 1 & 1 \\ 0 & -2 \end{vmatrix*}, \begin{vmatrix*}[r] 1 & 1 \\ -2 & -2 \end{vmatrix*} \right) \\[9pt] &= (2,-2,0). \end{align*}$
We compute
$\begin{align*} (A \times B) \times (A \times C) &= (-2,3,-1) \times (-4,4,-2) \\[9pt] &= \left( \begin{vmatrix*} 3 & -1 \\ 4 & -2 \end{vmatrix*}, \begin{vmatrix*}[r] -1 & -2 \\ -2 & -4 \end{vmatrix*}, \begin{vmatrix*}[r] -2 & 3 \\ -4 & 4 \end{vmatrix*} \right) \\[9pt] &= (-2,0,4). \end{align*}$