Proof that the inverse of 3x3 matrix is adjugate matrix divided by it's determinant

Question

Okay so I need to proof that $$A^{-1}=\frac{1}{det(A)}\begin{bmatrix}(x_2 \times x_3)^T\\(x_3 \times x_1)^T\\(x_1 \times x_2)^T\\\end{bmatrix}$$ Where $x$:s are column vectors. I really don't know where to start, but I know that $det(A)=x_1(x_2 \times x_3)$ if it helps. I will appreciate any help.

Answer 1

\begin{align*} \mathbb{A} &= \begin{pmatrix} \mathbf{x}_1 & \mathbf{x}_2 & \mathbf{x}_3 \end{pmatrix} \\[5pt] \mathbb{B} &= \begin{pmatrix} (\mathbf{x}_2 \times \mathbf{x}_3)^T \\ (\mathbf{x}_3 \times \mathbf{x}_1)^T \\ (\mathbf{x}_1 \times \mathbf{x}_2)^T \end{pmatrix} \\[5pt] \mathbb{BA} &= \begin{pmatrix} \mathbf{x}_{2} \times \mathbf{x}_3 \cdot \mathbf{x}_1 & \mathbf{x}_{2} \times \mathbf{x}_3 \cdot \mathbf{x}_2 & \mathbf{x}_{2} \times \mathbf{x}_3 \cdot \mathbf{x}_3 \\ \mathbf{x}_{3} \times \mathbf{x}_1 \cdot \mathbf{x}_1 & \mathbf{x}_{3} \times \mathbf{x}_1 \cdot \mathbf{x}_2 & \mathbf{x}_{3} \times \mathbf{x}_1 \cdot \mathbf{x}_3 \\ \mathbf{x}_{1} \times \mathbf{x}_2 \cdot \mathbf{x}_1 & \mathbf{x}_{1} \times \mathbf{x}_2 \cdot \mathbf{x}_2 & \mathbf{x}_{1} \times \mathbf{x}_2 \cdot \mathbf{x}_3 \\ \end{pmatrix} \\[5pt] &= \begin{pmatrix} \mathbf{x}_{2} \times \mathbf{x}_3 \cdot \mathbf{x}_1 & 0 & 0 \\ 0 & \mathbf{x}_{3} \times \mathbf{x}_1 \cdot \mathbf{x}_1 & 0 \\ 0 & 0 & \mathbf{x}_{1} \times \mathbf{x}_2 \cdot \mathbf{x}_3 \end{pmatrix} \\[5pt] &= (\det \mathbb{A}) \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \end{align*}