I am trying to get from a $2\times2$ determinant to a $3\times3$ determinant.
$$\left|\begin{array}{c1 c2 c3} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right| $$ How does one get to
$$ \det(A)=\sum_{j=1}^3 a_{j1} \; \det(e_j, a_2, a_3) $$
and then end up with
$$ \det(A)=\sum_{i,j,k=1}^3 a_{i1}a_{j2}a_{k3} \; \det(e_i, e_j, e_k) $$
Consider the given determinant as $\begin{vmatrix}\mathbf{a}_1 & \mathbf{a}_2 &\mathbf{a}_3 \end{vmatrix}$, where $\mathbf{a}_j$ is the $j-$th column.
By multi-linearity of the determinants (in particular keeping the second and third columns the same), we get $$\begin{vmatrix}\color{red}{\mathbf{u} +\lambda\mathbf{b}} & \mathbf{v} &\mathbf{w} \end{vmatrix}=\begin{vmatrix}\color{red}{\mathbf{u}} & \mathbf{v} &\mathbf{w} \end{vmatrix}+\begin{vmatrix}\color{red}{\lambda\mathbf{b}} & \mathbf{v} &\mathbf{w} \end{vmatrix}=\begin{vmatrix}\color{red}{\mathbf{u}} & \mathbf{v} &\mathbf{w} \end{vmatrix}+\color{red}{\lambda}\begin{vmatrix}\color{red}{\mathbf{b}} & \mathbf{v} &\mathbf{w} \end{vmatrix}$$
Observe that $$\mathbf{a}_1=a_{11}\mathbf{e}_1+a_{21}\mathbf{e}_2+a_{31}\mathbf{e}_3= \sum_{j=1}^3a_{j1}\mathbf{e}_j$$ Using multi-linearity of the determinants, we get \begin{align*} \begin{vmatrix}\mathbf{a}_1 & \mathbf{a}_2 &\mathbf{a}_3 \end{vmatrix}&=\begin{vmatrix}\sum_{j=1}^3a_{j1}\mathbf{e}_j & \mathbf{a}_2 &\mathbf{a}_3 \end{vmatrix}\\ &=\sum_{j=1}^3\begin{vmatrix}a_{j1}\mathbf{e}_j & \mathbf{a}_2 &\mathbf{a}_3 \end{vmatrix}\\ &=\sum_{j=1}^3a_{j1}\begin{vmatrix}\mathbf{e}_j & \mathbf{a}_2 &\mathbf{a}_3 \end{vmatrix}. \end{align*} Hope you can now proceed from here.
2x2to3x3expression. – Veak Dec 28 '22 at 17:52