$3\times3$ determinant using standard basis 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
$$ \det(A)=\sum_{i,j,k=1}^3 a_{i1}a_{j2}a_{k3} \; \det(e_i, e_j, e_k) $$
 A: 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.
