Weird proof with matrix determinant.

Question

Let $a_1,a_2,a_3,a_4,b \in \Bbb R^4$ and $t \in \Bbb R$.

Show that

$$\frac{d}{dt} \det(a_1, a_2 + tb, a_3, a_4) = \det(a_1, b, a_3, a_4)$$

I have absolutely no idea how to solve this. How can you derive a determinant?

Answer 1

To get you started:\begin{align} \frac{d}{dt} \det\begin{bmatrix} a_1 , a_2+tb,a_3, a_4\end{bmatrix} &=\frac{d}{dt} \det\begin{bmatrix} a_1 , a_2,a_3, a_4\end{bmatrix} + \frac{d}{dt} \det\begin{bmatrix} a_1 , tb,a_3, a_4\end{bmatrix} \\ &= \frac{d}{dt}\det \begin{bmatrix} a_1 , tb,a_3, a_4\end{bmatrix} \\ \end{align}

Note that $\det \begin{bmatrix} a_1 , tb,a_3, a_4\end{bmatrix}$ is linear in $t$.

Edit:

Note that $$\det\begin{bmatrix} a_1, bt, a_3, a_4\end{bmatrix}=t\det\begin{bmatrix} a_1, b, a_3, a_4\end{bmatrix}$$