I came across a question:
Using the formula for the determinant, find the constant $C$ such that
$$\begin{vmatrix} 2a_1 & 2a_2 & 2a_3 \\ 3b_1+4c_1 & 3b_2+4c_1 & 3b_3+4c_1 \\ 5c_1 & 5c_2 & 5c_3 \\ \end{vmatrix}= C\begin{vmatrix} a_1&a_2&a_3 \\ b_1&b_2&b_3 \\ c_1&c_2&c_3 \\ \end{vmatrix}$$
The determinant formula to be used is $$\sum_{\sigma\in S_3}sgn(\sigma)\prod_{i=1}^3a_{\sigma(i),i}$$
I'm not sure if the middle row in the LHS determinant should have all $c_1$'s. Even if the middle row has $c_1,c_2,c_3$ in the LHS, I'm still not sure how to handle the middle row terms in the determinant formula, i.e., $\prod_{i=1}^3a_{2,\sigma^{-1}(2)}$. Need help with this.
Edit: Preliminary approach provided the equation is:
$$\begin{vmatrix} 2a_1 & 2a_2 & 2a_3 \\ 3b_1+4c_1 & 3b_2+4c_2 & 3b_3+4c_3 \\ 5c_1 & 5c_2 & 5c_3 \\ \end{vmatrix}= C\begin{vmatrix} a_1&a_2&a_3 \\ b_1&b_2&b_3 \\ c_1&c_2&c_3 \\ \end{vmatrix}$$
Denote left matrix by $A$ and right matrix by $B$; and $\tau(i)=\sigma^{-1}(i)$. Then
$$det(A)=\sum_{\sigma\in S_3}sgn(\sigma)\prod_{i=1}^3A_{\sigma(i),i} \\=\sum_{\sigma\in S_3}sgn(\sigma)\prod_{\sigma^{-1}(i)=1}^3A_{i,\sigma^{-1}(i)} \\=\sum_{\tau\in S_3}sgn(\tau)\prod_{i=1}^3A_{i,\tau(i)}$$
I'm pretty sure I haven't properly justified the last equality, where I change from summation of $\sigma$ to $\tau$ and the product of $\sigma^{-1}(i)$ to $i$. Nevertheless, continuing from that point,
$$det(A)=\sum_{\tau\in S_3}sgn(\tau)A_{1,\tau(1)}A_{2,\tau(2)}A_{3,\tau(3)} \\=\sum_{\tau\in S_3}sgn(\tau)\times2B_{1,\tau(1)}\times(3B_{2,\tau(2)}+4B_{3,\tau(2)})\times5B_{3,\tau(3)} \\=30\sum_{\tau\in S_3}sgn(\tau)B_{1,\tau(1)}B_{2,\tau(2)}B_{3,\tau(3)} +40\sum_{\tau\in S_3}sgn(\tau)B_{1,\tau(1)}B_{3,\tau(2)}B_{3,\tau(3)} \\=30det(B)+40\sum_{\tau\in S_3}sgn(\tau)B_{1,\tau(1)}B_{3,\tau(2)}B_{3,\tau(3)}$$
(Thanks to @LordSharkTheUnknown for suggesting corrections)
For the second term in the last expression, given any permutation $\tau$, we can get another one by interchanging $\tau(2)$ and $\tau(3)$ to get $\tau'$. Then the sum can be split into pairs like:
$$sgn(\tau)B_{1,\tau(1)}B_{3,\tau(2)}B_{3,\tau(3)}+sgn(\tau')B_{1,\tau'(1)}B_{3,\tau'(2)}B_{3,\tau'(3)} \\=sgn(\tau)B_{1,\tau(1)}B_{3,\tau(2)}B_{3,\tau(3)}-sgn(\tau)B_{1,\tau(1)}B_{3,\tau(3)}B_{3,\tau(2)}=0$$
since $\tau'(1)=\tau(1)$, $\tau'(2)=\tau(3)$ and $\tau'(3)=\tau(2)$.
So the second term in the last expression vanishes and we're done. The only thing I'd like clarification on is why this equality holds:
$$\sum_{\sigma\in S_3}sgn(\sigma)\prod_{\sigma^{-1}(i)=1}^3A_{i,\sigma^{-1}(i)} =\sum_{\tau\in S_3}sgn(\tau)\prod_{i=1}^3A_{i,\tau(i)}$$