Row operations in determinants

Question

How to prove that by adding to a row a multiple of another row, the determinant remains the same ?

Answer 1

We have that $$\Delta = \begin{vmatrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\end{vmatrix}\tag{1}$$ $$\Rightarrow \text{ (Performing }R_1 \to R_1 + kR_2) \, \, \Delta_1 = \begin{vmatrix} a_1 + kb_1 & a_2+kb_2 & a_3 + kb_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\end{vmatrix} \text{ (say) }$$ $$\Rightarrow \Delta_1 = \begin{vmatrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\end{vmatrix} + \begin{vmatrix} kb_1 & kb_2 & kb_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\end{vmatrix}$$ $$\Rightarrow \Delta_1 = \begin{vmatrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\end{vmatrix} + k\begin{vmatrix} b_1 & b_2 & b_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\end{vmatrix}$$ $$\Rightarrow \Delta_1 = \begin{vmatrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\end{vmatrix} + 0$$ $$\Rightarrow \Delta_1 = \Delta$$

Similarly, we can show the same for any column operation. Thus, for any such column and row operation, the determinant remains the same. Hope it helps.