Suppose $\Delta=\begin{vmatrix} a_1&b_1&c_1 \\ a_2&b_2&c_2 \\ a_3&b_3&c_3 \end{vmatrix}$ and $\Delta *=\begin{vmatrix} a_1+pb_1&b_1+qc_1&c_1+ra_1\\\ a_2+pb_2&b_2+qc_2&c_2+ra_2\\a_3+pb_3&b_3+qc_3&c_3+ra_3\end {vmatrix}$, then prove that $\Delta*=\Delta (1+pqr)$
The solution given to me went along the following lines
They initially split the determinant $\Delta*$ in sum of two determinants along $C_1$
$$\Delta* =\begin {vmatrix} a_1&b_1+qc_1&c_1+ra_1\\\ a_2&b_2+qc_2&c_2+ra_2\\a_3&b_3+qc_3&c_3+ra_3\end {vmatrix} + \begin{vmatrix} pb_1&b_1+qc_1&c_1+ra_1\\\ pb_2&b_2+qc_2&c_2+ra_2\\pb_3&b_3+qc_3&c_3+ra_3\end {vmatrix}$$
In the first determinant apply $C_3\rightarrow C_3-aC_1$ and then $C_2\rightarrow C_2-rC_3$
In the second determinant take $p$ common, then apply $C_3\rightarrow C_3-C_2$ and then take $r$ common from $C_3$
I performed all those given operations, but I didn’t find them very useful because I still ended up with a lot of letters and I couldn’t see how it was getting to the answer. Is the given solution wrong, if so, then how should I solve it?
The idea is to be able to avoid explicitly calculating any determinants at all, and instead use manipulations to express $\Delta^*$ in terms of $\Delta.$
For the first determinant, if you instead use $C_3\to C_3-rC_1$ and then $C_2\to C_2-qC_3,$ then you'll see exactly $\Delta$ for the first determinant.
For the second, you'll start as recommended (by taking $p$ common), then do $C_2\to C_2-C_1.$ Do you think you can take it from there?