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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?

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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?

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  • $\begingroup$ So the given steps were wrong? $\endgroup$ – Aditya May 21 at 16:42
  • $\begingroup$ The given steps only make things more complicated and then force explicit determinant calculation. They aren't "wrong," per se, but they aren't helpful. $\endgroup$ – Cameron Buie May 21 at 17:12
  • $\begingroup$ Multiplication by $a$ in this case means introducing a new number right? $\endgroup$ – Aditya May 21 at 17:36
  • $\begingroup$ It seems so, yes. $\endgroup$ – Cameron Buie May 21 at 17:38
  • $\begingroup$ Well, that’s just dumb 😂 $\endgroup$ – Aditya May 21 at 17:56
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In the second determinant, factor out $p$, then $C_2\to C_2-C_1$. Factor out $q$ and $C_3\to C_3-C_2$. Factor out $r$. Now you have $pqr$ times $\Delta$.

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