# From the given data, prove the following relation for the determinants

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?

• So the given steps were wrong? Commented May 21, 2020 at 16:42
• 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. Commented May 21, 2020 at 17:12
• Multiplication by $a$ in this case means introducing a new number right? Commented May 21, 2020 at 17:36
• It seems so, yes. Commented May 21, 2020 at 17:38
• Well, that’s just dumb 😂 Commented May 21, 2020 at 17:56

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 $$\Delta$$.