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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS3 Marks
Question
Using properties of determinants, prove that:
$\begin{vmatrix}1&1+\text{p}&1+\text{p}+\text{q}\\2&3+2\text{p}&4+3\text{p}+\text{2q}\\3&6+3\text{p}&10+6\text{p}+3\text{q}\end{vmatrix}=1$
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Answer

$\triangle = \begin{vmatrix}1&1+\text{p}&1+\text{p}+\text{q}\\2&3+2\text{p}&4+3\text{p}+\text{2q}\\3&6+3\text{p}&10+6\text{p}+3\text{q}\end{vmatrix}$
Applying $R_2 → R_2 - 2R_1$ and $R_3 → R_3 - 3R_1$, we have:
$\triangle=\begin{vmatrix}1&1+\text{p}&1+\text{p}+\text{q}\\0&1&2+\text{p}\\0&3&7+3\text{p}\end{vmatrix}$
Applying $R_3 → R_3 - 3R_2$, we have:
$\triangle=\begin{vmatrix}1&1+\text{p}&1+\text{p}+\text{q}\\0&1&2+\text{p}\\0&0&1\end{vmatrix}$
Expanding along $C_1$, we have:
$\triangle=1\begin{vmatrix}1&2+\text{p}\\0&1\end{vmatrix}=1(1-0)=1$
Hence, the given result is proved.

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