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Determinants — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDeterminants5 Marks
Question
Prove that:
$\begin{vmatrix}1&1+\text{p}&1+\text{p}+\text{q}\\2&3+2\text{p}&4+3\text{p}+2\text{p}\\3&6+3\text{p}&10+6\text{p}+3\text{q}\end{vmatrix}=1$
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Answer

$=\begin{vmatrix}1&1&1\\2&3&4\\3&2&10\end{vmatrix}+\begin{vmatrix}1&1&\text{p}\\2&3&3\text{p}\\3&6&6\text{p}\end{vmatrix}+(\text{pq})\begin{vmatrix}1&1&1\\2&2&2\\3&3&3\end{vmatrix}$
$=\begin{vmatrix}1&1&1\\2&3&4\\3&2&10\end{vmatrix}+(\text{p})\begin{vmatrix}1&1&\text{p}\\2&3&3\\3&6&6\end{vmatrix}+0$
$=\begin{vmatrix}1&1&1\\2&3&4\\3&2&10\end{vmatrix}+0$
$[\because$ Value of determinant with two identical columns is zero$]$
$=\begin{vmatrix}1&0&0\\2&1&2\\3&3&7\end{vmatrix}$ [Applying $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_1]$
$=\left\{1\times\begin{vmatrix}1&2\\3&7\end{vmatrix}\right\}$ [Expanding along $R_1]$
$=7-6$
$=1$
$=\text{R.H.S}$

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