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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS5 Marks
Question
Prove that: $\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\\(\text{a}+2)(\text{a}+3)&\text{a}+3&1\\(\text{a}+3)(\text{a}+4)&\text{a}+4 &1\end{vmatrix}=-2$
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Answer

$\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\\(\text{a}+2)(\text{a}+3)&\text{a}+3&1\\(\text{a}+3)(\text{a}+4)&\text{a}+4 &1\end{vmatrix}=-2$
$\text{L.H.S}=\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\\(\text{a}+2)(\text{a}+3)&\text{a}+3&1\\(\text{a}+3)(\text{a}+4)&\text{a}+4 &1\end{vmatrix}$ Apply $R_3 → R_3 - R_2$
$=\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\\(\text{a}+2)(\text{a}+3)&\text{a}+3&1\\(\text{a}+3)2&1&0\end{vmatrix}$ Apply $R_2 → R_2 - R_1$
$=\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\\(\text{a}+2)2&1&0\\(\text{a}+3)2&1&0\end{vmatrix}$
$=[(2\text{a}+4)(1)-(1)(2\text{a}+6)]$
$=-2$
$=\text{R.H.S}$

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