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Determinants — MATHS STD 12 Science — Question

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

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