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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants4 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 R3 → R3 - R2
$\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\$\text{a}+2)(\text{a}+3)&\text{a}+3&1\$\text{a}+3)2&1&0\end{vmatrix}$
Apply R2 → R2 - R1
$\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\$\text{a}+2)2&1&0\$\text{a}+3)2&1&0\end{vmatrix}$
$=[(2\text{a}+4)(1)-(1)(2\text{a}+6)]$
$=-2$
$=\text{R.H.S}$

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