The value of

Question:

The value of $\left|\begin{array}{lll}(a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1\end{array}\right|$ is

  1. $(a+2)(a+3)(a+4)$

  2. $-2$

  3. $(a+1)(a+2)(a+3)$

  4. 0


Correct Option: , 2

Solution:

$\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}$ and $\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}$

$\Delta=\left|\begin{array}{ccc}(a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3-a-1) & 1 & 0 \\ a^{2}+7 a+12-a^{2}-3 a-2 & 2 & 0\end{array}\right|$

$=\left|\begin{array}{ccc}a^{2}+3 a+2 & a+2 & 1 \\ 2(a+2) & 1 & 0 \\ 4 a+10 & 2 & 0\end{array}\right|$

$=4(a+2)-4 a-10$

$=4 a+8-4 a-10=-2$

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