Value of left|begin{array}{lll}(a+1)(a+2) & a+2 & 1 (a+2)(a+3) & a+3 & 1 (a+3)(a+4)

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3 and 4 .Determinants and Matrices

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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

A

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

B

$-2$

C

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

D

$0$

(JEE MAIN-2021)

Solution

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

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

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

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

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

Standard 12

Mathematics

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