Prove that.
$\left|\begin{array}{llll}(a+1) & (a+2) & a+2 & 1 \\ (a+2) & (a+3) & a+3 & 1 \\ (a+3) & (a+4) & a+4 & 1\end{array}\right|=-2$
Let LHS $=\Delta=\mid(a+1)(a+2) \quad a+2 \quad 1$
$(a+2)(a+3) \quad a+3 \quad 1$
$(a+3)(a+4) \quad a+4 \quad 1 \mid$
$\mid(a+1)(a+2)-(a+2)(a+3) \quad(a+2)-(a+3) \quad 0(a+2)(a+3)-(a+3)(a+4) \quad(a+3)-(a+4) \quad 0(a+3)(a+4)$ $(a+4)$ $1 \mid$
[Applying $\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{2}$ and $\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{3}$ ]
$\begin{array}{lll}=\mid-2(a+2) & -1 & 0 \\ -2(a+3) & -1 & 0\end{array}$
$\begin{aligned}&(a+3)(a+4) \quad(a+4) \quad 1 \\&=\{1 \times \mid-2(a+2)\end{aligned} \quad-1$
$-2(\mathrm{a}+3) \quad-1 \mid\} \quad\left[\right.$ Expanding along $\left.\mathrm{C}_{3}\right]$
$=4+2 \mathrm{a}-2 \mathrm{a}-6$
$=-2$
$=$ RHS
Hence proved.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.