Question

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

• A. $-2$
• B. $(a+1)(a+2)(a+3)$
• C. $0$
• D. $(a+2)(a+3)(a+4)$

Answer 1

The correct option is A

$-2$

Explanation for the correct option:

Solve the given determinant

Given, $\left|\begin{array}{ccc}(a+1)(a+2)& a+2& 1\\ (a+2)(a+3)& a+3& 1\\ (a+3)(a+4)& a+4& 1\end{array}\right|$

$$\begin{gathered} C_1\to C_1-C_2 \\ △=\left|\begin{array}{ccc}(a+2)a& a+2& 1\\ (a+3)(a+1)& a+3& 1\\ (a+4)(a+2)& a+4& 1\end{array}\right| \\ {C}_2\to C_2-C_3 \\ △=\left|\begin{array}{ccc}(a+2)a& a+1& 1\\ (a+3)(a+1)& a+2& 1\\ (a+4)(a+2)& a+3& 1\end{array}\right| \\ {R}_2\to R_2-R_1,R_3\to R_3-R_1 \\ △=\left|\begin{array}{ccc}(a^2+2a)& a+1& 1\\ (2a+3)& 1& 0\\ (4a+8)& 2& 0\end{array}\right| \\ =6-8=-2 \end{gathered}$$

Hence, the correct answer is option (A).