First slide

Introduction to Determinants

Question

If $a_{1}, a_{2}, \dots, a_{n}, \dots$ form a G.P. then $|\begin{array}{lll} \log a_{n} \log a_{n + 1} \log a_{n + 2} \\ \log a_{n + 3} \log a_{n + 4} \log a_{n + 5} \\ \log a_{n + 6} \log a_{n + 7} \log a_{n + 8} \end{array}|$ is equal to = , (where $a_{i} > 0,$ for all $i \geq 1$ )

Moderate

A

0
B

1
C

2
D

3

Solution

We have,
${(a_{n + 1})}^{2} = a_{n} a_{n + 2}$
$\Rightarrow 2 \log a_{n + 1} = \log a_{n} + \log a_{n + 2}$
Similarly,
$\begin{matrix} 2 \log a_{n + 4} = \log a_{n + 3} + \log a_{n + 5} \\ 2 \log a_{n + 7} = \log a_{n + 6} + \log a_{n + 8} \end{matrix}$
Substituting these values in second column of determinant, we get
$\begin{matrix} Δ = \frac{1}{2} |\begin{array}{ccc} \log a_{n} & \log a_{n} + \log a_{n + 2} & \log a_{n + 2} \\ \log a_{n + 3} & \log a_{n + 3} + \log a_{n + 5} & \log a_{n + 5} \\ \log a_{n + 6} & \log_{n + 6} + \log a_{n + 8} & \log a_{n + 8} \end{array}| \\ = \frac{1}{2} (0) = 0 [Using C_{2} \to C_{2} - C_{1} - C_{3}] \end{matrix}$

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