Question

If a, b, c > 1, then $\Delta = \begin{vmatrix} \log_a (abc) & \log_a b & \log_a c \\ \log_b (abc) & 1 & \log_b c \\ \log_c (abc) & \log_c b & 1 \end{vmatrix}$ equals to ____.

• A. 0
• B. $\log_a b + \log_b c + \log_c a$
• C. $\log_{abc} (a + b + c)$
• D. none of these

Answer 1

Answer: (A)

0

Answer 2

Let the given determinant be $\Delta$.

$\Delta = \begin{vmatrix} \log_a (abc) & \log_a b & \log_a c \\ \log_b (abc) & 1 & \log_b c \\ \log_c (abc) & \log_c b & 1 \end{vmatrix}$

We use the property of logarithms $\log_x (yz) = \log_x y + \log_x z$ and $\log_x x = 1$. The elements in the first column can be simplified:

$\log_a (abc) = \log_a a + \log_a b + \log_a c = 1 + \log_a b + \log_a c$

$\log_b (abc) = \log_b a + \log_b b + \log_b c = \log_b a + 1 + \log_b c$

$\log_c (abc) = \log_c a + \log_c b + \log_c c = \log_c a + \log_c b + 1$

Substitute these into the determinant:

$\Delta = \begin{vmatrix} 1 + \log_a b + \log_a c & \log_a b & \log_a c \\ \log_b a + 1 + \log_b c & 1 & \log_b c \\ \log_c a + \log_c b + 1 & \log_c b & 1 \end{vmatrix}$

Let $C_1, C_2, C_3$ denote the first, second, and third columns, respectively. Apply the column operation $C_1 \to C_1 - C_2 - C_3$. The new elements in the first column are:

Row 1: $(1 + \log_a b + \log_a c) - \log_a b - \log_a c = 1$

Row 2: $(\log_b a + 1 + \log_b c) - 1 - \log_b c = \log_b a$

Row 3: $(\log_c a + \log_c b + 1) - \log_c b - 1 = \log_c a$

The determinant becomes:

$\Delta = \begin{vmatrix} 1 & \log_a b & \log_a c \\ \log_b a & 1 & \log_b c \\ \log_c a & \log_c b & 1 \end{vmatrix}$

Now, we can use the change of base formula for logarithms, $\log_x y = \frac{\ln y}{\ln x}$. Let $\ln a = \alpha$, $\ln b = \beta$, $\ln c = \gamma$. Since $a, b, c > 1$, $\alpha, \beta, \gamma > 0$.

$\log_a b = \frac{\ln b}{\ln a} = \frac{\beta}{\alpha}$

$\log_a c = \frac{\ln c}{\ln a} = \frac{\gamma}{\alpha}$

$\log_b a = \frac{\ln a}{\ln b} = \frac{\alpha}{\beta}$

$\log_b c = \frac{\ln c}{\ln b} = \frac{\gamma}{\beta}$

$\log_c a = \frac{\ln a}{\ln c} = \frac{\alpha}{\gamma}$

$\log_c b = \frac{\ln b}{\ln c} = \frac{\beta}{\gamma}$

Substitute these into the determinant:

$\Delta = \begin{vmatrix} 1 & \beta/\alpha & \gamma/\alpha \\ \alpha/\beta & 1 & \gamma/\beta \\ \alpha/\gamma & \beta/\gamma & 1 \end{vmatrix}$

Let $R_1, R_2, R_3$ denote the first, second, and third rows, respectively. Apply the row operations $R_2 \to R_2 - \frac{\alpha}{\beta} R_1$ and $R_3 \to R_3 - \frac{\alpha}{\gamma} R_1$.

For $R_2 \to R_2 - \frac{\alpha}{\beta} R_1$:

Row 2, element 1: $\frac{\alpha}{\beta} - \frac{\alpha}{\beta} \times 1 = 0$

Row 2, element 2: $1 - \frac{\alpha}{\beta} \times \frac{\beta}{\alpha} = 1 - 1 = 0$

Row 2, element 3: $\frac{\gamma}{\beta} - \frac{\alpha}{\beta} \times \frac{\gamma}{\alpha} = \frac{\gamma}{\beta} - \frac{\gamma}{\beta} = 0$

The new second row is $(0, 0, 0)$.

For $R_3 \to R_3 - \frac{\alpha}{\gamma} R_1$:

Row 3, element 1: $\frac{\alpha}{\gamma} - \frac{\alpha}{\gamma} \times 1 = 0$

Row 3, element 2: $\frac{\beta}{\gamma} - \frac{\alpha}{\gamma} \times \frac{\beta}{\alpha} = \frac{\beta}{\gamma} - \frac{\beta}{\gamma} = 0$

Row 3, element 3: $1 - \frac{\alpha}{\gamma} \times \frac{\gamma}{\alpha} = 1 - 1 = 0$

The new third row is $(0, 0, 0)$.

The determinant becomes:

$\Delta = \begin{vmatrix} 1 & \beta/\alpha & \gamma/\alpha \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{vmatrix}$

A determinant with a row of all zeros is equal to zero.

Thus, $\Delta = 0$.