For positive numbers x, y and z, the numerical value of the determinant` |[1,log_xy,log_xz],[log_yx,1,log_yz],[log_zx,log_zy,1]|`is

Options

A. 0
B. log x log y log z
C. 1
D. 8

Answer

To find the numerical value of the determinant

\[ D = \begin{vmatrix} 1 & \log_x y & \log_x z \\ \log_y x & 1 & \log_y z \\ \log_z x & \log_z y & 1 \end{vmatrix} \]

we will use properties of logarithms and determinants.

### Step-by-step Solution:

**Step 1: Rewrite the logarithms using the change of base formula.**

Using the property of logarithms, we can rewrite the logarithms in terms of natural logarithms (or any common base):

\[ \log_x y = \frac{\log y}{\log x}, \quad \log_x z = \frac{\log z}{\log x}, \quad \log_y z = \frac{\log z}{\log y}, \quad \log_z x = \frac{\log x}{\log z}, \quad \log_z y = \frac{\log y}{\log z} \]

Substituting these into the determinant, we get:

\[ D = \begin{vmatrix} 1 & \frac{\log y}{\log x} & \frac{\log z}{\log x} \\ \frac{\log x}{\log y} & 1 & \frac{\log z}{\log y} \\ \frac{\log x}{\log z} & \frac{\log y}{\log z} & 1 \end{vmatrix} \]

**Step 2: Calculate the determinant using cofactor expansion.**

We can expand the determinant along the first row:

\[ D = 1 \cdot \begin{vmatrix} 1 & \frac{\log z}{\log y} \\ \frac{\log y}{\log z} & 1 \end{vmatrix} - \frac{\log y}{\log x} \cdot \begin{vmatrix} \frac{\log x}{\log y} & \frac{\log z}{\log y} \\ \frac{\log x}{\log z} & 1 \end{vmatrix} + \frac{\log z}{\log x} \cdot \begin{vmatrix} \frac{\log x}{\log y} & 1 \\ \frac{\log x}{\log z} & \frac{\log y}{\log z} \end{vmatrix} \]

**Step 3: Calculate the 2x2 determinants.**

1. First determinant:

\[ \begin{vmatrix} 1 & \frac{\log z}{\log y} \\ \frac{\log y}{\log z} & 1 \end{vmatrix} = 1 \cdot 1 - \frac{\log z}{\log y} \cdot \frac{\log y}{\log z} = 1 - 1 = 0 \]

2. Second determinant:

\[ \begin{vmatrix} \frac{\log x}{\log y} & \frac{\log z}{\log y} \\ \frac{\log x}{\log z} & 1 \end{vmatrix} = \frac{\log x}{\log y} \cdot 1 - \frac{\log z}{\log y} \cdot \frac{\log x}{\log z} = \frac{\log x}{\log y} - \frac{\log x}{\log y} = 0 \]

3. Third determinant:

\[ \begin{vmatrix} \frac{\log x}{\log y} & 1 \\ \frac{\log x}{\log z} & \frac{\log y}{\log z} \end{vmatrix} = \frac{\log x}{\log y} \cdot \frac{\log y}{\log z} - 1 \cdot \frac{\log x}{\log z} = \frac{\log x}{\log z} - \frac{\log x}{\log z} = 0 \]

**Step 4: Substitute back into the determinant expression.**

Now substituting back into our determinant expression:

\[ D = 1 \cdot 0 - \frac{\log y}{\log x} \cdot 0 + \frac{\log z}{\log x} \cdot 0 = 0 \]

Thus, the numerical value of the determinant is

\[ \boxed{0} \]