`|(1,log_xy,log_xz),(log_yx,1,log_yz),(log_zx,log_zy,1)|x,y,z` being +ive

Options

A. `log_yx`
B. `log_zy`
C. `log_(x)z`
D. 0

Answer

To solve 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 the properties of logarithms and determinants.

### Step 1: Rewrite the logarithms using the change of base formula
We know that

\[ \log_a b = \frac{\log b}{\log a} \]

Using this property, we can rewrite the logarithms in the determinant:

- \(\log_x y = \frac{\log y}{\log x}\)
- \(\log_x z = \frac{\log z}{\log x}\)
- \(\log_y x = \frac{\log x}{\log y}\)
- \(\log_y z = \frac{\log z}{\log y}\)
- \(\log_z x = \frac{\log x}{\log z}\)
- \(\log_z y = \frac{\log y}{\log z}\)

Thus, we can express the determinant as:

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

### Step 2: Multiply each row by the corresponding logarithm
Now, we will multiply the first row by \(\log x\), the second row by \(\log y\), and the third row by \(\log z\):

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

### Step 3: Factor out the common logarithms
Now we can factor out \(\log x\), \(\log y\), and \(\log z\) from each row:

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

### Step 4: Evaluate the determinant of the identity matrix
The determinant of a matrix where all rows are identical is zero:

\[ \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix} = 0 \]

### Step 5: Conclude the value of the determinant
Thus, we have:

\[ D = \log x \cdot \log y \cdot \log z \cdot 0 = 0 \]

So, the value of the determinant is:

\[ \boxed{0} \]