If xyz=m and det `p=|{:(x,y,z),(z,x,y),(y,z,x):}|` , where p is an orthogonal matrix.
The value of `x^(-1)+y^(-1)+z^(-1)` is

Options

A. `pm""m`
B. `pm1`
C. 0
D. `m^2`

Answer

To solve the problem step by step, we will analyze the given determinant and the properties of the orthogonal matrix.

### Step 1: Understand the determinant
We are given the determinant \( p = \begin{vmatrix} x & y & z \\ z & x & y \\ y & z & x \end{vmatrix} \). We need to evaluate this determinant.

### Step 2: Calculate the determinant
Using the determinant formula for a 3x3 matrix, we have:
\[ \text{det}(p) = x \begin{vmatrix} x & y \\ z & x \end{vmatrix} - y \begin{vmatrix} z & y \\ y & x \end{vmatrix} + z \begin{vmatrix} z & x \\ y & z \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} x & y \\ z & x \end{vmatrix} = x^2 - yz \)
2. \( \begin{vmatrix} z & y \\ y & x \end{vmatrix} = zx - y^2 \)
3. \( \begin{vmatrix} z & x \\ y & z \end{vmatrix} = z^2 - xy \)

Substituting these back into the determinant:
\[ \text{det}(p) = x(x^2 - yz) - y(zx - y^2) + z(z^2 - xy) \]
\[ = x^3 - xyz - (yzx - y^3) + z^3 - xyz \]
\[ = x^3 + y^3 + z^3 - 3xyz \]

### Step 3: Use the orthogonal matrix property
Since \( p \) is an orthogonal matrix, we know that \( \text{det}(p) = \pm 1 \). Therefore, we have:
\[ x^3 + y^3 + z^3 - 3xyz = \pm 1 \]

### Step 4: Express \( x^3 + y^3 + z^3 \)
Using the identity for the sum of cubes, we can express \( x^3 + y^3 + z^3 \) in terms of \( xyz \):
\[ x^3 + y^3 + z^3 = (x+y+z)(x^2 + y^2 + z^2 - xy - xz - yz) + 3xyz \]
Let \( s_1 = x+y+z \), \( s_2 = xy + xz + yz \), and \( s_3 = xyz \). We can rewrite:
\[ x^3 + y^3 + z^3 = s_1(s_1^2 - 3s_2) + 3s_3 \]

### Step 5: Set up the equation
From the orthogonal condition, we have:
\[ s_1(s_1^2 - 3s_2) + 3s_3 - 3s_3 = \pm 1 \]
This simplifies to:
\[ s_1(s_1^2 - 3s_2) = \pm 1 \]

### Step 6: Find \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \)
We need to find \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \):
\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{xy + xz + yz}{xyz} = \frac{s_2}{s_3} \]
Given \( xyz = m \) and from our earlier calculations, we found \( s_2 = 0 \). Thus:
\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{0}{m} = 0 \]

### Final Answer
The value of \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \) is \( 0 \).