The system of equations `x+py=0, y+pz=0 and z+px=0` has infinitely many solutions for

Options

A. `p=1`
B. `p=0`
C. `p=-1`
D. no real value of p

Answer

To determine the value of \( p \) for which the system of equations

\[ \begin{align*} 1. & \quad x + py = 0 \\ 2. & \quad y + pz = 0 \\ 3. & \quad z + px = 0 \end{align*} \]

has infinitely many solutions, we can use the concept of determinants.

### Step 1: Write the system in matrix form
The given equations can be represented in matrix form as:

\[ \begin{bmatrix} 1 & p & 0 \\ 0 & 1 & p \\ p & 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

### Step 2: Find the determinant of the coefficient matrix
Let \( D \) be the determinant of the coefficient matrix:

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

### Step 3: Calculate the determinant
Using the determinant formula for a 3x3 matrix, we have:

\[ D = 1 \cdot \begin{vmatrix} 1 & p \\ 0 & 1 \end{vmatrix} - p \cdot \begin{vmatrix} 0 & p \\ p & 1 \end{vmatrix} + 0 \cdot \begin{vmatrix} 0 & 1 \\ p & 0 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 1 & p \\ 0 & 1 \end{vmatrix} = 1 \cdot 1 - p \cdot 0 = 1 \)
2. \( \begin{vmatrix} 0 & p \\ p & 1 \end{vmatrix} = 0 \cdot 1 - p \cdot p = -p^2 \)

Now substituting back into the determinant \( D \):

\[ D = 1 \cdot 1 - p \cdot (-p^2) = 1 + p^3 \]

### Step 4: Set the determinant to zero
For the system to have infinitely many solutions, we need:

\[ D = 0 \]

Thus, we set:

\[ 1 + p^3 = 0 \]

### Step 5: Solve for \( p \)
Rearranging gives:

\[ p^3 = -1 \]

Taking the cube root of both sides, we find:

\[ p = -1 \]

### Conclusion
The value of \( p \) for which the system of equations has infinitely many solutions is:

\[ \boxed{-1} \]