Let a, x, y, z be real numbers satisfying the equations
`ax+ay=z`
`x+ay=z`
`x+ay=az,` where x, y, z are not all zero, then the number of the possible values of a is

Answer

To solve the problem, we need to analyze the given equations and determine the possible values of \( a \). The equations provided are:

1. \( ax + ay = z \)
2. \( x + ay = z \)
3. \( x + ay = az \)

We can rewrite these equations in a standard form suitable for matrix representation.

### Step 1: Rewrite the equations
From the equations, we can express them in the form of a matrix equation \( A \mathbf{v} = \mathbf{0} \), where \( A \) is the coefficient matrix and \( \mathbf{v} \) is the vector of variables.

The equations can be rewritten as:
1. \( ax + ay - z = 0 \)
2. \( x + ay - z = 0 \)
3. \( x + ay - az = 0 \)

### Step 2: Form the coefficient matrix
The coefficient matrix \( A \) corresponding to the system of equations is:

\[ A = \begin{pmatrix} a & a & -1 \\ 1 & a & -1 \\ 1 & a & -a \end{pmatrix} \]

### Step 3: Set up the determinant
For the system to have non-trivial solutions (not all \( x, y, z \) are zero), the determinant of the coefficient matrix must be zero:

\[ \text{det}(A) = 0 \]

### Step 4: Calculate the determinant
We can calculate the determinant using the formula for a 3x3 matrix:

\[ \text{det}(A) = a \begin{vmatrix} a & -1 \\ a & -a \end{vmatrix} - a \begin{vmatrix} 1 & -1 \\ 1 & -a \end{vmatrix} - 1 \begin{vmatrix} 1 & a \\ 1 & a \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} a & -1 \\ a & -a \end{vmatrix} = a(-a) - (-1)(a) = -a^2 + a = a - a^2 \)

2. \( \begin{vmatrix} 1 & -1 \\ 1 & -a \end{vmatrix} = 1(-a) - (-1)(1) = -a + 1 = 1 - a \)

3. \( \begin{vmatrix} 1 & a \\ 1 & a \end{vmatrix} = 1(a) - 1(a) = 0 \)

Putting it all together:

\[ \text{det}(A) = a(a - a^2) - a(1 - a) - 0 \]
\[ = a^2 - a^3 - a + a^2 \]
\[ = 2a^2 - a^3 - a \]

### Step 5: Set the determinant to zero
Now we set the determinant to zero:

\[ -a^3 + 2a^2 - a = 0 \]

Factoring out \( -a \):

\[ -a(a^2 - 2a + 1) = 0 \]
\[ -a(a - 1)^2 = 0 \]

### Step 6: Solve for \( a \)
From this equation, we have:

1. \( -a = 0 \) → \( a = 0 \)
2. \( (a - 1)^2 = 0 \) → \( a = 1 \)

### Conclusion
The possible values of \( a \) are \( 0 \) and \( 1 \). Therefore, the number of possible values of \( a \) is \( 2 \).