Consider the system of equations `ax+y+bz=0, bx+y+az=0 and ax+by+abz=0` where `a, b in {0, 1, 2, 3, 4}`. The number of ordered pairs (a, b) for which the system has non - trivial solutions is

Answer

To determine the number of ordered pairs \((a, b)\) for which the given system of equations has non-trivial solutions, we start by analyzing the system of equations:

1. \( ax + y + bz = 0 \)
2. \( bx + y + az = 0 \)
3. \( ax + by + abz = 0 \)

### Step 1: Write the system in matrix form
We can represent the system of equations in matrix form as follows:

\[ \begin{bmatrix} a & 1 & b \\ b & 1 & a \\ a & b & ab \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

### Step 2: Find the determinant of the coefficient matrix
For the system to have non-trivial solutions, the determinant of the coefficient matrix must be zero. We need to compute the determinant:

\[ \Delta = \begin{vmatrix} a & 1 & b \\ b & 1 & a \\ a & b & ab \end{vmatrix} \]

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

\[ \Delta = a \begin{vmatrix} 1 & a \\ b & ab \end{vmatrix} - 1 \begin{vmatrix} b & a \\ a & ab \end{vmatrix} + b \begin{vmatrix} b & 1 \\ a & b \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \(\begin{vmatrix} 1 & a \\ b & ab \end{vmatrix} = 1 \cdot ab - a \cdot b = ab - ab = 0\)

2. \(\begin{vmatrix} b & a \\ a & ab \end{vmatrix} = b \cdot ab - a \cdot a = ab^2 - a^2\)

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

Substituting these back into the determinant:

\[ \Delta = a(0) - 1(ab^2 - a^2) + b(b^2 - a) \]
\[ = -ab^2 + a^2 + b^3 - ab \]
\[ = a^2 - ab^2 + b^3 - ab \]

### Step 4: Set the determinant to zero
We want to find when \(\Delta = 0\):

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

### Step 5: Factor the equation
Rearranging gives:

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

Factoring out common terms, we can rewrite it as:

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

This gives us two cases:

1. \(a = 0\)
2. \(a = b^2\)

### Step 6: Analyze the cases
- **Case 1:** \(a = 0\)

If \(a = 0\), then \(b\) can take any value from \{0, 1, 2, 3, 4\}. This gives us the pairs:
\((0, 0), (0, 1), (0, 2), (0, 3), (0, 4)\) → 5 pairs.

- **Case 2:** \(a = b^2\)

Now we check for values of \(b\) from \{0, 1, 2, 3, 4\}:
- If \(b = 0\), \(a = 0^2 = 0\) → (0, 0)
- If \(b = 1\), \(a = 1^2 = 1\) → (1, 1)
- If \(b = 2\), \(a = 2^2 = 4\) → (4, 2)
- If \(b = 3\), \(a = 3^2 = 9\) → Not valid since \(a\) must be ≤ 4.
- If \(b = 4\), \(a = 4^2 = 16\) → Not valid since \(a\) must be ≤ 4.

From this case, we have the pairs:
\((0, 0), (1, 1), (4, 2)\) → 3 additional pairs.

### Step 7: Count the total pairs
Combining both cases, we have:
- From Case 1: 5 pairs
- From Case 2: 3 pairs

Total ordered pairs = \(5 + 3 = 8\).

### Final Answer
The number of ordered pairs \((a, b)\) for which the system has non-trivial solutions is **8**.