Show that in general there are three values of t for which the following system of equations has a non- trival solution (a-t)x+by+cz=0, bx+(c-t)y+az=0 and cx+ay+(b-t)z=0.
Express the product of these values of t in the form of a determinant.

Answer

To solve the problem, we need to show that the system of equations has three values of \( t \) for which it has a non-trivial solution and express the product of these values in the form of a determinant.

### Step-by-Step Solution:

1. **Write the System of Equations in Matrix Form**:
The given system of equations is:
\[ (a - t)x + by + cz = 0 \]
\[ bx + (c - t)y + az = 0 \]
\[ cx + ay + (b - t)z = 0 \]
We can express this system in matrix form as:
\[ \begin{bmatrix} a - t & b & c \\ b & c - t & a \\ c & a & b - t \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

2. **Condition for Non-Trivial Solutions**:
For the system to have a non-trivial solution (i.e., not all \( x, y, z \) are zero), the determinant of the coefficient matrix must be zero:
\[ \det \begin{bmatrix} a - t & b & c \\ b & c - t & a \\ c & a & b - t \end{bmatrix} = 0 \]

3. **Calculate the Determinant**:
We need to compute the determinant:
\[ D(t) = \det \begin{bmatrix} a - t & b & c \\ b & c - t & a \\ c & a & b - t \end{bmatrix} \]
Using the determinant formula for a \( 3 \times 3 \) matrix, we can expand this determinant.

4. **Expand the Determinant**:
Expanding the determinant, we have:
\[ D(t) = (a - t)((c - t)(b - t) - a^2) - b(b(c - t) - ac) + c(b(a - t) - bc) \]
This will yield a polynomial in \( t \) of degree 3.

5. **Form of the Cubic Equation**:
The determinant \( D(t) \) can be expressed in the form:
\[ D(t) = \alpha t^3 + \beta t^2 + \gamma t + \delta \]
where \( \alpha, \beta, \gamma, \delta \) are constants dependent on \( a, b, c \).

6. **Roots of the Cubic Equation**:
The cubic equation will have three roots (values of \( t \)). The product of the roots of a cubic equation \( at^3 + bt^2 + ct + d = 0 \) is given by:
\[ \text{Product of roots} = -\frac{d}{a} \]
In our case, we need to express this product in terms of the determinant.

7. **Substituting \( t = 0 \)**:
To find \( \delta \) (the constant term), we substitute \( t = 0 \) into the determinant:
\[ D(0) = \det \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix} \]

8. **Final Expression**:
Therefore, the product of the values of \( t \) for which the system has a non-trivial solution can be expressed as:
\[ \text{Product of roots} = -\frac{D(0)}{\alpha} \]
where \( D(0) \) is the determinant calculated above.