The system of homogenous equations
`tx+(t+1)y+(t-1)z=0`, `(t+1)x+ty+(t+2)z=0`, `(t-1)x+(t+2)y+tz=0` has a non trivial solution for

Options

A. exactly three real values of `t`
B. exactly two real values of `t`
C. exactly one real values of `t`
D. infinite number of values of `t`

Explanation

To determine the values of \( t \) for which the system of homogeneous equations has a non-trivial solution, we need to set the determinant of the coefficient matrix to zero. The given equations are:

1. \( tx + (t + 1)y + (t - 1)z = 0 \)
2. \( (t + 1)x + ty + (t + 2)z = 0 \)
3. \( (t - 1)x + (t + 2)y + tz = 0 \)

### Step 1: Write the Coefficient Matrix
The coefficient matrix \( A \) for the system can be written as:

\[ A = \begin{bmatrix} t & t + 1 & t - 1 \\ t + 1 & t & t + 2 \\ t - 1 & t + 2 & t \end{bmatrix} \]

### Step 2: Calculate the Determinant
To find the values of \( t \) for which the determinant of \( A \) is zero, we compute \( \det(A) \).

\[ \det(A) = \begin{vmatrix} t & t + 1 & t - 1 \\ t + 1 & t & t + 2 \\ t - 1 & t + 2 & t \end{vmatrix} \]

### Step 3: Expand the Determinant
We can expand the determinant using cofactor expansion along the first row:

\[ \det(A) = t \begin{vmatrix} t & t + 2 \\ t + 2 & t \end{vmatrix} - (t + 1) \begin{vmatrix} t + 1 & t + 2 \\ t - 1 & t \end{vmatrix} + (t - 1) \begin{vmatrix} t + 1 & t \\ t - 1 & t + 2 \end{vmatrix} \]

### Step 4: Calculate the 2x2 Determinants
Calculating the 2x2 determinants:

1. \( \begin{vmatrix} t & t + 2 \\ t + 2 & t \end{vmatrix} = t^2 - (t + 2)(t + 2) = t^2 - (t^2 + 4t + 4) = -4t - 4 \)

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

3. \( \begin{vmatrix} t + 1 & t \\ t - 1 & t + 2 \end{vmatrix} = (t + 1)(t + 2) - t(t - 1) = (t^2 + 3t + 2) - (t^2 - t) = 4t + 2 \)

### Step 5: Substitute Back into the Determinant
Substituting these back into the determinant expression:

\[ \det(A) = t(-4t - 4) - (t + 1)(2t - 1) + (t - 1)(4t + 2) \]

### Step 6: Simplify the Expression
Now we simplify:

\[ = -4t^2 - 4t - (2t^2 + t - 2t - 1) + (4t^2 + 2t - 4t - 2) \]
\[ = -4t^2 - 4t - 2t^2 + 1 + 4t^2 - 2 \]
\[ = -2t^2 - 4t - 1 \]

### Step 7: Set the Determinant to Zero
Setting the determinant equal to zero for a non-trivial solution:

\[ -2t^2 - 4t - 1 = 0 \]

### Step 8: Solve the Quadratic Equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = -2, b = -4, c = -1 \):

\[ t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(-2)(-1)}}{2(-2)} \]
\[ = \frac{4 \pm \sqrt{16 - 8}}{-4} \]
\[ = \frac{4 \pm \sqrt{8}}{-4} \]
\[ = \frac{4 \pm 2\sqrt{2}}{-4} \]
\[ = -1 \mp \frac{\sqrt{2}}{2} \]

### Conclusion
Thus, there are exactly two real values of \( t \) for which the system has a non-trivial solution.

### Final Answer
The correct option is **exactly two real values of \( t \)**.

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