The roots of the equation `|(1,t-1,1),(t-1,1,1),(1,1,t-1)| = 0` are

Options

A. A) 1, 2
B. B) -1, 2
C. C) 1, -2
D. D) -1, -2

Answer

To solve the equation given by the determinant \(|(1,t-1,1),(t-1,1,1),(1,1,t-1)| = 0\), we will follow these steps:

### Step 1: Write the Determinant
We start with the determinant:
\[ D = \begin{vmatrix} 1 & t-1 & 1 \\ t-1 & 1 & 1 \\ 1 & 1 & t-1 \end{vmatrix} \]

### Step 2: Expand the Determinant
We will expand the determinant using the first row:
\[ D = 1 \cdot \begin{vmatrix} 1 & 1 \\ 1 & t-1 \end{vmatrix} - (t-1) \cdot \begin{vmatrix} t-1 & 1 \\ 1 & t-1 \end{vmatrix} + 1 \cdot \begin{vmatrix} t-1 & 1 \\ 1 & 1 \end{vmatrix} \]

### Step 3: Calculate the 2x2 Determinants
Now we calculate the 2x2 determinants:
1. \(\begin{vmatrix} 1 & 1 \\ 1 & t-1 \end{vmatrix} = 1(t-1) - 1(1) = t-1 - 1 = t - 2\)
2. \(\begin{vmatrix} t-1 & 1 \\ 1 & t-1 \end{vmatrix} = (t-1)(t-1) - 1(1) = (t-1)^2 - 1 = t^2 - 2t + 1 - 1 = t^2 - 2t\)
3. \(\begin{vmatrix} t-1 & 1 \\ 1 & 1 \end{vmatrix} = (t-1)(1) - 1(1) = t - 1 - 1 = t - 2\)

### Step 4: Substitute Back into the Determinant
Now substituting back into the determinant:
\[ D = 1(t-2) - (t-1)(t^2 - 2t) + 1(t-2) \]
\[ D = (t-2) - (t-1)(t^2 - 2t) + (t-2) \]
\[ D = 2(t-2) - (t-1)(t^2 - 2t) \]

### Step 5: Simplify the Expression
Now we simplify:
\[ D = 2(t-2) - (t^3 - 2t^2 - t^2 + 2t) = 2(t-2) - (t^3 - 3t^2 + 2t) \]
\[ D = 2t - 4 - t^3 + 3t^2 - 2t = -t^3 + 3t^2 - 4 \]

### Step 6: Set the Determinant to Zero
Now we set the determinant equal to zero:
\[ -t^3 + 3t^2 - 4 = 0 \]
Multiplying through by -1 gives:
\[ t^3 - 3t^2 + 4 = 0 \]

### Step 7: Factor the Polynomial
We can factor this polynomial. We can check for rational roots using the Rational Root Theorem. Testing \(t = 2\):
\[ 2^3 - 3(2^2) + 4 = 8 - 12 + 4 = 0 \]
So \(t = 2\) is a root. We can factor out \(t - 2\):
\[ t^3 - 3t^2 + 4 = (t - 2)(t^2 - t - 2) \]

### Step 8: Factor the Quadratic
Now we factor the quadratic:
\[ t^2 - t - 2 = (t - 2)(t + 1) \]
Thus, we have:
\[ t^3 - 3t^2 + 4 = (t - 2)^2(t + 1) \]

### Step 9: Find the Roots
Setting each factor to zero gives:
1. \(t - 2 = 0 \Rightarrow t = 2\) (with multiplicity 2)
2. \(t + 1 = 0 \Rightarrow t = -1\)

### Final Answer
The roots of the equation are \(t = 2\) (with multiplicity 2) and \(t = -1\).

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