The roots of the equation `|{:(x-1,1,1),(1,x-1,1),(1,1,x-1):}|=0` are

Options

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

Answer

To solve the equation given by the determinant \( |{:(x-1,1,1),(1,x-1,1),(1,1,x-1):}|=0 \), we will follow a step-by-step approach.

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

### Step 2: Simplify the determinant
We can simplify the determinant using row operations. Let's add the second row to the third row:
\[ R_2 + R_3 \rightarrow R_3 \]
This gives us:
\[ D = \begin{vmatrix} x-1 & 1 & 1 \\ 1 & x-1 & 1 \\ 2 & 2 & x-2 \end{vmatrix} \]

### Step 3: Expand the determinant
Now, we will expand the determinant along the first row:
\[ D = (x-1) \begin{vmatrix} x-1 & 1 \\ 2 & x-2 \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 2 & x-2 \end{vmatrix} + 1 \begin{vmatrix} 1 & x-1 \\ 2 & 2 \end{vmatrix} \]

### Step 4: Calculate the 2x2 determinants
Calculating the first 2x2 determinant:
\[ \begin{vmatrix} x-1 & 1 \\ 2 & x-2 \end{vmatrix} = (x-1)(x-2) - 2 = x^2 - 3x + 2 \]

Calculating the second 2x2 determinant:
\[ \begin{vmatrix} 1 & 1 \\ 2 & x-2 \end{vmatrix} = 1(x-2) - 2 = x - 4 \]

Calculating the third 2x2 determinant:
\[ \begin{vmatrix} 1 & x-1 \\ 2 & 2 \end{vmatrix} = 1 \cdot 2 - 2(x-1) = 2 - 2x + 2 = 4 - 2x \]

### Step 5: Substitute back into the determinant
Substituting these values back into the determinant expression:
\[ D = (x-1)(x^2 - 3x + 2) - (x - 4) + (4 - 2x) \]
This simplifies to:
\[ D = (x-1)(x^2 - 3x + 2) - x + 4 + 4 - 2x \]
\[ D = (x-1)(x^2 - 3x + 2) - 3x + 8 \]

### Step 6: Expand and simplify
Now, we expand \( (x-1)(x^2 - 3x + 2) \):
\[ D = x^3 - 3x^2 + 2x - x^2 + 3x - 2 - 3x + 8 \]
\[ D = x^3 - 4x^2 + 6 \]

### Step 7: Set the determinant to zero
Setting the determinant equal to zero gives us:
\[ x^3 - 4x^2 + 6 = 0 \]

### Step 8: Find the roots
To find the roots of the cubic equation \( x^3 - 4x^2 + 6 = 0 \), we can use numerical methods or factorization. Trying possible rational roots, we find:
1. \( x = 2 \) is a root.
2. We can factor \( (x - 2) \) out to find the other roots.

Using synthetic division or polynomial long division, we can find:
\[ x^3 - 4x^2 + 6 = (x - 2)(x^2 - 2x - 3) \]
Factoring \( x^2 - 2x - 3 \) gives:
\[ (x - 3)(x + 1) \]

### Final Roots
Thus, the roots of the equation are:
\[ x = 2, \quad x = 3, \quad x = -1 \]

### Summary of Roots
The roots of the equation \( |{:(x-1,1,1),(1,x-1,1),(1,1,x-1):}|=0 \) are \( x = 2, -1, \text{ and } 3 \).

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