Let `A=[a_(ij)]_(3xx3)` be a matrix, where `a_(ij)={{:(x,inej),(1,i=j):} Aai, j in N & i,jle2.` If `C_(ij)` be the cofactor of `a_(ij) and C_(12)+C_(23)+C_(32)=6`, then the number of value(s) of `x(AA x in R)` is (are)

Answer

To solve the problem, we need to analyze the given matrix \( A \) and its cofactors. The matrix \( A \) is defined as follows:

\[ A = \begin{bmatrix} 1 & x & x \\ x & 1 & x \\ x & x & 1 \end{bmatrix} \]

### Step 1: Calculate the cofactors \( C_{12}, C_{23}, C_{32} \)

1. **Cofactor \( C_{12} \)**:
- The cofactor \( C_{12} \) is calculated by removing the first row and second column from matrix \( A \):
\[ C_{12} = (-1)^{1+2} \det\begin{bmatrix} x & x \\ x & 1 \end{bmatrix} = -\det\begin{bmatrix} x & x \\ x & 1 \end{bmatrix} \]
- The determinant is calculated as:
\[ \det\begin{bmatrix} x & x \\ x & 1 \end{bmatrix} = x \cdot 1 - x \cdot x = x - x^2 \]
- Therefore,
\[ C_{12} = -(x - x^2) = x^2 - x \]

2. **Cofactor \( C_{23} \)**:
- The cofactor \( C_{23} \) is calculated by removing the second row and third column:
\[ C_{23} = (-1)^{2+3} \det\begin{bmatrix} 1 & x \\ x & x \end{bmatrix} = \det\begin{bmatrix} 1 & x \\ x & x \end{bmatrix} \]
- The determinant is:
\[ \det\begin{bmatrix} 1 & x \\ x & x \end{bmatrix} = 1 \cdot x - x \cdot x = x - x^2 \]
- Therefore,
\[ C_{23} = x - x^2 \]

3. **Cofactor \( C_{32} \)**:
- The cofactor \( C_{32} \) is calculated by removing the third row and second column:
\[ C_{32} = (-1)^{3+2} \det\begin{bmatrix} 1 & x \\ x & x \end{bmatrix} = -\det\begin{bmatrix} 1 & x \\ x & x \end{bmatrix} \]
- The determinant is the same as above:
\[ C_{32} = -(x - x^2) = x^2 - x \]

### Step 2: Sum of the cofactors

Now we sum the cofactors:
\[ C_{12} + C_{23} + C_{32} = (x^2 - x) + (x - x^2) + (x^2 - x) \]
This simplifies to:
\[ C_{12} + C_{23} + C_{32} = x^2 - x + x - x^2 + x^2 - x = x^2 - x \]

### Step 3: Set the equation to 6

According to the problem, we have:
\[ C_{12} + C_{23} + C_{32} = 6 \]
Thus, we set up the equation:
\[ x^2 - x = 6 \]
Rearranging gives:
\[ x^2 - x - 6 = 0 \]

### Step 4: Solve the quadratic equation

We can factor this equation:
\[ (x - 3)(x + 2) = 0 \]
Thus, the solutions are:
\[ x = 3 \quad \text{or} \quad x = -2 \]

### Conclusion

The number of values of \( x \) is 2.