Let `A=a_(ij)` be a matrix of order 3, where `a_(ij)={{:(x,,if i=j","x in R),(1,,if |i-j|=1),(0,,"otherwise"):}` then which of the following hold good :

Options

A. for x=2, A is a diagonal matrix
B. A is a symmetric matrix
C. for x=2, detA has the value equal to 6
D. Let f(x)=detA , then the function f(x) has both the maxima and minima

Answer

To solve the problem, we need to construct the matrix \( A \) based on the given conditions and analyze its properties.

### Step 1: Construct the Matrix \( A \)

Given the conditions:
- \( a_{ij} = x \) if \( i = j \)
- \( a_{ij} = 1 \) if \( |i - j| = 1 \)
- \( a_{ij} = 0 \) otherwise

For a \( 3 \times 3 \) matrix, we can write:

\[ A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \]

Now, substituting the values based on the conditions:

- \( a_{11} = x \) (since \( i = j \))
- \( a_{12} = 1 \) (since \( |1 - 2| = 1 \))
- \( a_{13} = 0 \) (since \( |1 - 3| \neq 1 \))
- \( a_{21} = 1 \) (since \( |2 - 1| = 1 \))
- \( a_{22} = x \) (since \( i = j \))
- \( a_{23} = 1 \) (since \( |2 - 3| = 1 \))
- \( a_{31} = 0 \) (since \( |3 - 1| \neq 1 \))
- \( a_{32} = 1 \) (since \( |3 - 2| = 1 \))
- \( a_{33} = x \) (since \( i = j \))

Thus, the matrix \( A \) becomes:

\[ A = \begin{bmatrix} x & 1 & 0 \\ 1 & x & 1 \\ 0 & 1 & x \end{bmatrix} \]

### Step 2: Check if \( A \) is Symmetric

A matrix is symmetric if \( A = A^T \). The transpose of \( A \) is:

\[ A^T = \begin{bmatrix} x & 1 & 0 \\ 1 & x & 1 \\ 0 & 1 & x \end{bmatrix} \]

Since \( A = A^T \), the matrix \( A \) is symmetric.

### Step 3: Check if \( A \) is a Diagonal Matrix

A diagonal matrix has non-zero elements only on the diagonal. In our case, \( A \) has off-diagonal elements (1's), hence \( A \) cannot be a diagonal matrix.

### Step 4: Calculate the Determinant of \( A \)

To find the determinant of \( A \):

\[ \text{det}(A) = \begin{vmatrix} x & 1 & 0 \\ 1 & x & 1 \\ 0 & 1 & x \end{vmatrix} \]

Using the determinant formula for a \( 3 \times 3 \) matrix:

\[ \text{det}(A) = x \begin{vmatrix} x & 1 \\ 1 & x \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 0 & x \end{vmatrix} + 0 \]

Calculating the minors:

\[ \begin{vmatrix} x & 1 \\ 1 & x \end{vmatrix} = x^2 - 1 \]

\[ \begin{vmatrix} 1 & 1 \\ 0 & x \end{vmatrix} = x \]

Thus, we have:

\[ \text{det}(A) = x(x^2 - 1) - x = x^3 - x - x = x^3 - 2x \]

### Step 5: Analyze the Determinant

The determinant \( \text{det}(A) = x^3 - 2x \) can be factored as:

\[ \text{det}(A) = x(x^2 - 2) \]

Setting this equal to zero gives us the critical points:

\[ x = 0 \quad \text{or} \quad x^2 - 2 = 0 \quad \Rightarrow \quad x = \pm \sqrt{2} \]

### Conclusion

From the analysis:
1. \( A \) is symmetric.
2. \( A \) cannot be a diagonal matrix.
3. The determinant is \( x^3 - 2x \).