If all elements of a third order determinant are equal to 1 or `-1`. then determinant itself is (i) An Odd Integer (ii) An Even Integer (iii) An Imaginary Number (iv) multiple of 3

Options

A. An odd integer
B. An even iterger
C. An imaginary number
D. A real number

Answer

To solve the problem, we need to analyze the determinant of a 3x3 matrix where all elements are either 1 or -1. Let's denote the determinant as \( D \).

### Step 1: Define the Determinant
Let \( D \) be the determinant of a 3x3 matrix \( A \) where each element \( a_{ij} \) is either 1 or -1. The general form of the determinant can be expressed as:

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

### Step 2: Calculate the Determinant
We can compute the determinant using the formula for a 3x3 matrix:

\[ D = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31}) \]

### Step 3: Analyze the Elements
Since each \( a_{ij} \) can be either 1 or -1, the possible combinations of the elements will lead to various sums and products.

### Step 4: Consider Different Cases
Let's consider a few specific cases:

1. **All elements are 1:**
\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix} = 0 \]

2. **Two elements are 1 and one is -1:**
\[ D = \begin{vmatrix} 1 & 1 & -1 \\ 1 & -1 & 1 \\ -1 & 1 & 1 \end{vmatrix} \]
After calculating, we find \( D = 4 \).

3. **One element is 1 and two are -1:**
\[ D = \begin{vmatrix} 1 & -1 & -1 \\ -1 & 1 & -1 \\ -1 & -1 & 1 \end{vmatrix} \]
After calculating, we find \( D = -4 \).

### Step 5: General Conclusion
From the calculations, we observe that the determinant \( D \) can take values like 0, 4, or -4, which are all even integers.

### Final Answer
Thus, the determinant itself is an even integer.