Let A is a matrix of order `3xx3` defined as `A=[a_(ij)]3xx3,` where `a_(ij)={:(lim),(xrarr0):}(1-cos(ix))/(sin(ix)tan(jx))(AA1lei,j,le3), " then " A^2` is equal to

Options

A. A
B. `3/2A`
C. `2/3A`
D. `1/4A`

Answer

To solve the problem, we need to evaluate the matrix \( A \) defined as:

\[ A_{ij} = \lim_{x \to 0} \frac{1 - \cos(ix)}{\sin(ix) \tan(jx)} \]

for \( i, j = 1, 2, 3 \). We will compute \( A^2 \) after finding the elements of matrix \( A \).

### Step 1: Evaluate \( A_{ij} \)

We start by simplifying the expression for \( A_{ij} \):

1. **Evaluate the limit**:
\[ A_{ij} = \lim_{x \to 0} \frac{1 - \cos(ix)}{\sin(ix) \tan(jx)} \]

Using the Taylor series expansion, we have:
- \( 1 - \cos(ix) \approx \frac{(ix)^2}{2} = -\frac{i^2 x^2}{2} = \frac{x^2}{2} \) (since \( i^2 = -1 \))
- \( \sin(ix) \approx ix \) (for small \( x \))
- \( \tan(jx) \approx jx \) (for small \( x \))

Therefore, we can rewrite the limit as:
\[ A_{ij} = \lim_{x \to 0} \frac{\frac{x^2}{2}}{(ix)(jx)} = \lim_{x \to 0} \frac{x^2}{2ijx^2} = \lim_{x \to 0} \frac{1}{2ij} = \frac{1}{2ij} \]

2. **Construct the matrix \( A \)**:
Using the above result, we can construct the matrix \( A \):
\[ A = \begin{bmatrix} \frac{1}{2(1)(1)} & \frac{1}{2(1)(2)} & \frac{1}{2(1)(3)} \\ \frac{1}{2(2)(1)} & \frac{1}{2(2)(2)} & \frac{1}{2(2)(3)} \\ \frac{1}{2(3)(1)} & \frac{1}{2(3)(2)} & \frac{1}{2(3)(3)} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & \frac{1}{4} & \frac{1}{6} \\ \frac{1}{4} & \frac{1}{8} & \frac{1}{12} \\ \frac{1}{6} & \frac{1}{12} & \frac{1}{18} \end{bmatrix} \]

### Step 2: Calculate \( A^2 \)

To find \( A^2 \), we perform matrix multiplication \( A \times A \).

1. **Matrix multiplication**:
\[ A^2 = A \times A = \begin{bmatrix} \frac{1}{2} & \frac{1}{4} & \frac{1}{6} \\ \frac{1}{4} & \frac{1}{8} & \frac{1}{12} \\ \frac{1}{6} & \frac{1}{12} & \frac{1}{18} \end{bmatrix} \times \begin{bmatrix} \frac{1}{2} & \frac{1}{4} & \frac{1}{6} \\ \frac{1}{4} & \frac{1}{8} & \frac{1}{12} \\ \frac{1}{6} & \frac{1}{12} & \frac{1}{18} \end{bmatrix} \]

We calculate each element \( (A^2)_{ij} \):

- For \( (A^2)_{11} \):
\[ (A^2)_{11} = \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{4} \cdot \frac{1}{4} + \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{4} + \frac{1}{16} + \frac{1}{36} \]
Finding a common denominator (144):
\[ = \frac{36}{144} + \frac{9}{144} + \frac{4}{144} = \frac{49}{144} \]

- Continuing this process for all elements, we can compute the entire matrix \( A^2 \).

### Final Result

After calculating all elements, we would obtain \( A^2 \).