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 find the square of the matrix \( A \) defined as:

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

### Step 1: Simplify the Limit Expression

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

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

Using the known limits:

- \( \lim_{x \to 0} \frac{1 - \cos(x)}{x^2} = \frac{1}{2} \)
- \( \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \)
- \( \lim_{x \to 0} \frac{\tan(x)}{x} = 1 \)

We can rewrite the limit as follows:

\[ A_{ij} = \lim_{x \to 0} \frac{1 - \cos(ix)}{(ix)^2} \cdot \frac{(ix)^2}{\sin(ix) \tan(jx)} \]

### Step 2: Substitute the Limits

Substituting the limits, we have:

\[ A_{ij} = \frac{1}{2} \cdot \frac{i^2 x^2}{(ix)(jx)} = \frac{1}{2} \cdot \frac{i^2}{ij} = \frac{i}{2j} \]

### Step 3: Construct the Matrix \( A \)

Now we can construct the matrix \( A \):

\[ A = \begin{bmatrix} \frac{1}{2 \cdot 1} & \frac{1}{2 \cdot 2} & \frac{1}{2 \cdot 3} \\ \frac{2}{2 \cdot 1} & \frac{2}{2 \cdot 2} & \frac{2}{2 \cdot 3} \\ \frac{3}{2 \cdot 1} & \frac{3}{2 \cdot 2} & \frac{3}{2 \cdot 3} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & \frac{1}{4} & \frac{1}{6} \\ 1 & \frac{1}{2} & \frac{1}{3} \\ \frac{3}{2} & \frac{3}{4} & \frac{1}{2} \end{bmatrix} \]

### Step 4: Calculate \( A^2 \)

Now we will calculate \( A^2 = A \cdot A \):

\[ A^2 = \begin{bmatrix} \frac{1}{2} & \frac{1}{4} & \frac{1}{6} \\ 1 & \frac{1}{2} & \frac{1}{3} \\ \frac{3}{2} & \frac{3}{4} & \frac{1}{2} \end{bmatrix} \cdot \begin{bmatrix} \frac{1}{2} & \frac{1}{4} & \frac{1}{6} \\ 1 & \frac{1}{2} & \frac{1}{3} \\ \frac{3}{2} & \frac{3}{4} & \frac{1}{2} \end{bmatrix} \]

Calculating the elements of \( A^2 \):

- First row, first column:
\[ \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{4} \cdot 1 + \frac{1}{6} \cdot \frac{3}{2} = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4} \]

- First row, second column:
\[ \frac{1}{2} \cdot \frac{1}{4} + \frac{1}{4} \cdot \frac{1}{2} + \frac{1}{6} \cdot \frac{3}{4} = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{3}{8} \]

- Continuing this process for all elements, we find:

\[ A^2 = \frac{3}{2} A \]

### Final Result

Thus, the final result is:

\[ A^2 = \frac{3}{2} A \]