Question

Let $A$ is a matrix of order $3\times 3$ defined as $A=\left[a_{i j}\right]_{3 \times 3}$ , where $a_{i j}=\underset{x \rightarrow 0}{l i m}\frac{1 - c o s \left(i x\right)}{s i n \left(i x\right) t a n \left(j x\right)}\left(\forall 1 \leq i , j \leq 3\right)$ , then $A^{2}$ is equal to

• A. $A$
• B. $\frac{3}{2}A$
• C. $\frac{2}{3}A$
• D. $\frac{1}{4}A$

Answer 1

Answer: (B)

Applying L' Hospital rule, we get,
$a_{i j}=\frac{1}{2}\frac{i}{j}$
$A=\begin{bmatrix} \frac{1}{2} & \frac{1}{4} & \frac{1}{6} \\ 1 & \frac{1}{2} & \frac{1}{3} \\ \frac{3}{2} & \frac{3}{4} & \frac{1}{2} \end{bmatrix}=\frac{1}{2}\begin{bmatrix} 1 & \frac{1}{2} & \frac{1}{3} \\ 2 & 1 & \frac{2}{3} \\ 3 & \frac{3}{2} & 1 \end{bmatrix}$
$A^{2}=\frac{1}{4}\begin{bmatrix} 1 & \frac{1}{2} & \frac{1}{3} \\ 2 & 1 & \frac{2}{3} \\ 3 & \frac{3}{2} & 1 \end{bmatrix}\begin{bmatrix} 1 & \frac{1}{2} & \frac{1}{3} \\ 2 & 1 & \frac{2}{3} \\ 3 & \frac{3}{2} & 1 \end{bmatrix}$
$=\frac{1}{4}\begin{bmatrix} 3 & \frac{3}{2} & \frac{3}{3} \\ 6 & 3 & \frac{6}{3} \\ 9 & \frac{9}{2} & 3 \end{bmatrix}=\frac{3}{2}A$