Question

Let $A$ is a matrix of order $3\times 3$ defined as $A={\left[a_{ij}\right]}_{3\times 3}$ , where $a_{ij}=\underset{x\to 0}{\lim }\frac{1-cos\left(ix\right)}{sin\left(ix\right)tan\left(jx\right)}\left(\forall 1\le i,j\le 3\right)$ , 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,
$a_{ij}=\frac{1}{2}\frac{i}{j}$
$A=\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{4}& \frac{1}{6}\\ 1& \frac{1}{2}& \frac{1}{3}\\ \frac{3}{2}& \frac{3}{4}& \frac{1}{2}\end{array}\right]=\frac{1}{2}\left[\begin{array}{ccc}1& \frac{1}{2}& \frac{1}{3}\\ 2& 1& \frac{2}{3}\\ 3& \frac{3}{2}& 1\end{array}\right]$
$A^2=\frac{1}{4}\left[\begin{array}{ccc}1& \frac{1}{2}& \frac{1}{3}\\ 2& 1& \frac{2}{3}\\ 3& \frac{3}{2}& 1\end{array}\right]\left[\begin{array}{ccc}1& \frac{1}{2}& \frac{1}{3}\\ 2& 1& \frac{2}{3}\\ 3& \frac{3}{2}& 1\end{array}\right]$
$=\frac{1}{4}\left[\begin{array}{ccc}3& \frac{3}{2}& \frac{3}{3}\\ 6& 3& \frac{6}{3}\\ 9& \frac{9}{2}& 3\end{array}\right]=\frac{3}{2}A$