Solve this

`
Question:

If $A=\left[a_{i j}\right]$ is a square matrix of even order such that $a_{i j}=i^{2}-j^{2}$, then

(a) $A$ is a skew-symmetric matrix and $|A|=0$

(b) $A$ is symmetric matrix and $|A|$ is a square

(c) $A$ is symmetric matrix and $|A|=0$

(d) none of these.

Solution:

(d) none of these

Given: $A$ is a square matrix of even order.

Let $A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]$

$\Rightarrow A=\left[\begin{array}{cc}0 & -3 \\ 3 & 0\end{array}\right] \quad\left[\because a_{i j}=i^{2}-j^{2}\right]$

So, it is a skew-symmetric matrix as $a_{i j}=-a_{j i}$. Now,

$|A|=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]=\left[a_{11} a_{22}-a_{21} a_{12}\right]=[0-(-9)]=9$

Leave a comment