If A is skew-symmetric matrix of order 2 and B, C are matrices $\begin{bmatrix}1&4\\2&9\end{bmatrix}$ and $\begin{bmatrix}9&-4\\-2&1\end{bmatrix}$ respectively. Then $A^3BC + A^5 (B^2C^2) + A^7 (B^3C^3) + ... + A^{2n+1}B^nC^n$, is

a skew-symmetric matrix

We observed that $BC=I=CB$.

$∴B^nC^n =C^nB^n = I$ for all $n ∈ N$

Let $X=A^3BC + A^5 (B^2C^2) + A^7 (B^3C^3) +...+ A^{2n+1}B^nC^n$. Then

$X=A^3I+A^5I + A^7I+ ... + A^{2n+1}I$  [Using (i)]

$⇒X^T = A^3 + A^5 +A^7+ ... + A^{2n+1}$

$⇒X^T = (A^3 + A^5 +A^7+ ... + A^{2n+1})^T$

$⇒X^T = (A^3)^T +(A^5)^T+(A^7)^T +...+(A^{2n+1})^T$

$⇒X^T = (A^T)^3 + (A^T)^5+ (A^T)^7+ .... + (A^T)^{2n+1}$

$⇒X^T=(-A)^3 + (-A)^5 +(-A)^7+...+(-A)^{2n+1}$   $[∵ A^T=-A]$

$⇒X^T=-X⇒X$ is a skew-symmetric matrix.