# Matrices-orthogonal, anti symmetric, singular

P and Q are non-zero $3$X$3$ matrices and satisfy the equation

$(PQ)^T+Q^{-1}P=0$

(i) Prove that if Q is orthogonal, then, P is skew symmetric.

(ii) Without assuming Q is orthogonal, prove that P is singular.

Part (i) is correct for me. Using cyclic properties of Transpose matrix, $(PQ)^T=Q^TP^T$.

Then, since Q is orthogonal,$Q^T=Q^{-1}$.

Using associative law for matrix operation,

$Q^{-1}[P^T+P]=0$

Pre- multiplying by Q on both sides ends up giving $P=-P^T$

For part (ii), we need to show $det(P)=0$ for singular matrices

$|Q^TP^T|+|Q^{-1}P|=|0|$

Using properties that $|Q^{-1}|=\frac{1}{|Q|}$

We end up with $|Q||P^T|+\frac{1}{|Q|}|P|=0$

How to show $|P|=0$ from here.

$$Q^tP^t+Q^{-1}P=0\Rightarrow Q^tP^t=-Q^{-1}P\Rightarrow QQ^tP^t=-P$$
Witch means $$\tag 1 det(QQ^t)det(P^t)=-det(P)$$
Now $det(QQ^t)$ is always positive (since $QQ^t$ is positive definite, because Q is invertible). So we must have $det(P)=0$ for equation (1) to hold.
PS: I think the equation $|Q^TP^T|+|Q^{-1}P|=|0|$ you used in your question is incorrect.