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


(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,


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


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.


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