For $$3 \times 3$$ matrix M, let $$\mid M \mid$$ denote the determinant of M. Let
$$E = \begin{bmatrix}1 & 2 & 3 \\2 & 3 & 4 \\ 8 & 13 & 18 \end{bmatrix}, P = E = \begin{bmatrix}1 & 0 & 0 \\0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$$ and $$F = \begin{bmatrix}1 & 3 & 2 \\8 & 18 & 13 \\ 2 & 4 & 3 \end{bmatrix}$$
If Q is a nonsingular matrix of order $$3 \times 3$$, then which of the following statements is (are) TRUE ?

Let us examine each statement one by one.

Option A
First compute $$P^2$$ :

$$P \;=\;\begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix},\qquad P^2\;=\;P\!\cdot\!P \;=\;\begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix} \begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix} \;=\;\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$$

Hence $$P^2$$ is indeed the identity matrix of order 3.

Next, verify $$F = PEP$$ :

• Right-multiplying $$E$$ by $$P$$ swaps its 2nd and 3rd columns.
$$E\,P=\begin{bmatrix}1&3&2\\2&4&3\\8&18&13\end{bmatrix}$$

• Left-multiplying the above result by $$P$$ swaps its 2nd and 3rd rows.
$$P\,(E\,P)=\begin{bmatrix}1&3&2\\8&18&13\\2&4&3\end{bmatrix}=F$$

Both parts of Option A are therefore true, so Option A itself is correct.

Option B
The determinant is not an additive function in general, i.e. $$\mid A+B\mid\neq\mid A\mid+\mid B\mid$$ for arbitrary square matrices $$A,B$$. Hence the asserted equality $$\mid EQ + PFQ^{-1} \mid = \mid EQ \mid + \mid PFQ^{-1} \mid$$ cannot hold for all non-singular $$Q$$; a single counter-example suffices to falsify the claim. Therefore Option B is FALSE.

Option C
Since $$\mid E\mid=0$$ and $$\mid F\mid=0$$, we have $$\mid EF\mid=\mid E\mid\mid F\mid=0$$ and consequently $$\mid(EF)^3\mid=\bigl(\mid EF\mid\bigr)^{3}=0,\qquad \bigl(\mid EF\mid\bigr)^{2}=0.$$ Thus $$\mid(EF)^3\mid\gt\mid EF\mid^{2}$$ is impossible. Option C is FALSE.

Option D
The trace (sum of diagonal entries) is similarity-invariant: $$\operatorname{tr}(P^{-1}EP)=\operatorname{tr}(E),\qquad \operatorname{tr}(P^{-1}FP)=\operatorname{tr}(F).$$ Hence $$\operatorname{tr}\bigl(P^{-1}EP+F\bigr) =\operatorname{tr}(P^{-1}EP)+\operatorname{tr}(F) =\operatorname{tr}(E)+\operatorname{tr}(F),$$ while $$\operatorname{tr}\bigl(E+P^{-1}FP\bigr) =\operatorname{tr}(E)+\operatorname{tr}(P^{-1}FP) =\operatorname{tr}(E)+\operatorname{tr}(F).$$ The two traces coincide, so the statement in Option D is wrong. Option D is FALSE.

Only Option A is correct.

Final Answer : Option A which is: $$F = PEP$$ and $$P^2 = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$$.