Evaluate Statements on Matrix Multiplication Trace and Determinants

To determine which of the given options are correct, we analyze each statement individually using matrix algebra, properties of determinants, and trace of matrices.

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Step 1: Evaluation of Option (A)

We are given the matrices:

$$P = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}, \quad E = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 8 & 13 & 18 \end{bmatrix}, \quad F = \begin{bmatrix} 1 & 3 & 2 \\ 8 & 18 & 13 \\ 2 & 4 & 3 \end{bmatrix}$$

First, let's calculate $P^2$:

$$P^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$$

Next, we calculate $PEP$:

$$PE = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 8 & 13 & 18 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 \\ 8 & 13 & 18 \\ 2 & 3 & 4 \end{bmatrix}$$

Multiplying $PE$ by $P$ on the right:

$$PEP = (PE)P = \begin{bmatrix} 1 & 2 & 3 \\ 8 & 13 & 18 \\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 3 & 2 \\ 8 & 18 & 13 \\ 2 & 4 & 3 \end{bmatrix} = F$$

Thus, $F = PEP$ and $P^2 = I$.

Statement (A) is TRUE.

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Step 2: Evaluation of Option (B)

First, let's compute the determinant of matrix $E$:

$$|E| = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 8 & 13 & 18 \end{vmatrix}$$

Using row operations $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 8R_1$:

$$|E| = \begin{vmatrix} 1 & 2 & 3 \\ 0 & -1 & -2 \\ 0 & -3 & -6 \end{vmatrix} = 1 \cdot ((-1)(-6) - (-2)(-3)) = 6 - 6 = 0$$

Since $|E| = 0$: $|F| = |PEP| = |P||E||P| = 0$

Now, using $F = PEP$ and $P^2 = I$, we get $PF = P(PEP) = P^2 EP = EP$. Thus, for the given non-singular matrix $Q$: $EQ + PFQ^{-1} = EQ + EPQ^{-1} = E(Q + PQ^{-1})$

Taking the determinant on both sides: $|EQ + PFQ^{-1}| = |E(Q + PQ^{-1})| = |E| \cdot |Q + PQ^{-1}| = 0 \cdot |Q + PQ^{-1}| = 0$

Also: $|EQ| = |E||Q| = 0 \cdot |Q| = 0$ $|PFQ^{-1}| = |P||F||Q^{-1}| = |P| \cdot 0 \cdot |Q^{-1}| = 0$

Therefore: $|EQ + PFQ^{-1}| = 0 = 0 + 0 = |EQ| + |PFQ^{-1}|$

Statement (B) is TRUE.

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Step 3: Evaluation of Option (C)

Using the multiplicative property of determinants: $|(EF)^3| = |EF|^3 = (|E||F|)^3 = (0 \cdot 0)^3 = 0$ $|EF|^2 = (|E||F|)^2 = (0 \cdot 0)^2 = 0$

Since $0 > 0$ is mathematically false, $|(EF)^3| > |EF|^2$ does not hold.

Statement (C) is FALSE.

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Step 4: Evaluation of Option (D)

The sum of diagonal entries of any square matrix $M$ is defined as its trace, $\text{Tr}(M)$. Using the properties of trace:

1. $\text{Tr}(A + B) = \text{Tr}(A) + \text{Tr}(B)$
2. $\text{Tr}(AB) = \text{Tr}(BA) \implies \text{Tr}(P^{-1}EP) = \text{Tr}(EPP^{-1}) = \text{Tr}(E)$

Evaluating the sum of the diagonal entries for both expressions: $\text{Tr}(P^{-1}EP + F) = \text{Tr}(P^{-1}EP) + \text{Tr}(F) = \text{Tr}(E) + \text{Tr}(F)$ $\text{Tr}(E + P^{-1}FP) = \text{Tr}(E) + \text{Tr}(P^{-1}FP) = \text{Tr}(E) + \text{Tr}(F)$

Since both expressions evaluate to $\text{Tr}(E) + \text{Tr}(F)$, they are equal.

Statement (D) is TRUE.

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Conclusion

The correct options are A, B, and D.