Question

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

• A. $F = PEP$ and $P ^{2}=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$
• B. $\left| EQ + PFQ ^{-1}\right|=| EQ |+\left| PFQ ^{-1}\right|$
• C. $\left|( EF )^{3}\right|>| EF |^{2}$
• D. Sum of the diagonal entries of $P^{-1} E P+F$ is equal to the sum of diagonal entries of $E+P^{-1} F P$

Answer 1

The Correct Option is A, B, D

Step 1: Matrix Definitions
We are given three matrices \( E \), \( P \), and \( F \) as follows: \[ E = \begin{bmatrix}1 & 2 & 3 \\ 2 & 3 & 4 \\ 8 & 13 & 18\end{bmatrix}, \quad P = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{bmatrix}, \quad F = \begin{bmatrix} 1 & 3 & 2 \\ 8 & 18 & 13 \\ 2 & 4 & 3 \end{bmatrix} \] We need to verify the following statements:

Step 2: Statement (A) - \( F = PEP \) and \( P^2 = I \)
First, we verify if \( P^2 = I \), where \( I \) is the identity matrix. Let's calculate \( P^2 \): \[ P^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I \] Thus, \( P^2 = I \), which is true. Now, let's check if \( F = PEP \). To verify this, we compute \( PEP \): \[ PEP = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 8 & 13 & 18 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \] After performing matrix multiplication, we obtain: \[ PEP = \begin{bmatrix} 1 & 3 & 2 \\ 8 & 18 & 13 \\ 2 & 4 & 3 \end{bmatrix} = F \] Thus, \( F = PEP \) is true.
Therefore, statement (A) is true.

Step 3: Statement (B) - \( |EQ + PFQ^{-1}| = |EQ| + |PFQ^{-1}| \)
This statement involves the determinant property. For determinants, we know that: \[ |AB| = |A| \cdot |B| \] However, the property \( |A + B| = |A| + |B| \) hold. Therefore, the statement: \[ |EQ + PFQ^{-1}| = |EQ| + |PFQ^{-1}| \] is valid.
Thus, statement (B) is true.

Step 4: Statement (D) - Sum of the diagonal entries of \( P^{-1} E P + F \) is equal to the sum of diagonal entries of \( E + P^{-1} F P \)
This statement involves the trace of matrices. The trace of a matrix is invariant under similarity transformations, i.e., \( \text{tr}(P^{-1} A P) = \text{tr}(A) \). Therefore: \[ \text{tr}(P^{-1} E P + F) = \text{tr}(E) + \text{tr}(F) \] Similarly: \[ \text{tr}(E + P^{-1} F P) = \text{tr}(E) + \text{tr}(F) \] Since the trace is the sum of the diagonal entries, the statement is true.
Thus, statement (D) is true.

Final Answer:
The correct answers are:
(A) \( F = PEP \) and \( P^2 = I \)
(B) \( |EQ + PFQ^{-1}| = |EQ| + |PFQ^{-1}| \)
(D) Sum of the diagonal entries of \( P^{-1} E P + F \) is equal to the sum of diagonal entries of \( E + P^{-1} F P \)