For any 3 × 3 matrix M, let |M| denote the determinant of M. Let E = [[1, 2, 3], [2, 3, 4], [8, 13,

Q: Wait, why is $|E| = 0$ so important? Does it make the whole problem trivial?

It is not trivial, but it is a massive hint! When you find $|E| = 0$, it acts as a key that unlocks the determinant-based options. Since $|AB| = |A| \cdot |B|$, if one matrix has a determinant of zero, the determinant of any product involving it will also be zero. This allows you to quickly evaluate complex-looking expressions like $|EQ + PFQ^{-1}|$ without needing to compute the inverse of $Q$.

Q: Can I just multiply the matrices out for $PEP$ instead of using the row/column swap trick?

You absolutely can, and you will get the correct answer! However, in JEE Advanced, time is your most precious resource. Recognizing $P$ as a permutation matrix that swaps rows (when pre-multiplying) and columns (when post-multiplying) saves you minutes of tedious arithmetic and reduces the chance of a calculation error. Always look for the structural 'shortcut' before diving into brute-force multiplication.

Q: Is $Tr(P^{-1}EP)$ always equal to $Tr(E)$? How do I prove that?

Yes, it is! This is a direct application of the cyclic property of the trace: $Tr(ABC) = Tr(BCA)$. Here, let $A = P^{-1}$, $B = E$, and $C = P$. Then $Tr(P^{-1}EP) = Tr(EPP^{-1})$. Since $PP^{-1} = I$, this simplifies to $Tr(EI)$, which is just $Tr(E)$. It is a powerful invariant property that makes trace problems much simpler than they appear.

Question

For any $3 	imes 3$ 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 nonsingular matrix of order $3 	imes 3$, then which of the following statements is (are) TRUE ?

Options:

(A) $F = PEP$ and $P^2 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$

(B) $|EQ + PFQ^{-1}| = |EQ| + |PFQ^{-1}|$

(C) $|(EF)^3| > |EF|^2$

(D) Sum of the diagonal entries of $P^{-1}EP + F$ is equal to the sum of diagonal entries of $E + P^{-1}FP$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We begin by examining the matrix $P = \begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 \end{bmatrix}$. This is a **permutation matrix**, which acts as an operator that swaps the second and third rows of any matrix it pre-multiplies.

When we pre-multiply a matrix $E$ by $P$, the rows of $E$ are swapped. Conversely, post-multiplying by $P$ performs the same operation on the columns. This relationship is central to understanding the transformation $F = PEP$.

### The Determinant Trap

Next, we calculate the determinant of matrix $E$. Expanding along the first row, we have:

$$|E| = 1(54 - 52) - 2(36 - 32) + 3(26 - 24)$$

Performing the arithmetic:

$$|E| = 1(2) - 2(4) + 3(2) = 2 - 8 + 6 = 0$$

Since $|E| = 0$, the matrix $E$ is **singular**. Because $F = PEP$, the determinant of $F$ is given by $|F| = |P| \cdot |E| \cdot |P|$. Since $|P| = -1$, we find $|F| = (-1) \cdot 0 \cdot (-1) = 0$.

### Unraveling the Options

With $|E| = 0$ and $|F| = 0$, we evaluate the given options. For Option B, we consider the expression $|EQ + PFQ^{-1}|$.

Given $F = PEP$, we substitute to find $PF = P(PEP) = P^2EP$. Since $P^2 = I$, this simplifies to $PF = EP$.

The expression becomes $|EQ + EPQ^{-1}|$, which factors to $|E(Q + PQ^{-1})|$. Because $|E| = 0$, the entire determinant is $0$. The right-hand side $|EQ| + |PFQ^{-1}|$ also evaluates to $0 + 0 = 0$, confirming that **Option B is correct**.

### Evaluating Further Claims

Option C claims that $|(EF)^3| > |EF|^2$. Since $|EF| = |E| \cdot |F| = 0 \cdot 0 = 0$, the inequality reduces to $0 > 0$, which is **false**.

Finally, we examine the trace in Option D. Using the cyclic property of the trace, $Tr(ABC) = Tr(BCA)$, we observe:

$$Tr(P^{-1}EP) = Tr(EPP^{-1}) = Tr(E)$$

Comparing $Tr(P^{-1}EP + F)$ and $Tr(E + P^{-1}FP)$, both sides simplify to $Tr(E) + Tr(F)$. This confirms the symmetry of the trace property.

In linear algebra, recognizing these structural patterns is far more powerful than brute-force calculation.

Select Answer:

Visualized Solution

Analyze Matrix $P$

Compute $P^{2}$

Compute $P E P$

Verify Option A

Determinant of $E$

Evaluate $∣ E ∣$ and $∣ F ∣$

Analyze Option B (LHS)

Analyze Option B (RHS)

Verify Option C

Trace of a Matrix

Verify Option D

Final Conclusion

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Determinant Trap

Unraveling the Options

Evaluating Further Claims

Similar Questions

For any $3 \times 3$ matrix $M$ , let $∣ M ∣$ denote the determinant of $M$ . Let $I$ be the $3 \times 3$ identity matrix. Let $E$ and $F$ be two $3 \times 3$ matrices such that $(I - E F)$ is invertible. If $G = (I - E F)^{- 1}$ , then which of the following statements is (are) TRUE ?

Let $P = 1416014001$ and $I$ be the identity matrix of order 3. If $Q = [q_{ij}]$ is a matrix such that $P^{50} - Q = I$ , then $\frac{q _{31} + q _{32}}{q _{21}}$ equals

Let M be a $3 \times 3$ matrix satisfying $M 010 = - 1 23, M 1 - 1 0 = 11 - 1$ , and $M 111 = 0012$ . Then the sum of the diagonal entries of M is

Let $P = 139013001$ and $Q = [q_{ij}]$ be two $3 \times 3$ matrices such that $Q - P^{5} = I_{3}$ . Then $\frac{q _{21} + q _{31}}{q _{32}}$ is equal to:

Let $P_{1} = I = 100010001, P_{2} = 100001010, P_{3} = 010100001, P_{4} = 001100010, P_{5} = 010001100, P_{6} = 001010100$ and $X = \sum_{k = 1}^{6} P_{k} 213102321 P_{k}^{T}$ where $P_{k}^{T}$ denotes the transpose of the matrix $P_{k}$ . Then which of the following options is/are correct?

Let $A$ and $B$ be two $3 \times 3$ real matrices such that $(A^{2} - B^{2})$ is invertible matrix. If $A^{5} = B^{5}$ and $A^{3} B^{2} = A^{2} B^{3}$ , then the value of the determinant of the matrix $A^{3} + B^{3}$ is equal to :

Let $A = [a_{ij}]$ be a square matrix of order 3 such that $a_{ij} = 2^{j - i}$ , for all $i, j = 1, 2, 3$ . Then, the matrix $A^{2} + A^{3} + \dots + A^{10}$ is equal to

Let $M$ and $N$ be two $3 \times 3$ non-singular skew-symmetric matrices such that $M N = N M$ . If $P^{T}$ denotes the transpose of $P$ , then $M^{2} N^{2} (M^{T} N)^{- 1} (M N^{- 1})^{T}$ is equal to

Let $A = [a_{ij}]$ be $3 \times 3$ matrix such that $A 010 = 001$ , $A 413 = 010$ and $A 212 = 100$ , then $a_{23}$ equals:

Which of the following is(are) NOT the square of a $3 \times 3$ matrix with real entries ?

Select Answer:

Visualized Solution

Analyze Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Compute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P2

Compute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PEP