Let P₁ = I = [[1,0,0],[0,1,0],[0,0,1]], P₂ = [[1,0,0],[0,0,1],[0,1,0]], P₃ = [[0,1,0],[1,0,0],[0,0,1

Q: Why is $X$ symmetric? I thought matrix multiplication wasn't commutative.

Great observation! Matrix multiplication is indeed not commutative, but here we are looking at the transpose of the entire sum. Because $Q$ is symmetric ($Q^T = Q$) and we are summing over all permutations, the transpose operation maps the sum back onto itself. It's a beautiful property of the set of permutation matrices.

Q: Why is $P_k^T R = R$?

Think about what a permutation matrix does. It swaps rows. If you have a column vector where every entry is $1$, swapping the rows doesn't change the vector at all! It's like shuffling a deck of cards where every card is an Ace of Spades; the deck looks exactly the same.

Q: Why is $X - 30I$ not invertible?

Recall that a matrix $A$ is invertible if and only if its determinant is non-zero. We found that $XR = 30R$, which means $(X - 30I)R = 0$. Since $R$ is a non-zero vector, this implies that $(X - 30I)$ has a non-trivial null space, meaning its determinant must be zero. Therefore, it cannot be invertible.

Question

Let $P_1 = I = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}, P_2 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 \end{bmatrix}, P_3 = \begin{bmatrix} 0 & 1 & 0 \ 1 & 0 & 0 \ 0 & 0 & 1 \end{bmatrix}, P_4 = \begin{bmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0 \end{bmatrix}, P_5 = \begin{bmatrix} 0 & 0 & 1 \ 1 & 0 & 0 \ 0 & 1 & 0 \end{bmatrix}, P_6 = \begin{bmatrix} 0 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 0 \end{bmatrix}$ and $X = \sum_{k=1}^{6} P_k \begin{bmatrix} 2 & 1 & 3 \ 1 & 0 & 2 \ 3 & 2 & 1 \end{bmatrix} P_k^T$ where $P_k^T$ denotes the transpose of the matrix $P_k$. Then which of the following options is/are correct?

Options:

(A) $X - 30I$ is an invertible matrix

(B) The sum of diagonal entries of $X$ is 18

(C) If $X \begin{bmatrix} 1 \ 1 \ 1 \ \end{bmatrix} = \alpha \begin{bmatrix} 1 \ 1 \ 1 \ \end{bmatrix}$, then $\alpha = 30$

(D) $X$ is a symmetric matrix

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, JEE warriors! Today, we are going to dissect a problem that looks like a chaotic mess of matrices but is actually a masterclass in symmetry and structural beauty. We are dealing with a sum of matrices defined as:

$$X = \sum_{k=1}^{6} P_k Q P_k^T$$

At first glance, this looks like a nightmare of computation. But pause. In JEE Advanced, when you see a structure like this, it is rarely a brute-force calculation; it is a puzzle waiting for you to find the pattern.

### The Symmetry of $X$

To check if $X$ is symmetric, we apply the transpose operator to the entire sum. Using the reversal law of transposes, $(ABC)^T = C^T B^T A^T$, the transpose of each term $P_k Q P_k^T$ becomes:

$$(P_k^T)^T Q^T P_k^T = P_k Q^T P_k^T$$

Since our central matrix $Q$ is given by:

$$Q = \begin{bmatrix} 2 & 1 & 3 \ 1 & 0 & 2 \ 3 & 2 & 1 \end{bmatrix}$$

It is clearly symmetric ($Q^T = Q$). Consequently, the entire sum $X$ remains unchanged under transposition, meaning $X^T = X$. Symmetry is confirmed, making the symmetry-related option a solid victory.

### The Trace and the Cyclic Property

Next, we calculate the trace of $X$. We know that $	ext{Trace}(X) = \sum 	ext{Trace}(P_k Q P_k^T)$.

Invoking the powerful cyclic property of the trace, $	ext{Trace}(ABC) = 	ext{Trace}(CAB)$, we cycle $P_k^T$ to the front:

$$	ext{Trace}(P_k^T P_k Q)$$

Since $P_k$ are permutation matrices, they are orthogonal, meaning $P_k^T P_k = I$. The expression collapses to $	ext{Trace}(I \cdot Q) = 	ext{Trace}(Q)$.

The trace of $Q$ is the sum of its diagonal elements: $2 + 0 + 1 = 3$. Since we are summing this constant over six terms, the total trace is:

$$6 	imes 3 = 18$$

### The Eigenvalue Mystery

Now, let us evaluate the eigenvalue equation $XR = \alpha R$, where $R = [1, 1, 1]^T$. Substituting our definition of $X$:

$$XR = \sum_{k=1}^{6} P_k Q P_k^T R$$

Multiplying a vector of ones by a permutation matrix does not change the vector, so $P_k^T R = R$. This simplifies our expression to $XR = (\sum P_k) (QR)$.

Calculating $QR$ by summing the rows of $Q$ yields the vector $[6, 3, 6]^T$. Since there are $3! = 6$ permutation matrices, and each position $(i, j)$ contains a $1$ exactly twice, the sum of all six permutation matrices is a matrix where every entry is $2$.

Multiplying this matrix of twos by $[6, 3, 6]^T$ gives:

$$XR = \begin{bmatrix} 2 & 2 & 2 \ 2 & 2 & 2 \ 2 & 2 & 2 \end{bmatrix} \begin{bmatrix} 6 \ 3 \ 6 \end{bmatrix} = \begin{bmatrix} 30 \ 30 \ 30 \end{bmatrix} = 30R$$

Thus, $\alpha = 30$.

### The Final Verdict

Finally, we determine if $X - 30I$ is invertible. We have shown that $XR = 30R$, which implies $(X - 30I)R = 0$.

Since $R$ is a non-zero vector, the matrix $(X - 30I)$ maps a non-zero vector to the zero vector. This is the definition of a singular matrix, meaning its determinant is zero.

Therefore, $(X - 30I)$ is **not invertible**. We have navigated the complexity and found the truth; remember, the math is always simpler than it appears.

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Visualized Solution

Defining $Q$ and Checking Symmetry

Trace of $X$ and Cyclic Property

Orthogonality of Permutation Matrices

The Eigenvalue Equation Setup

Simplifying $X R$ and Calculating $QR$

Sum of Permutation Matrices

Final Calculation for $α$

Invertibility of $X - 30 I$

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Symmetry of $X$

The Trace and the Cyclic Property

The Eigenvalue Mystery

The Final Verdict

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Visualized Solution

Defining .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q and Checking Symmetry

Trace of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X and Cyclic Property

Orthogonality of Permutation Matrices

The Eigenvalue Equation Setup

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } XR and Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } QR

Sum of Permutation Matrices

Final Calculation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Invertibility of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X−30I