JEE Advanced 2019 · Paper 2 · Mathematics Q1

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MathematicsMultiple correct+4 / -1Question 37 of 54

Let

$P_1 = I = \begin{bmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix},\quad P_2 = \begin{bmatrix}1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0\end{bmatrix},\quad P_3 = \begin{bmatrix}0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1\end{bmatrix},$

$P_4 = \begin{bmatrix}0 & 1 & 0\\ 0 & 0 & 1\\ 1 & 0 & 0\end{bmatrix},\quad P_5 = \begin{bmatrix}0 & 0 & 1\\ 1 & 0 & 0\\ 0 & 1 & 0\end{bmatrix},\quad P_6 = \begin{bmatrix}0 & 0 & 1\\ 0 & 1 & 0\\ 1 & 0 & 0\end{bmatrix}$

and

$X = \sum_{k=1}^6P_k\begin{bmatrix}2 & 1 & 3\\ 1 & 0 & 2\\ 3 & 2 & 1\end{bmatrix}P_k^T$

where $P_k^T$ denotes the transpose of the matrix $P_k$ . Then which of the following options is/are correct?

X\begin{bmatrix}1\\ 1\\ 1\end{bmatrix} = \alpha\begin{bmatrix}1\\ 1\\ 1\end{bmatrix}

, then

\alpha = 30

X

is a symmetric matrix

The sum of diagonal entries of

X

is 18

X - 30I

is an invertible matrix

Answer and worked solution

Official answer: A, B, C

The six $P_k$ are all the $3\times3$ permutation matrices. Let $A$ be the given matrix, so each term is $P_kAP_k^T$ .

(B) $A$ is symmetric and $(P_kAP_k^T)^T=P_kA^TP_k^T=P_kAP_k^T$ , so every term, and hence $X$ , is symmetric. True.

Entries of $X$ . Summing over all permutations:

each diagonal entry of $X$ is $2(2+0+1)=6$ ;
each off-diagonal entry is $2(1+3+2)=12$ .

So $X=6I+12(J-I)$ , where $J$ is the all-ones matrix.

(C) $\operatorname{tr}X=18$ . True.

(A) Each row sums to $6+24=30$ , so $X(1,1,1)^T=30(1,1,1)^T$ and $\alpha=30$ . True.

(D) $(1,1,1)^T$ is in the kernel of $X-30I$ , so it is singular. False.

(A), (B), (C)

Topics:Special matrices Matrix algebra

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JEE Advanced 2019 · Paper 2, Mathematics question 1