JEE Advanced 2021 · Paper 1 · Mathematics Q11

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MathematicsMultiple correct+4 / -2Question 49 of 57

For any $3\times3$ matrix $M$ , let $|M|$ denote the determinant of $M$ . Let
$E = \begin{bmatrix}1 & 2 & 3\\ 2 & 3 & 4\\ 8 & 13 & 18\end{bmatrix},\quad P = \begin{bmatrix}1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0\end{bmatrix}\quad\text{and}\quad F = \begin{bmatrix}1 & 3 & 2\\ 8 & 18 & 13\\ 2 & 4 & 3\end{bmatrix}$
If $Q$ is a nonsingular matrix of order $3\times3$ , then which of the following statements is (are) TRUE?

F = PEP

and

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

|EQ + PFQ^{-1}| = |EQ| + |PFQ^{-1}|

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

Sum of the diagonal entries of

P^{-1}EP + F

is equal to the sum of diagonal entries of

E + P^{-1}FP

Answer and worked solution

Official answer: A, B, D

Preliminaries. $P$ swaps rows 2 and 3 (on the left) or columns 2 and 3 (on the right), and $P^2=I$ . Also $|E|=1(2)-2(4)+3(2)=0$ .

Check each option.

(A) $PEP$ swaps rows 2, 3 and then columns 2, 3 of $E$ . This gives exactly $F$ , and $P^2=I$ . True.
(B) $PF=P(PEP)=EP$ , so $EQ+PFQ^{-1}=E(Q+PQ^{-1})$ , whose determinant is 0. Both terms on the right are also 0, since $|E|=|F|=0$ . True.
(C) $|EF|=0$ , so the claim reads $0>0$ . False.
(D) The trace is unchanged by similarity: $\operatorname{tr}(P^{-1}EP+F)=\operatorname{tr}E+\operatorname{tr}F=\operatorname{tr}(E+P^{-1}FP)$ . True.

(A), (B), (D)

Topics:Matrix algebra Determinants and properties

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JEE Advanced 2021 · Paper 1, Mathematics question 11