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JEE Advanced 2021 · Determinants

For any 3×33 \times 3 matrix MM, let ∣M∣|M| denote the determinant of MM. Let E=[12323481318]E = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 8 & 13 & 18 \end{bmatrix}, P=[100001010]P = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} and F=[13281813243]F = \begin{bmatrix} 1 & 3 & 2 \\ 8 & 18 & 13 \\ 2 & 4 & 3 \end{bmatrix} If QQ is a non-singular matrix of order 3×33 \times 3, then which of the following statements is (are) TRUE?

Answers: A. F=PEPF = PEP and P2=[100010001]P^{2} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} · B. ∣EQ+PFQ−1∣=∣EQ∣+∣PFQ−1∣|EQ + PFQ^{-1}| = |EQ| + |PFQ^{-1}| · D. Sum of the diagonal entries of P−1EP+FP^{-1} EP + F is equal to the sum of diagonal entries of E+P−1FPE + P^{-1}FP

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Written solution

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

Which of (1)–(4) are TRUE?\text{Which of (1)--(4) are TRUE?}

  1. verify PEP = F and P²=I

2. find ∣E∣ and ∣F∣2.\ \text{find } |E| \text{ and } |F|

3. test (2), (3), (4)3.\ \text{test (2), (3), (4)}

P: swap rows/cols 2↔3⇒P2=I,P−1=PP: \text{ swap rows/cols } 2 \leftrightarrow 3 \Rightarrow P^{2} = I,\quad P^{-1} = P

PEP: swap R₂,R₃ then C₂,C₃ of E ⇒ F

P2=I⇒Statement (1) TRUEP^{2}=I \Rightarrow \text{Statement (1) TRUE}

R3=2R1+3R2⇒∣E∣=0R_3 = 2R_1 + 3R_2 \Rightarrow |E| = 0

F = PEP ⇒ |F| = |P| |E| |P| = 0 PF = EP ⇒ |EQ + PFQ⁻¹| = |E(Q+PQ⁻¹)| = |E|·|Q+PQ⁻¹| = 0 |EQ| + |PFQ⁻¹| = 0 ⇒ Statement (2) TRUE

∣EF∣=∣E∣ ∣F∣=0⇒∣(EF)3∣=0=∣EF∣2|EF| = |E|\,|F| = 0 \Rightarrow |(EF)^3| = 0 = |EF|^2

0>0 is false⇒Statement (3) FALSE0 > 0 \text{ is false} \Rightarrow \text{Statement (3) FALSE}

tr⁡(P−1EP)=tr⁡(E),tr⁡(P−1FP)=tr⁡(F)\operatorname{tr}(P^{-1}EP) = \operatorname{tr}(E),\quad \operatorname{tr}(P^{-1}FP) = \operatorname{tr}(F)

both sides =tr⁡(E)+tr⁡(F)⇒Statement (4) TRUE\text{both sides } = \operatorname{tr}(E) + \operatorname{tr}(F) \Rightarrow \text{Statement (4) TRUE}

Statements (1), (2) and (4) are TRUE\boxed{\text{Statements (1), (2) and (4) are TRUE}}

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