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Mathematics · Matrices

JEE Advanced 2025 — Paper 1 — Question 18

Consider the matrix P=(200020003)P=\left(\begin{array}{lll} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{array}\right) Let the transpose of a matrix XX be denoted by XT\mathrm{X}^{T}. Then the number of 3×33 \times 3 invertible matrices Q with integer entries, such that Q−1=QT and PQ=QPQ^{-1}=Q^{T} \text { and } P Q=Q P is

  1. Option A:

    32

  2. Option B:

    8

  3. Option C:

    16

    Correct
  4. Option D:

    24

Answer: C

Step-by-step solution

Q is orthogonal matrix

Q=[a1b1c1a2b2c2a3b3c3]Q = \begin{bmatrix} a_1 & b_1 & c_1 \\[4pt] a_2 & b_2 & c_2 \\[4pt] a_3 & b_3 & c_3 \end{bmatrix} PQ=QP⇒[2a12b12c12a22b22c23a33b33c3]=[2a12b13c12a22b23c22a32b33c3]PQ = QP \Rightarrow \begin{bmatrix} 2a_1 & 2b_1 & 2c_1 \\[4pt] 2a_2 & 2b_2 & 2c_2 \\[4pt] 3a_3 & 3b_3 & 3c_3 \end{bmatrix} = \begin{bmatrix} 2a_1 & 2b_1 & 3c_1 \\[4pt] 2a_2 & 2b_2 & 3c_2 \\[4pt] 2a_3 & 2b_3 & 3c_3 \end{bmatrix} ⇒  c1=0,  c2=0,  a3=0,  b3=0\Rightarrow \; c_1 = 0, \; c_2 = 0, \; a_3 = 0, \; b_3 = 0 ∴  Q=[a1b10a2b2000c3]\therefore \; Q = \begin{bmatrix} a_1 & b_1 & 0 \\[4pt] a_2 & b_2 & 0 \\[4pt] 0 & 0 & c_3 \end{bmatrix} Since Q is orthogonal, QQT=I\text{Since } Q \text{ is orthogonal, } QQ^{\mathrm{T}} = I ⇒{a1a2+b1b2=0,a12+b12=1,a22+b22=1,c32=1\Rightarrow \begin{cases} a_1 a_2 + b_1 b_2 = 0, \\[4pt] a_1^2 + b_1^2 = 1, \\[4pt] a_2^2 + b_2^2 = 1, \\[4pt] c_3^2 = 1 \end{cases} Q=[a1b10a2b2000c3],a1a2+b1b2=0,  a12+b12=1,  a22+b22=1,  c32=1\boxed{ Q = \begin{bmatrix} a_1 & b_1 & 0 \\[4pt] a_2 & b_2 & 0 \\[4pt] 0 & 0 & c_3 \end{bmatrix}, \quad a_1 a_2 + b_1 b_2 = 0,\; a_1^2 + b_1^2 = 1,\; a_2^2 + b_2^2 = 1,\; c_3^2 = 1 }
a1{{a}_{1}}b1{{b}_{1}}a2{{a}_{2}}b2{{b}_{2}}c3{{c}_{3}}
1001, -11 , -1
-1001 , -11, -1
011, -101, -1
0-11, -101, -1

Total 16 matrices

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2025
Paper
Paper 1
Subject
Mathematics
Chapter
Matrices
Topic
Introduction to Matrices