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Mathematics · Matrices

JEE Advanced 2024 — Paper 1 — Question 14

Let α\alpha and β\beta be the distinct roots of the equation x2+x−1=0x^{2}+x-1=0. Consider the set T={1,α,β}T=\{1, \alpha, \beta\}. For a 3×33 \times 3 matrix M=(aij)3×3M=\left(a_{i j}\right)_{3 \times 3}, define Ri=ai1+ai2+ai3R_{i}=a_{i 1}+a_{i 2}+a_{i 3} and Cj=a1j+a2j+a3jC_{j}=a_{1 j}+a_{2 j}+a_{3 j} for i=1,2i=1,2, 3 and j=1,2,3\mathrm{j}=1,2,3.

Match each entry in List-I to the correct entries in List-II.

List - IList - II
(P) The number of matrices M=(aij)3×3\mathrm{M}=\left(\mathrm{a}_{\mathrm{ij}}\right)_{3 \times 3} with all entries in T such that Ri=Cj=0R_{i}=C_{j}=0 for all i,ji, j, is(1) 1
(Q) The number of symmetric matrices M=(ai)3×3\mathrm{M}=\left(\mathrm{a}_{\mathrm{i}}\right)_{3 \times 3} with all entries in TT such that Cj=0\mathrm{C}_{\mathrm{j}}=0 for all j , is(2) 12
(R) Let M=(ai)3×3M=\left(a_{\mathrm{i}}\right)_{3 \times 3} be a skew symmetric matrix such that aij∈T\mathrm{a}_{\mathrm{ij}} \in \mathrm{T} for i>j\mathrm{i}>\mathrm{j}. Then the number of elements in the set {(xyz):x,y,z∈R,M(xyz)=(a120−a23)}\left\{\left(\begin{array}{l}x\\ y\\ z\end{array}\right): x, y, z \in R, M\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}a_{12} 0 -a_{23}\end{array}\right)\right\} is(3) infinite
(S) Let M=(aij)3×3M=\left(a_{i j}\right)_{3 \times 3} be a matrix with all entries in TT such that Ri=0R_{i}=0 for all ii. Then the absolute value of determinant of MM is(4) 6
(5) 0
  1. Option A:

    (P) →\rightarrow (4)(Q) →\rightarrow (2)(R)→(\mathrm{R}) \rightarrow (5)(S) →\rightarrow (1)

  2. Option B:

    (P) →(2)\rightarrow(2) (Q)→(4)(\mathrm{Q}) \rightarrow(4) (R)→(1)(S)→(5)(\mathrm{R}) \rightarrow(1) \quad(\mathrm{S}) \rightarrow(5)

  3. Option C:

    (P)→(2)(\mathrm{P}) \rightarrow(2) (Q)→(4)(Q) \rightarrow(4) (R)→(3)(S)→(5)(R) \rightarrow(3) \quad(S) \rightarrow(5)

    Correct
  4. Option D:

    (P)→(1)(\mathrm{P}) \rightarrow(1) (Q)→(5)(\mathrm{Q}) \rightarrow(5) (R)→(3)(S)→(4)(\mathrm{R}) \rightarrow(3) \quad(\mathrm{S}) \rightarrow(4)

Answer: C

Step-by-step solution

α+β=−1 \alpha+\beta=-1 (P) Only possible when each row and each column of MM is made of 1,α,β1, \alpha, \beta. Number of ways for first row =3×2×1=6=3 \times 2 \times 1=6. Number of ways for second row =2×1×1=2=2 \times 1 \times 1=2 Number of ways for third row =1×1×1=1=1 \times 1 \times 1=1 Number of such matrices =6×2×1=12=6 \times 2 \times 1=12 (Q) Each column should be made of 1,α,β1, \alpha, \beta.

Since matrix is symmetric, after making column 1, other entries would be fixed by default. Number of ways =6=6 (R) Let a21=a,a31=b,a32=ca_{21}=a, a_{31}=b, a_{32}=c, where {a,b,c}∈T\{a, b, c\} \in T M=[0−a−ba0−cbc0]M=\left[\begin{array}{ccc}0 & -a & -b \\ a & 0 & -c\\ b & c & 0\end{array}\right] Then -ay −bz=−a⇒ay+bz=a-\mathrm{bz}=-\mathrm{a} \Rightarrow \mathrm{ay}+\mathrm{bz}=\mathrm{a} ax−cz=0⇒ax=cz\mathrm{ax}-\mathrm{cz}=0 \Rightarrow \mathrm{ax}=\mathrm{cz} and bx+cy=c\mathrm{bx}+\mathrm{cy}=\mathrm{c} If a=b=ca=b=c, then y+z=1x=z,x+y=1y+z=1 x=z, x+y=1 This has infinite solutions. (S) Each row is made of {1,α,β}\{1, \alpha, \beta\}

Also, 1+α+β=01+\alpha+\beta=0 C1→C1+C2+C3\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3} makes every element of column 1 as ' 0 '. Hence determinant is 0 .

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2024
Paper
Paper 1
Subject
Mathematics
Chapter
Matrices
Topic
Types of matrices & its properties