Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Matrices and Determinants question

2025 · 4 Apr · Shift 1 · Q48
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Matrices and Determinants
  5. /2025 · 4 Apr · Shift 1 · Q48

Matrices and Determinants question

2025 · 4 Apr · Shift 1 · Q48

JEE MainMathematicsMatrices and DeterminantsNumerical+4 / −1
Let A=[cos⁡θ0−sin⁡θ010sin⁡θ0cos⁡θ]A=\left[\begin{array}{ccc}\cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta\end{array}\right]A=​cosθ0sinθ​010​−sinθ0cosθ​​. If for some θ∈(0,π),A2=AT\theta \in(0, \pi), A^2=A^Tθ∈(0,π),A2=AT, then the sum of the diagonal elements of the matrix (A+I)3+(A−I)3−6 A(\mathrm{A}+\mathrm{I})^3+(\mathrm{A}-\mathrm{I})^3-6 \mathrm{~A}(A+I)3+(A−I)3−6 A is equal to ‾\underline{\hspace{2cm}}​ .
Numerical answer
View written solutionFree

Correct answer: 6

  1. Identify the matrix form

Given

A=[cos⁡θ0−sin⁡θ010sin⁡θ0cos⁡θ]A=\begin{bmatrix} \cos\theta & 0 & -\sin\theta\\ 0&1&0\\ \sin\theta&0&\cos\theta \end{bmatrix}A=​cosθ0sinθ​010​−sinθ0cosθ​​

Let

Then

A=[c0−s010s0c].A=\begin{bmatrix} c&0&-s\\ 0&1&0\\ s&0&c \end{bmatrix}.A=​c0s​010​−s0c​​.

This is an orthogonal matrix, so AT=A−1.A^T=A^{-1}.AT=A−1. Given A2=AT,A^2=A^T,A2=AT, we get A2=A−1.A^2=A^{-1}.A2=A−1. Multiplying both sides by AAA, A3=I.A^3=I.A3=I.

  1. Use the condition to find possible θ\thetaθ

Since AAA represents rotation by angle θ\thetaθ about the yyy-axis, A3A^3A3 corresponds to rotation by angle 3θ3\theta3θ. Thus A3=I  ⟹  3θ=2kπ.A^3=I \implies 3\theta=2k\pi.A3=I⟹3θ=2kπ. Because θ∈(0,π)\theta\in(0,\pi)θ∈(0,π), the only possibility is θ=2π3.\theta=\frac{2\pi}{3}.θ=32π​. Hence

cos⁡θ=−12,sin⁡θ=32.\cos\theta=-\frac12, \qquad \sin\theta=\frac{\sqrt3}{2}.cosθ=−21​,sinθ=23​​.
  1. Simplify the required matrix expression

We need the sum of diagonal elements, i.e. the trace, of

(A+I)3+(A−I)3−6A.(A+I)^3+(A-I)^3-6A.(A+I)3+(A−I)3−6A.

Expand:

(A+I)3=A3+3A2+3A+I,(A+I)^3=A^3+3A^2+3A+I,(A+I)3=A3+3A2+3A+I, (A−I)3=A3−3A2+3A−I.(A-I)^3=A^3-3A^2+3A-I.(A−I)3=A3−3A2+3A−I.

Adding,

(A+I)3+(A−I)3=2A3+6A.(A+I)^3+(A-I)^3=2A^3+6A.(A+I)3+(A−I)3=2A3+6A.

So the given matrix becomes

2A3+6A−6A=2A3.2A^3+6A-6A=2A^3.2A3+6A−6A=2A3.

Since A3=IA^3=IA3=I,

(A+I)3+(A−I)3−6A=2I.(A+I)^3+(A-I)^3-6A=2I.(A+I)3+(A−I)3−6A=2I.
  1. Find the sum of diagonal elements

The trace of 2I32I_32I3​ is

2+2+2=6.2+2+2=6.2+2+2=6.

Therefore, the required integer is 6.\boxed{6}.6​.

PreviousNext

More from Matrices and Determinants

  • Let the matrix A=​110​001​010​​ satisfy An=An−2+A2−I for n⩾3. Then the sum of all the elements of A50 is :2025 · MCQ
  • Let A be a 3×3 matrix such that ∣adj(adj(adjA))∣=81. If S={n∈Z:(∣adj(adjA)∣)2(n−1)2​=∣A∣(3n2−5n−4)}…2025 · MCQ
  • Let the system of equations : ​2x+3y+5z=97x+3y−2z=812x+3y−(4+λ)z=16−μ​ have infinitely many solutions. Then the radius of the circle centred at (λ,μ) and…2025 · MCQ
  • The number of singular matrices of order 2 , whose elements are from the set {2,3,6,9}, is ​.2025 · Numerical
  • Let the system of equations x + 5y - z = 1 4x + 3y - 3z = 7 24x + y + λz = μ λ, μ ∈ ℝ, have infinitely many solutions. Then the number of the solutions of this system, if x, y, z are integers and satisfy 7 ≤ x + y + z ≤ 77, is :2025 · MCQ
  • Let α be a solution of x2+x+1=0, and for some a and b in R,[4​a​b​]​1−1−2​16−1−14​132−8​​=[0​0​0​]…2025 · MCQ
  • Let A=​246​2+p6+2p12+3p​2+p+q8+3p+2q20+6p+3q​​. If det(adj(adj(3A)))=2m⋅3n, m,n∈N, then m+n is equal to2025 · MCQ
  • Let A be a square matrix of order 3 such that det(A)=−2 and det(3adj(−6adj(3A)))=2m+n⋅3mn,m>n. Then 4m+2n is equal to ​.2025 · Numerical