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NDA 2 2025 Maths, question 40

Topic: Matrices and determinants

Passage · Q39 to Q40

Let A=[cos⁡θsin⁡θ−sin⁡θcos⁡θ]A=\begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}

Q40

What is [adjA]−1[\text{adj}A]^{-1} equal to?

  1. (a) −A-A
  2. (b) −AT-A^{T}
  3. (c) AACorrect answer
  4. (d) ATA^{T}

Solution.

Answer: (c) AA

  1. For any invertible A, adj(A)=det(A)A⁻¹.
  2. Here det(A)=1, so adj(A)=A⁻¹ and [adj(A)]⁻¹=A.