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Saturday, 31 October

JEE Main 2024 · MathsMultiple choiceSingle correctEasyCalculation

JEE Main 1 February 2024, Shift 1, Maths Q3

Question 3 of 90 in this shift, Maths question 3 of 30, Section A.

If A=[21−12]A=\begin{bmatrix}\sqrt{2}&1\\-1&\sqrt{2}\end{bmatrix}, B=[1011]B=\begin{bmatrix}1&0\\1&1\end{bmatrix}, C=ABATC=ABA^T and X=ATC2AX=A^TC^2A, then det⁡X\det X is equal to :
  1. (1)27
  2. (2)729Official answer
  3. (3)891
  4. (4)243

Official answer

Option 2

NTA final key.