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JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant

Let \(A\) be a square matrix such that \(A A^T=I\). Then \(\frac{1}{2} A\left[\left(A+A^T\right)^2+\left(A-A^T\right)^2\right]\) is equal to

  1. A  \(A^2+I\)
  2. B \(A^3+I\)
  3. C  \(A^2+A^T\)
  4. D  \(A^3+A^T\)
Verified Solution

Answer & Solution

Correct Answer

(D)  \(A^3+A^T\)

Step-by-step Solution

Detailed explanation

\(\mathrm{AA}^{\mathrm{T}}=\mathrm{I}=\mathrm{A}^{\mathrm{T}} \mathrm{A}\) On solving given expression, we get \( \frac{1}{2} A\left[A^2+\left(A^T\right)^2+2 A A^T+A^2+\left(A^T\right)^2-2 A A^T\right] \) \( =A\left[A^2+\left(A^T\right)^2\right]=A^3+A^T\)