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
- A \(A^2+I\)
- B \(A^3+I\)
- C \(A^2+A^T\)
- D \(A^3+A^T\)
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 \( \frac{1}{2} A\left[A^2+\left(A^T\right)^2+2 A A^T+A^2+\left(A^T\right)^2-2 A A^T\right] \) \( =A\left[A^2+\left(A^T\right)^2\right]=A^3+A^T\)
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