Let A be a square matrix such that A A T = I . Then 1 2 A [ ( A + A T ) 2 + ( A - A T ) 2 ] is equal to [2024]

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Solution

Q.
Let $A$ be a square matrix such that $A A^{T} = I .$ Then $\frac{1}{2} A [{(A + A^{T})}^{2} + {(A - A^{T})}^{2}]$ is equal to [2024]

1 $A^{2} + A^{T}$

2 $A^{2} + I$

3 $A^{3} + I$

4 $A^{3} + A^{T}$

Ans.
(4)

We have, $\frac{1}{2} A [{(A + A^{T})}^{2} + {(A - A^{T})}^{2}]$

$= \frac{1}{2} A [(A^{2} + {(A^{T})}^{2}) + 2 A A^{T} + A^{2} + {(A^{T})}^{2} - 2 A A^{T}]$

$= \frac{1}{2} A [2 A^{2} + 2 {(A^{T})}^{2}] = \frac{1}{2} \times 2 A [A^{2} + {(A^{T})}^{2}]$

$= A^{3} + A A^{T} \cdot A^{T}$

$= A^{3} + I \cdot A^{T}$ [Given, $A A^{T} = I$ ]

$= A^{3} + A^{T}$

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