Question
Assertion $(A)$ : If $A=\frac{1}{3}\left[\begin{array}{ccc}1 & -2 & 2 \\ -2 & 1 & 2 \\ -2 & -2 & -1\end{array}\right]$, then $A\left(A^T\right)=I$
Reason $(R)$ : For any square matrix $A,\left(A^T\right)^T=A$

Answer

$\because A A^T=\frac{1}{3}\left[\begin{array}{ccc}1 & -2 & 2 \\ -2 & 1 & 2 \\ -2 & -2 & -1\end{array}\right] \cdot \frac{1}{3}\left[\begin{array}{ccc}1 & -2 & -2 \\ -2 & 1 & -2 \\ 2 & 2 & -1\end{array}\right] $
$ =\frac{1}{9}\left[\begin{array}{lll}9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=I$

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