The number of all 3 × 3 matrices A, with enteries from the set {–1, 0, 1} such that the sum of the diagonal elements of AA^T is 3, is

Let A = $$\left[ {\matrix{ {{a_{11}}} &amp; {{a_{12}}} &amp; {{a_{13}}} \cr {{a_{21}}} &amp; {{a_{22}}} &amp; {{a_{23}}} \cr {{a_{31}}} &amp; {{a_{32}}} &amp; {{a_{33}}} \cr } } \right]$$ <br><br>$\therefore$ A^T = $$\left[ {\matrix{ {{a_{11}}} &amp; {{a_{21}}} &amp; {{a_{31}}} \cr {{a_{12}}} &amp; {{a_{22}}} &amp; {{a_{32}}} \cr {{a_{13}}} &amp; {{a_{23}}} &amp; {{a_{33}}} \cr } } \right]$$ <br><br>diagonal elements of AA^T are $a_{11}^2 + a_{12}^2 + a_{13}^2$ , <br>$a_{21}^2 + a_{22}^2 + a_{23}^2$ , $a_{31}^2 + a_{32}^2 + a_{33}^2$ <br><br>Given Sum = ($a_{11}^2 + a_{12}^2 + a_{13}^2$) + <br>($a_{21}^2 + a_{22}^2 + a_{23}^2$) + ($a_{31}^2 + a_{32}^2 + a_{33}^2$) = 3 <br><br>This is only possible when three enteries must be either 1 or – 1 and all other six enteries are 0. <br><br>$\therefore$ Number of matrices = ⁹C₃ $\times$ 2 $\times$ 2 $\times$ 2 <br><br>= 672