Let S be the sample space of all 3 × 3 matrices with entries from...

For sum to be seven, we need seven $1s$ and two $0s$
Number of different possible matrices= $= \dfrac{9}{7 2} = 36$
Eg. \left[\begin{cases}1 & 1 & 1 \\ 1 & 1 & 1 \\ 0 & 0 & 1\end{cases}\right] {arrangement of seven 1s and two 0s }$<br />Now, for determined of matrix to be zero, both the zero must be in same row or in same column.<br />Eg.$\left[$\begin{cases}1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{cases}$\right]$Now, arrangement of 0s and$1s = \mathop{6}_{$$\begin{array}{ll}Selectionof \\ acolumn / row\end{array}$$} \times \mathop{3}_{$$\begin{array}{ll}arrangement \\ oftwo0s \\ andone1in \\ arow / column\end{array}$$} = 18$<br />Hence, required probability$= \dfrac{18}{36} = \dfrac{1}{2}<br />(Note: If two zero's lies on different row/column then \left|\begin{cases}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1\end{cases}\right| = + 1 \ne 0 )