Let $A$ be a square matrix such that ${a_{ij}} \in \left\{ { – 1,0,1} \right\}\forall\,\, i,j$ and it has only one non-zero entry in each row as well as in each column, then

If $A = \left[ {\begin{array}{*{20}{c}}1&0&1\\0&1&1\\1&0&0\end{array}} \right]$, then $A$ is

If $B = \left[ {\begin{array}{*{20}{c}}
5&{2\alpha }&1\\
0&2&1\\
\alpha &3&{ – 1}
\end{array}} \right]$ is the inverse of a $3 \times 3$ matrix $A$, then the sum of all values of $\alpha $ for which $det\, (A) + 1 = 0$, is

If $A = \left[ {\begin{array}{*{20}{c}}{{{\cos }^2}\theta }&{\sin \theta \cos \theta }\\{\sin \theta \cos \theta }&{{{\sin }^2}\theta }\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{c}}{{{\cos }^2}\phi }&{\sin \phi \cos \phi }\\{\sin \phi \cos \phi }&{{{\sin }^2}\phi }\end{array}} \right]$ and $\theta $ and $\phi $ differs by $\frac{\pi }{2},$ then $AB = $