Mathematics - Determinant Question with Solution | TestHub

$|A|=-(e-d)+c(b-a)=\pm 1$ Case (i) : $c=0\Rightarrow (e,d)=(1,0),(0,1)\to 2$ ways b and a can be each 2 ways $\Rightarrow$ Total $=8$ ways Case (ii) : $c\ne 0\Rightarrow c=1$ $\Rightarrow d-e+b-a=\pm 1$ $\begin{array}{llll}1& 1& 1& 0\\ 1& 1& 0& 1\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\}\to 4\times 2=8$ ways Total $=16$ ways