Mathematics - Probability Question with Solution | TestHub

$A={\left[a_y\right]}_{2\times 2}$ and entries are 0 or 1. $\begin{array}{rl}& \therefore \left|\begin{array}{ll}a& b\\ c& d\end{array}\right|=0\\ & \Rightarrow ad-bc=0\end{array}$ Case I: $ad=bc=1$ $\therefore a=b=c=d=1$ Case II: $ad=bc=0$ $\begin{array}{ll}a=0,d=0& b=0,c=0\\ a=0,d=1& b=0,c=1\\ a=1,d=0& b=1,c=0\end{array}$ $\therefore$ Total 10 cases when matrix is non invertible Total possible matrix $=2^4=16$ Required probability of invertible $=\frac{16-10}{16}=\frac{6}{16}=\frac{3}{8}$