JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Probability

A $2\times 2$ matrix with entries from $\{0,1\}$ has $2^4=16$ possible matrices, since each of the 4 entries can be chosen independently in 2 ways.

The matrix $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ is invertible exactly when $\det A=ad-bc\ne 0$. With entries restricted to 0 or 1, $ad-bc$ can only take the values $-1,0,1$.

$\det A=1$ requires $ad=1$ and $bc=0$. Since $a,d\in \{0,1\}$, $ad=1$ forces $a=d=1$. Then $bc=0$ rules out only $b=c=1$, leaving $(b,c)\in \{(0,0),(0,1),(1,0)\}$ — 3 matrices.

$\det A=-1$ requires $ad=0$ and $bc=1$. Here $bc=1$ forces $b=c=1$, and $ad=0$ rules out only $a=d=1$, leaving $(a,d)\in \{(0,0),(0,1),(1,0)\}$ — 3 matrices.

So the number of invertible matrices is $3+3=6$ out of the 16 total, giving