1. Total number of possible matrices

A $2\times 2$ matrix with entries either $0$ or $1$ has 4 entries, and each entry has 2 choices.

So total number of matrices is $2^4=16.$

2. General form of the matrix

Let $A=\begin{pmatrix} a & b \\ c & d \end{pmatrix},$ where $a,b,c,d\in\{0,1\}$.

A matrix is invertible iff its determinant is non-zero.

So we need $\det(A)=ad-bc \neq 0.$

Since $a,b,c,d\in\{0,1\}$, both $ad$ and $bc$ are either $0$ or $1$. Thus $ad-bc\in\{-1,0,1\}.$

So invertibility occurs when:

• $ad=1,\ bc=0$, giving determinant $1$, or
• $ad=0,\ bc=1$, giving determinant $-1$.

3. Case 1: $ad=1$ and $bc=0$

For $ad=1$, we must have $a=1,\ d=1.$ For $bc=0$, at least one of $b,c$ is $0$.

Possible pairs $(b,c)$ are: $(0,0),\ (0,1),\ (1,0).$

So this case gives $3$ matrices.

4. Case 2: $ad=0$ and $bc=1$

For $bc=1$, we must have $b=1,\ c=1.$ For $ad=0$, at least one of $a,d$ is $0$.

Possible pairs $(a,d)$ are: $(0,0),\ (0,1),\ (1,0).$

So this case also gives $3$ matrices.

5. Total invertible matrices

Hence number of invertible matrices is $3+3=6.$

Therefore, $P(E)=\frac{6}{16}=\frac{3}{8}.$

6. Checking options

• A: $\frac{3}{8}$ ✅
• B: $\frac{1}{8}$ ❌
• C: $\frac{3}{16}$ ❌
• D: $\frac{5}{8}$ ❌

Therefore, the correct answer is Option A.