1. Total number of matrices in the sample space

Each entry of the $2\times 2$ matrix can be chosen independently from $\{0,1,2\}$.

So, $|S|=3^4=81.$

We need the probability that a matrix is invertible.

2. Condition for invertibility

Let $M=\begin{pmatrix} a & b \\ c & d \end{pmatrix}, \qquad a,b,c,d\in\{0,1,2\}.$

Then $M$ is invertible iff $\det(M)=ad-bc \neq 0.$

So we count the number of matrices for which $ad-bc=0,$ and subtract from $81$.

3. Count singular matrices: $ad=bc$

We count the number of quadruples $(a,b,c,d)\in\{0,1,2\}^4$ satisfying $ad=bc.$

Since each product can only take values in $\{0,1,2,4\},$ we count according to the value of $ad$ and $bc$.

Step 3.1: Count ordered pairs $(x,y)\in\{0,1,2\}^2$ by product

Possible ordered pairs and their products:

• product $0$: $(0,0),(0,1),(0,2),(1,0),(2,0)$ → $5$ pairs
• product $1$: $(1,1)$ → $1$ pair
• product $2$: $(1,2),(2,1)$ → $2$ pairs
• product $4$: $(2,2)$ → $1$ pair

Thus for $(a,d)$ and separately for $(b,c)$, the counts are: $N_0=5,\quad N_1=1,\quad N_2=2,\quad N_4=1.$

Step 3.2: Equal products

For $ad=bc$, both products must be the same. Hence the number of singular matrices is $N_0^2+N_1^2+N_2^2+N_4^2=5^2+1^2+2^2+1^2=25+1+4+1=31.$

So, $\text{number of singular matrices}=31.$

4. Count invertible matrices

$|A|=81-31=50.$

Therefore, $P(A)=\frac{|A|}{|S|}=\frac{50}{81}.$

5. Check options

• A: $\frac{47}{81}$
• B: $\frac{49}{81}$
• C: $\frac{50}{81}$
• D: $\frac{16}{27}=\frac{48}{81}$

Hence the correct option is $\boxed{\text{C }\frac{50}{81}}.$

6. Comparison with stored correct answer

Stored correct answer: C

Derived answer: C

They agree.