Let $$S = M = a_{ij}$$, $$a_{ij} \in \{0, 1, 2\}$$, $$1 \leq i, j \leq 2$$ be a sample space and $$A = \{M \in S: M \text{ is invertible}\}$$ be an even. Then $$P(A)$$ is equal to

We need to find the probability that a randomly chosen 2x2 matrix with entries from $$\{0, 1, 2\}$$ is invertible.

Determine the sample space.

A 2x2 matrix $$M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ has 4 entries, each chosen from $$\{0, 1, 2\}$$. Total number of matrices = $$3^4 = 81$$.

Determine the condition for invertibility.

A 2x2 matrix is invertible if and only if its determinant is non-zero:

$$ \det(M) = ad - bc \neq 0 $$

So we need to count matrices where $$ad = bc$$ (non-invertible), then subtract from 81.

Count the number of ways each product value can occur.

For two numbers chosen from $$\{0, 1, 2\}$$, the possible products and the number of ordered pairs $$(x, y)$$ giving each product are:

- Product = 0: pairs where at least one is 0. These are: $$(0,0), (0,1), (0,2), (1,0), (2,0)$$ = 5 ways

- Product = 1: $$(1,1)$$ = 1 way

- Product = 2: $$(1,2), (2,1)$$ = 2 ways

- Product = 4: $$(2,2)$$ = 1 way

Verification: $$5 + 1 + 2 + 1 = 9 = 3^2$$ .

Count matrices with $$ad = bc$$ (determinant = 0).

For $$ad = bc$$, both products must equal the same value. The number of such matrices is:

$$ N(\det = 0) = 5 \times 5 + 1 \times 1 + 2 \times 2 + 1 \times 1 $$

$$ = 25 + 1 + 4 + 1 = 31 $$

Here, for each possible product value $$p$$, we multiply the number of ways to get $$ad = p$$ by the number of ways to get $$bc = p$$.

Calculate the number of invertible matrices and the probability.

Number of invertible matrices = $$81 - 31 = 50$$

$$ P(A) = \frac{50}{81} $$

The correct answer is Option 4: $$\dfrac{50}{81}$$.