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Step-by-Step Solution
Step 1: Determine the Size of the Sample Space
Each entry of the 2×2 matrix M can be 0, 1, or 2. Since there are 4 entries and each entry has 3 possible values, the total number of matrices in the sample space S is:
$3^4 = 81$.
Step 2: Invertibility Criterion
A 2×2 matrix
$
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
$
is invertible if and only if its determinant is nonzero, i.e.,
$
ad - bc \neq 0.
$
Step 3: Count the Matrices with Zero Determinant
We need to count how many matrices satisfy $ad - bc = 0,$ or equivalently $ad = bc.$ We will consider the possible products arising from pairs (a, d) and (b, c) in {0,1,2}.
Case I: $ad = bc = 0$
The product of two numbers is zero if at least one factor is 0.
Possible pairs (a, d) giving product 0 are: (0,0), (0,1), (0,2), (1,0), (2,0). There are 5 such pairs.
Similarly, possible pairs (b, c) giving product 0 are also 5 in number.
Hence, total possibilities for this case = $5 \times 5 = 25.$
Case II: $ad = bc = 1$
The only way to get a product 1 from {0,1,2} is (1,1). So for $ad=bc=1,$
a = 1, d = 1, b = 1, c = 1
$
\implies
\begin{bmatrix}
1 & 1 \\
1 & 1
\end{bmatrix}.
$
Hence, there is exactly 1 such matrix.
Case III: $ad = bc = 2$
Possible pairs that give product 2 are (1,2) and (2,1). There are 2 ways for (a, d) and 2 ways for (b, c). Thus, total possibilities for this case = $2 \times 2 = 4.$
Case IV: $ad = bc = 4$
The only way to get 4 from {0,1,2} is (2,2). So a = 2, d = 2, b = 2, c = 2, giving the matrix:
$
\begin{bmatrix}
2 & 2 \\
2 & 2
\end{bmatrix}.
$
There is exactly 1 such matrix satisfying $ad - bc = 0.$
The total number of matrices with determinant zero is therefore:
$25 + 1 + 4 + 1 = 31.$
Step 4: Calculate the Probability of an Invertible Matrix
Since there are 81 total matrices and 31 have determinant zero, the number of invertible matrices is $81 - 31 = 50.$ Therefore, the probability of randomly choosing an invertible matrix from S is:
$
P(\text{invertible}) = \frac{50}{81}.
$
Final Answer
$\frac{50}{81}$
