Question:

Step 1: Determine the total number of possible matrices. Since each entry $a_{ij} \in \{2, 3\}$ and the matrix is $2 \times 2$, there are 4 entries, each with 2 choices. Therefore, the total number of matrices is: $$\text{Total matrices} = 2^4 = 16$$ Step 2: Set up the general form and determinant formula. Let the matrix be: $$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$ where $a, b, c, d \in \{2, 3\}$. The determinant is: $$\det(A) = ad - bc$$ Step 3: Enumerate all 16 cases and calculate their determinants. We systematically list all combinations $(a, b, c, d)$ and compute $\det(A) = ad - bc$: | Case | $(a,b,c,d)$ | $ad$ | $bc$ | $\det(A)$ | Sign | |------|-------------|------|------|-----------|------| | 1 | $(2,2,2,2)$ | 4 | 4 | 0 | $\geq 0$ ✓ | | 2 | $(2,2,2,3)$ | 6 | 4 | 2 | $> 0$ ✓ | | 3 | $(2,2,3,2)$ | 4 | 6 | -2 | $< 0$ ✗ | | 4 | $(2,2,3,3)$ | 6 | 6 | 0 | $\geq 0$ ✓ | | 5 | $(2,3,2,2)$ | 4 | 6 | -2 | $< 0$ ✗ | | 6 | $(2,3,2,3)$ | 6 | 9 | -3 | $< 0$ ✗ | | 7 | $(2,3,3,2)$ | 4 | 9 | -5 | $< 0$ ✗ | | 8 | $(2,3,3,3)$ | 6 | 9 | -3 | $< 0$ ✗ | | 9 | $(3,2,2,2)$ | 6 | 4 | 2 | $> 0$ ✓ | | 10 | $(3,2,2,3)$ | 9 | 4 | 5 | $> 0$ ✓ | | 11 | $(3,2,3,2)$ | 6 | 6 | 0 | $\geq 0$ ✓ | | 12 | $(3,2,3,3)$ | 9 | 6 | 3 | $> 0$ ✓ | | 13 | $(3,3,2,2)$ | 6 | 6 | 0 | $\geq 0$ ✓ | | 14 | $(3,3,2,3)$ | 9 | 6 | 3 | $> 0$ ✓ | | 15 | $(3,3,3,2)$ | 6 | 9 | -3 | $< 0$ ✗ | | 16 | $(3,3,3,3)$ | 9 | 9 | 0 | $\geq 0$ ✓ | Step 4: Count cases where $\det(A) \geq 0$. From the enumeration above, the cases with $\det(A) \geq 0$ are: 1, 2, 4, 9, 10, 11, 12, 13, 14, 16. $$\text{Number of cases with } \det(A) \geq 0 = 10$$ Step 5: Count cases where $\det(A) > 0$ (invertible matrices). A matrix is invertible if and only if $\det(A) \neq 0$. Among the cases with $\det(A) \geq 0$, those with $\det(A) = 0$ are: 1, 4, 11, 13, 16 (5 cases). Therefore, cases with $\det(A) > 0$ are: 2, 9, 10, 12, 14. $$\text{Number of invertible cases (among } \det(A) \geq 0) = 5$$ Step 6: Calculate the conditional probability. We need to find the probability that a matrix is invertible given that its determinant is non-negative: $$P(\text{invertible} \mid \det(A) \geq 0) = \frac{\text{Number of cases with } \det(A) > 0}{\text{Number of cases with } \det(A) \geq 0} = \frac{5}{10} = \frac{1}{2}$$ **Final Answer:** The probability that the matrix is invertible given that its determinant is non-negative is $\boxed{\frac{1}{2}}$, which corresponds to **Option 1**.