Question:

<div class="solution"> <p><strong>Step 1:</strong> To find the probability that the matrix \(A\) is invertible given that its determinant is non-negative, we first need to understand what makes a matrix invertible. A matrix is invertible if its determinant is non-zero.</p> <p><strong>Step 2:</strong> The matrix \(A\) is a \(2 \times 2\) matrix where each element \(a_{ij}\) can be either 2 or 3. This gives us \(2^4 = 16\) possible combinations of elements for the matrix, since there are 4 elements in a \(2 \times 2\) matrix and each can be one of two values.</p> <p><strong>Step 3:</strong> The determinant of a \(2 \times 2\) matrix \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\) is given by \(ad - bc\). For the determinant to be non-negative, we must have \(ad - bc \geq 0\). Given \(a, b, c, d \in \{2, 3\}\), we can enumerate the cases where the determinant is non-negative and calculate the probability of the matrix being invertible (i.e., having a non-zero determinant) among these cases.</p> <p><strong>Step 4:</strong> Let's enumerate the cases where the determinant is non-negative and calculate the number of such cases. The possible values for \(ad\) are \(2*2 = 4\), \(2*3 = 6\), or \(3*3 = 9\). The possible values for \(bc\) are also \(4\), \(6\), or \(9\). The determinant is non-negative if \(ad \geq bc\). This includes cases where \(ad = bc\), but for the matrix to be invertible, we need \(ad > bc\). We will count the total number of matrices with non-negative determinants and then subtract the ones with zero determinant to find the number of invertible matrices.</p> <p><strong>Step 5:</strong> For \(ad = bc\), the only cases are when both pairs \((a, d)\) and \((b, c)\) are the same (either both are 2, both are 3, or both are mixed), leading to a determinant of 0. There are 4 such cases (when all elements are the same) and 4 cases when the pairs are mixed but equal (e.g., \(a = b = 2\) and \(c = d = 3\) or any other combination that results in \(ad = bc\)), totaling 8 cases where the determinant is 0. For non-zero determinants, we consider cases where \(ad > bc\), which includes various combinations of 2s and 3s that satisfy this inequality.</p> <p><strong>Step 6:</strong> Calculating the total number of matrices with non-negative determinants directly can be complex due to the need to consider all combinations of \(a, b, c,\) and \(d\). However, we know that for a \(2 \times 2\) matrix to have a non-negative determinant, the combinations must satisfy \(ad \geq bc\). Given the possible values, we can see that having all elements as 2 or all elements as 3 gives a determinant of 0. Mixed combinations where \(ad > bc\) will give positive determinants. Since we're looking for the probability of a matrix being invertible given it has a non-negative determinant, we focus on the ratio of matrices with positive determinants to those with non-negative determinants.</p> <p><strong>Step 7:</strong> To calculate the probability that the matrix is invertible given that its determinant is non-negative, we need to identify the total number of matrices with non-negative determinants and then find how many of those are invertible (i.e., have a non-zero determinant). Given the constraints and possible values, we can enumerate or logically deduce the number of such matrices. There are a total of 16 possible matrices. The non-negative determinant condition includes matrices with zero determinant, which are not invertible.</p> <p><strong>Step 8:</strong> After careful consideration, we find that there are 8 matrices with zero determinant (where \(ad = bc\)) and the rest have either positive or negative determinants. Since we are interested in non-negative determinants, we look at the cases where \(ad > bc\), which will be positive, and include the cases where \(ad = bc\), which are zero. The total number of matrices with non-negative determinants (including zero) is thus a portion of the 16 total matrices. We need to calculate