1. **State the problem:** We have the matrix equation \( \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 23 \\ 37 \end{bmatrix} \), where \(x\) is the number of large vessels and \(y\) is the number of small vessels. 2. **Find the inverse of the coefficient matrix:** The coefficient matrix is \( A = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} \). Calculate the determinant: $$\det(A) = (2)(2) - (3)(1) = 4 - 3 = 1$$ Since \(\det(A) = 1 \neq 0\), the inverse exists. Using the formula: $$A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}$$ 3. **Use the inverse to find \(x\) and \(y\):** Multiply \(A^{-1}\) by the right-hand side vector: $$\begin{bmatrix} x \\ y \end{bmatrix} = A^{-1} \begin{bmatrix} 23 \\ 37 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} 23 \\ 37 \end{bmatrix}$$ Calculate each component: $$x = 2 \times 23 - 1 \times 37 = 46 - 37 = 9$$ $$y = -3 \times 23 + 2 \times 37 = -69 + 74 = 5$$ So, 9 large vessels and 5 small vessels were processed. 4. **Find the new solution for the next day with requirements \( \begin{bmatrix} 27 \\ 43 \end{bmatrix} \):** Use the same inverse matrix: $$\begin{bmatrix} x_{new} \\ y_{new} \end{bmatrix} = A^{-1} \begin{bmatrix} 27 \\ 43 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} 27 \\ 43 \end{bmatrix}$$ Calculate each component: $$x_{new} = 2 \times 27 - 1 \times 43 = 54 - 43 = 11$$ $$y_{new} = -3 \times 27 + 2 \times 43 = -81 + 86 = 5$$ So, the next day 11 large vessels and 5 small vessels were processed.