91Ó°ÊÓ

#tag_title#Step 3: Equate corresponding elements and solve for A#tag_content# Equate the corresponding elements of the result of the matrix multiplication to that of the given second set of solutions: \(2t^2 - 1 = a_{11} + a_{21} \ln t + a_{31} t^2\) \(3 = a_{12} + a_{22} \ln t + a_{32} t^2\) \(\ln t^3 = a_{13} + a_{23} \ln t + a_{33} t^2\) Now, we can solve for the elements of matrix A. From the first equation: \(2t^2 - 1 = a_{11} + a_{21} \ln t + a_{31} t^2\) Comparing the constant, \(\ln t\), and \(t^2\) terms, we can deduce: \(a_{11} = -1\), \(a_{21} = 0\), \(a_{31} = 2\) From the second equation: \(3 = a_{12} + a_{22} \ln t + a_{32} t^2\) Comparing the constant, \(\ln t\), and \(t^2\) terms, we can deduce: \(a_{12} = 3\), \(a_{22} = 0\), \(a_{32} = 0\) From the third equation: \(\ln t^3 = a_{13} + a_{23} \ln t + a_{33} t^2\) Comparing the constant, \(\ln t\), and \(t^2\) terms, we can deduce: \(a_{13} = 0\), \(a_{23} = 3\), \(a_{33} = 0\) So, the matrix \(A\) is: \(A = \begin{bmatrix} -1 & 3 & 0\\ 0 & 0 & 3\\ 2 & 0 & 0 \end{bmatrix}\) #tag_title#Step 4: Determine whether the second set of solutions is fundamental#tag_content# To determine if the second set of solutions is also a fundamental set, we need to calculate the determinant of matrix \(A\): \(\det(A) = \begin{vmatrix} -1 & 3 & 0\\ 0 & 0 & 3\\ 2 & 0 & 0 \end{vmatrix} = 18\) Since the determinant of matrix \(A\) is non-zero, the second set of solutions \(\left\\{\bar{y}_{1}(t), \bar{y}_{2}(t), \bar{y}_{3}(t)\right\\}\) is also a fundamental set of solutions for the given differential equation. #Answer# The constant matrix A is: \(A = \begin{bmatrix} -1 & 3 & 0\\ 0 & 0 & 3\\ 2 & 0 & 0 \end{bmatrix}\) The second set of solutions \(\left\\{\bar{y}_{1}(t), \bar{y}_{2}(t), \bar{y}_{3}(t)\right\\}\) is also a fundamental set of solutions for the given differential equation since the determinant of matrix \(A\) is non-zero.