Consider the initial value problem for the vector-valued function x, x′=Ax, A=[1−225],x(0)=[1−1] Find the eigenvalues λ1,λ2 and their corresponding eigenvectors v1,v2 of the coefficient matrix A. (a) Eigenvalues: (if repeated, enter it twice separated by commas)

Step-by-step explanation:

Consider the matrix . We will calculate the correspondent eigenvalues and eigen vector of the matrix. REcall that, to calculate the eigenvalues of a square matrix A, we must solve the following equation where I is the identity matrix. In this case we have the following

which gives us the following polynomial (known as the characteristic polynomial of the matrix A).

.

Hence, the only eigenvalue of the given matrix is .

For us to calculate the eigenvalue, we want to find a base for the Kernel of matrix replacing the value of lambda with the value of the eigen value. REcall that finding the base for the Kernel is solving the associated homogeneus system of the matrix

which lead us to the equation or equivalently, . Hence, the solution of the system is of the form (x,y) = (x,-x) = x(1,-1). So  the correspondent eigenvector is the vector (1,-1).

95.03 is the answer. hope this !

600,000,000,000, or 6e14

step-by-step explanation: