Final answer:
To find the repeated eigenvalue of the coefficient matrix A(t), we solve the characteristic equation det(A(t) - λI) = 0. The repeated eigenvalue is λ = 4. To find an eigenvector for this eigenvalue, we solve (A(t) - λI)X = 0, which gives us X1 = [2, 1]. Finally, to solve the given initial-value problem X'(t) = (2 4 -1 6)X, X(0) = (-1 7), we use X(t) = e^(At)X(0), where e^(At) is the matrix exponential of A(t).
Step-by-step explanation:
To find the repeated eigenvalue, we need to find the eigenvalues of the coefficient matrix A(t). The eigenvalues are the solutions to the characteristic equation det(A(t) - λI) = 0, where I is the identity matrix.
In this case, the characteristic equation is:
|2-λ 4|
| -1 6-λ| = 0
Simplifying the determinant, we get:
(2-λ)(6-λ) - (4)(-1) = 0
Simplifying further, we get a quadratic equation:
λ^2 - 8λ + 14 = 0
Solving this quadratic equation, we find that the repeated eigenvalue is λ = 4.
To find an eigenvector for this eigenvalue, we need to solve the equation (A(t) - λI)X = 0, where X is the eigenvector.
In this case, the equation is:
|2-4 4|
| -1 6-4| |X1|
Simplifying this equation, we get:
| -2 4||X1| = 0
From this, we can see that the eigenvector is X1 = [2, 1].
Finally, to solve the given initial-value problem X′=(2 4 −1 6)X, X(0)=(−1 7), we can use the formula X(t)=e^(At)X(0), where e^(At) is the matrix exponential of A(t).
Using this formula, we find:
X(t) = e^(4t)|X1|
Substituting in the values we found earlier:
X(t) = e^(4t)|[2, 1]|
Simplifying further:
X(t) = |2e^(4t)|
|1e^(4t)|