Solve differential equations 𝑥 ᇱ ሺ𝑡ሻ ൌ 9𝑥 െ 2𝑦 and 𝑦 ᇱ ሺ𝑡ሻ ൌ െ2𝑥 ൅ 6𝑦 by writing them inmatrix form and then diagonalize the resulting square matrix. (20 marks)

The given system of differential equations is:

dx/dt = 9x - 2y dy/dt = -2x + 6y

We can write this system in matrix form as follows:

d/dt [x, y] = [9, -2; -2, 6] [x, y]

This is equivalent to dX/dt = AX, where X = [x, y] and A = [9, -2; -2, 6].

To diagonalize the matrix A, we first find its eigenvalues. The eigenvalues λ of a matrix A satisfy the characteristic equation det(A - λI) = 0, where I is the identity matrix. For the given matrix A, the characteristic equation is:

det([9 - λ, -2; -2, 6 - λ]) = 0 (9 - λ)(6 - λ) - ((-2)(-2)) = 0 λ^2 - 15λ + 52 = 0

Solving this quadratic equation gives the eigenvalues λ1 = 7 and λ2 = 8.

Next, we find the eigenvectors of A by plugging each eigenvalue back into the equation (A - λI)v = 0 and solving for v. For λ1 = 7, we get:

[9 - 7, -2; -2, 6 - 7] [v1, v2] = 0 [2, -2; -2, -1] [v1, v2] = 0

Solving this system gives the eigenvector v1 = [1, 1] for λ1 = 7.

Similarly, for λ2 = 8, we get:

[9 - 8, -2; -2, 6 - 8] [v1, v2] = 0 [1, -2; -2, -2] [v1, v2] = 0

Solving this system gives the eigenvector v2 = [2, 1] for λ2 = 8.

Therefore, the matrix A is diagonalizable and its diagonal form D is given by the matrix whose diagonal entries are the eigenvalues of A, and the change of basis matrix P is the matrix whose columns are the eigenvectors of A. In this case, we have:

D = [7, 0; 0, 8] P = [1, 2; 1, 1]

The solution to the system of differential equations is then given by X(t) = P e^(Dt) P^-1 X(0), where e^(Dt) is the matrix exponential of Dt, and X(0) is the initial condition.