Time left 0:11:05Question 2Tries remaining: 2Marked out of 1.00Flag questionQuestion textDiagonalize the matrix [−1−425][−12−45].That is, find a diagonal matrix D𝐷 and an invertible matrix P𝑃 such that A=P⋅D⋅P−1

To diagonalize a matrix, we need to find its eigenvalues and eigenvectors. The eigenvalues are the roots of the characteristic equation, which is obtained by subtracting the eigenvalue (λ) times the identity matrix from the original matrix and setting the determinant equal to zero. The eigenvectors are the null space of the matrix subtracted by the eigenvalue times the identity matrix.

Given matrix A = [−1−425][−12−45], let's find the eigenvalues first.

1. Set up the characteristic equation:

det(A - λI) = 0

where I is the identity matrix. For matrix A, this gives us:

det([−1-λ -4][2 5-λ]) = 0

2. Solve the equation:

(-1-λ)(5-λ) - (2*-4) = λ^2 + λ - 9 = 0

The roots of this equation are the eigenvalues of A. Solving this quadratic equation gives us λ1 = 3 and λ2 = -3.

3. Now, let's find the eigenvectors for each eigenvalue.

For λ1 = 3, we solve the system of equations:

(-1-3)x - 4y = 0 2x + (5-3)y = 0

Solving this system gives us the eigenvector v1 = [2, 1].

For λ2 = -3, we solve the system of equations:

(-1+3)x - 4y = 0 2x + (5+3)y = 0

Solving this system gives us the eigenvector v2 = [-1, 1].

4. The diagonal matrix D is formed by placing the eigenvalues on the diagonal:

D = [3 0][0 -3]

5. The invertible matrix P is formed by placing the eigenvectors as columns:

P = [2 -1][1 1]

6. Finally, we can check that A = PDP^-1.

So, the diagonal matrix D is [3 0][0 -3] and the invertible matrix P is [2 -1][1 1].