Question

Let A be a 2×2 matrix with eigenvalues 6 and 1/6 and corresponding eigenvectors

v₁ = [ 1 /1] and v₂ = [−1/1 ]. Let {xk} be a solution of the difference equation
xk+1 = Axk ,x₀ = [3/1]
a. Compute x₁ = Ax₀
.
​
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Answer 1

The question involves performing a matrix-vector multiplication to find the next term in a sequence generated by a difference equation. Given the initial vector and eigenvalues and eigenvectors of the matrix, the multiplication results in the first iteration of the sequence.

Step-by-step explanation:

The student has been provided with a 2×2 matrix A, its eigenvalues, and corresponding eigenvectors, and must compute the sequence {xk} generated by the given difference equation. Given initial vector x₀, we need to find x₁ = Ax₀. The matrix A acts on x₀ to generate a new vector x₁.

To solve for x₁, we perform the matrix-vector multiplication Ax₀. Given x₀ = [3/1], and assuming A is composed of eigenvectors [1 -1/1 1], we multiply A by x₀ to obtain:

• First row: (3 × eigenvalue corresponding to v₁) + (1 × eigenvalue corresponding to v₂)
• Second row: (3 × eigenvalue corresponding to v₁) + (1 × eigenvalue corresponding to v₂)

Resulting in x₁ which is the first iteration of the solution to the difference equation. Since the student has not provided the exact matrix A, but only its eigenvalues and eigenvectors, complete computation of x₁ requires constructing A from its eigenvalues and eigenvectors or directly using them in the calculation.