A Level Further Mathematics Core Pure Practice Test

For A = [[4,1],[0,2]], which vector is an eigenvector corresponding to λ = 2?

(1,0) (0,1) (1,1) (1,-2)

Correct Answer:

(1,-2)

Explanation:

To find an eigenvector for a given eigenvalue, solve Av = λv, which is the same as (A − λI)v = 0. Here A − 2I = [[4−2, 1], [0, 2−2]] = [[2, 1], [0, 0]]. The equation (A − 2I)v = 0 gives 2x + y = 0, so y = −2x. Thus any nonzero vector of the form (x, −2x) is an eigenvector for λ = 2, i.e., multiples of (1, −2). The vector (1, −2) satisfies A(1, −2) = (2, −4) = 2(1, −2), confirming it is an eigenvector with eigenvalue 2. Other listed vectors don’t satisfy Av = 2v.

Access more questions from this quiz

Continue to A Level Further Mathematics Core Pure Practice Test for more practice questions and the full quiz experience.

Access More Questions