Determining possible values of determinant A

Question

Let A be a 3 x 3 matrix and let 1/8A^3 - 1/2A = 0. What are all possible values of detA?

A) -8, 2, 0

B) -8, 8, 0

C) -4, 2

D) 4, 2

E) 2, 0

Answer 1

Answer: B

Explanation: Suppose A is a 3×3 matrix such that 1/8 A³ − 1/2 A = 0. Then

A³ – 4A = 0, so f(x) = x³ − 4x = x(x + 2)(x − 2) is an annihilating polynomial of A, i.e. f(A) = 0. Since every characteristic value of A must be a zero of any annihilating polynomial of A, all possible characteristic values of A are 0, −2, and 2. There are only the following two cases:

Case 1: One of the 3 characteristic values of A is 0: Since det A equals the product of all characteristic values of A, det A = 0..

Case 2: The 3 characteristic values contain only 2 or −2 or both. Then their product can only be 8 or −8. Thus det A = 8 or −8.

Answer 2

From the given equation (where "0" is the 3x3 zero matrix) we have A³ = 4A.

So, det(A³) = det(4A)

Recall that, det(XY) = (detX)(detY) assuming that X and Y are nxn matrices. So, det(A³) = (detA)³. Also, the 3x3 matrix 4A is obtained from the matrix A by multiplying all the entries in each row of A by 4. Each time a row of A is multiplied by 4, the determinant gets multiplied by 4. So, since A is 3x3 by assumption,

det(4A) = 4³(detA) = 64detA.

Therefore, (detA)³ = 64(detA).

(detA)³ - 64(detA) = 0

detA[(detA)² - 64) = 0

detA = 0 or (detA)² = 64

detA = 0, 8, or -8