Matrices & Determinants
Matrix multiplication
Grade Class 12
Question:
Let M and N be two 3 x 3 matrices such that MN = NM. Further, if M != N^2 and M^2 = N^4, then
(A) determinant of (M^2 + MN^2) is 0
(B) there is a 3 x 3 non-zero matrix U such that (M^2 + MN^2)U is zero matrix
(C) determinant of (M^2 + MN^2) >= 1
(D) for a 3 x 3 matrix U, if (M^2 + MN^2)U equals the zero matrix then U is the zero matrix
Step-by-Step Solution
Key Concept: Given M^2 = N^4, we have M^2 - N^4 = 0, which factors as (M - N^2)(M + N^2) = 0. Since MN = NM, the matrices commute. The expression M^2 + MN^2 = M(M + N^2). Since M^2 = N^4, M^2 - N^4 = (M - N^2)(M + N^2) = 0. If M != N^2, then M - N^2 is not necessarily zero, but the product is zero, implying the determinant of M^2 + MN^2 is 0.
Given M^2 = N^4 and MN = NM. Then M^2 - N^4 = 0, so (M - N^2)(M + N^2) = 0. Since M != N^2, the matrix M^2 + MN^2 = M(M + N^2) must have a determinant of 0 because the product of (M - N^2) and (M + N^2) is zero. Thus, there exists a non-zero matrix U such that (M^2 + MN^2)U = 0.
Correct Answer: 1, 2