Let M = \begin{bmatrix} a & -360 \\ b & c \end{bmatrix}, where a, b and c are integers. Find the smallest positive value of b such that M^2 = \mathbf{0}, where \mathbf{0} denotes 2 \times 2 null matrix.
Step-by-Step Solution
Key Concept: For a 2x2 matrix M, M^2 = 0 implies that the trace of M is 0 and the determinant of M is 0. Thus, a + c = 0 (so c = -a) and ac - (-360b) = 0. Substituting c = -a gives -a^2 + 360b = 0, or a^2 = 360b. Since 360 = 36 * 10 = 6^2 * 10, for b to be the smallest positive integer, we need 10b to be a perfect square. The smallest such b is 10.
Given M = \begin{bmatrix} a & -360 \\ b & c \end{bmatrix}. Since M^2 = 0, the characteristic equation is \lambda^2 - tr(M)\lambda + det(M) = 0. By Cayley-Hamilton theorem, M^2 - tr(M)M + det(M)I = 0. For M^2 = 0, we must have tr(M) = 0 and det(M) = 0. Thus, a + c = 0 \implies c = -a. Also, det(M) = ac - (-360b) = -a^2 + 360b = 0 \implies a^2 = 360b. We have 360 = 36 \times 10 = 6^2 \times 10. For a^2 = 6^2 \times 10 \times b to be a perfect square, b must be of the form 10k^2. The smallest positive integer b is obtained for k=1, so b = 10.
Correct Answer: 10