The trace of a square matrix is defined to be the sum of its diagonal entries. If A is a 2x2 matrix such that the trace of A is 3 and the trace of A^3 is -18, then the value of the determinant of A is
Step-by-Step Solution
Key Concept: Let A = [[a, b], [c, d]]. Trace(A) = a+d = 3. Characteristic equation is A^2 - Trace(A)A + Det(A)I = 0. A^2 = 3A - Det(A)I. Multiply by A: A^3 = 3A^2 - Det(A)A = 3(3A - Det(A)I) - Det(A)A = (9 - Det(A))A - 3Det(A)I. Trace(A^3) = (9 - Det(A))Trace(A) - 3Det(A)Trace(I) = (9 - Det(A))(3) - 3Det(A)(2) = 27 - 3Det(A) - 6Det(A) = 27 - 9Det(A). Given Trace(A^3) = -18, so 27 - 9Det(A) = -18 => 9Det(A) = 45 => Det(A) = 5.
Let A be a 2x2 matrix. Let its eigenvalues be \lambda1 and \lambda2. Trace(A) = \lambda1 + \lambda2 = 3. Trace(A^3) = \lambda1^3 + \lambda2^3 = -18. We know \lambda1^3 + \lambda2^3 = (\lambda1 + \lambda2)^3 - 3\lambda1\lambda2(\lambda1 + \lambda2). Substituting values: -18 = (3)^3 - 3(Det(A))(3) => -18 = 27 - 9Det(A) => 9Det(A) = 45 => Det(A) = 5.
Correct Answer: 4