If A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} and det. (A^n - I) = 1 - \lambda^n, n \in N then the value of \lambda, is -
Step-by-Step Solution
Key Concept: Calculate A^n for the given matrix A. A^2 = [[2, 2], [2, 2]] = 2A. By induction, A^n = 2^{n-1}A = [[2^{n-1}, 2^{n-1}], [2^{n-1}, 2^{n-1}]]. Then A^n - I = [[2^{n-1}-1, 2^{n-1}], [2^{n-1}, 2^{n-1}-1]]. The determinant is (2^{n-1}-1)^2 - (2^{n-1})^2 = 2^{2n-2} - 2*2^{n-1} + 1 - 2^{2n-2} = 1 - 2^n. Comparing with 1 - \lambda^n, we get \lambda = 2.
Given A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}. A^2 = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix} = 2A. A^3 = A^2 A = 2A^2 = 2(2A) = 4A = 2^2 A. By induction, A^n = 2^{n-1} A = \begin{pmatrix} 2^{n-1} & 2^{n-1} \\ 2^{n-1} & 2^{n-1} \end{pmatrix}. A^n - I = \begin{pmatrix} 2^{n-1}-1 & 2^{n-1} \\ 2^{n-1} & 2^{n-1}-1 \end{pmatrix}. det(A^n - I) = (2^{n-1}-1)^2 - (2^{n-1})^2 = (2^{n-1}-1 - 2^{n-1})(2^{n-1}-1 + 2^{n-1}) = (-1)(2 \cdot 2^{n-1} - 1) = -(2^n - 1) = 1 - 2^n. Given det(A^n - I) = 1 - \lambda^n, so \lambda = 2.
Correct Answer: 2