Matrices & Determinants
Matrix Polynomial / Eigenvalues
MJMT_Full_Test_07
Grade 12
Question:
Let $A = \begin{pmatrix}1&1\\1&1\end{pmatrix}$. If $\det(A^n - I) = 1 - \lambda^n$, find $\lambda$.
Step-by-Step Solution
Key Concept: Eigenvalues of $A$ are 0 and 2; $A^n$ has eigenvalues $0^n=0$ and $2^n$. $\det(A^n-I)=(0-1)(2^n-1)=1-2^n$.
$A^n = 2^{n-1}A = \begin{pmatrix}2^{n-1}&2^{n-1}\\2^{n-1}&2^{n-1}\end{pmatrix}$. $A^n - I = \begin{pmatrix}2^{n-1}-1&2^{n-1}\\2^{n-1}&2^{n-1}-1\end{pmatrix}$. $\det = (2^{n-1}-1)^2 - 2^{2(n-1)} = -2^n+1 = 1-2^n$. So $\lambda = 2$.
Correct Answer: 1