Matrices & Determinants
Matrix Powers
Grade 12
Question:
<p>If <span>A</span> = <span>\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\end{pmatrix}</span> and <span>\det(A^n - I) = 1 - \lambda^n, n \in \mathbb{N}</span>, then the value of <span>\lambda</span> is</p>
<p>(a) 1</p>
<p>(b) 2</p>
<p>(c) 3</p>
<p>(d) 4</p>
Step-by-Step Solution
Key Concept: Calculate powers of matrix A using the pattern that emerges, then compute the determinant of (A^n - I) to identify λ.
<p><strong>Solution:</strong></p><p>Given <span>A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}</span></p><p><span>A^2 = \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix} = 2A</span></p><p><span>A^3 = A^2 \cdot A = 2A \cdot A = 2A^2 = 2^2 A</span></p><p>Similarly, <span>A^n = 2^{n-1} A</span></p><p><span>A^n - I = \begin{pmatrix} 2^{n-1} - 1 & 2^{n-1} \\ 2^{n-1} & 2^{n-1} - 1 \end{pmatrix}</span></p><p><span>\det(A^n - I) = (2^{n-1} - 1)^2 - (2^{n-1})^2 = 1 - 2^n = 1 - \lambda^n</span></p><p>Therefore, <span>\lambda = 2</span></p><p>∴ Answer is <strong>(b) 2</strong></p>
Correct Answer: B