If A = $\begin{bmatrix} 1&4 \\ 1&1 \end{bmatrix} $, find $A^7 + 31A^2 + I$

Subject: Applied Mathematics 4

Topic: Matrix Theory

Difficulty: Medium

For characteristic equation |A - $\lambda$ I| = 0

$$ \begin{vmatrix} 1-\lambda&4 \\ 1&1-\lambda \end{vmatrix} = 0 $$

$ (1-\lambda)(1-\lambda) -4 \\ 1 -2 \lambda + \lambda^2 - 4 = 0 \\ \lambda^2 - 2 \lambda - 3 = 0 $

By Cayley-HAmilton theorem, $ A^2 - 2A - 3I = 0 $

$ \therefore A^7 + 31A^2 + I \\ = (A^2 - 2A - 3I)(A^5 + 2A^4 + 7A^3 + 20A^2 + 61A + 213I) + (509A + 639I) \\ = 509A + 639 I \\ = 509 \begin{bmatrix} 1 & 4 \\ 1 & 1 \end{bmatrix} + \begin{bmatrix} 639 & 0 \\ 0 & 639 \end{bmatrix} \\ = \begin{bmatrix} 1148 & 2036 \\ 509 & 1148 \end{bmatrix} $