Let \[ A=\begin{pmatrix} 3/2&1/2\\ -1/2&1/2 \end{pmatrix}. \] Calculate \(A^{100}\) and \(A^{-7}\).
Solution
Set \[ N=A-I = \frac12 \begin{pmatrix} 1&1\\ -1&-1 \end{pmatrix}. \]
1. One has\[ N^2=0. \]
Direct multiplication gives \[ \begin{pmatrix} 1&1\\ -1&-1 \end{pmatrix}^2 = \begin{pmatrix} 0&0\\ 0&0 \end{pmatrix}. \] Hence \(N^2=0\).
2. For every integer \(m\),\[ A^m=I+mN. \]
For \(m\geq0\), the binomial theorem and step 1 give \[ (I+N)^m=I+mN. \]
Also, \[ (I+N)(I-N)=I-N^2=I, \] so \[ A^{-1}=I-N. \] If \(m=-k<0\), then \[ A^m=(I-N)^k. \] Again using \(N^2=0\), \[ (I-N)^k=I-kN=I+mN. \] Thus the formula holds for every integer \(m\).
3. One has\[ \boxed{ A^{100} = \begin{pmatrix} 51&50\\ -50&-49 \end{pmatrix} }. \]
By step 2, \[ \begin{aligned} A^{100} &= I+100N\\ &= \begin{pmatrix} 1&0\\ 0&1 \end{pmatrix} + 50 \begin{pmatrix} 1&1\\ -1&-1 \end{pmatrix}\\ &= \begin{pmatrix} 51&50\\ -50&-49 \end{pmatrix}. \end{aligned} \]
4. One has\[ \boxed{ A^{-7} = \begin{pmatrix} -5/2&-7/2\\ 7/2&9/2 \end{pmatrix} }. \]
By step 2, \[ \begin{aligned} A^{-7} &= I-7N\\ &= \begin{pmatrix} 1&0\\ 0&1 \end{pmatrix} - \frac72 \begin{pmatrix} 1&1\\ -1&-1 \end{pmatrix}\\ &= \begin{pmatrix} -5/2&-7/2\\ 7/2&9/2 \end{pmatrix}. \end{aligned} \]
5. Q.E.D.
Steps 3 and 4 are the requested calculations.