- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
2.16 Exercises
9. $\quad$ Let $A = \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix}.$ Prove that $A^2 = 2A - I$ and compute $A^{100}.$
Solution. $\quad$ We have \begin{align*} A^2 &= \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix} \\ &= \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix} \\ &= \begin{bmatrix} 2 & 0 \\ -2 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ &= 2A - I \end{align*} To compute $A^{100},$ we will first compute $A^3$ and $A^4$ to determine a possible pattern applicable to all $n \geq 1.$ \begin{align*} A^3 &= \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix} \\ &= \begin{bmatrix} 1 & 0 \\ -3 & 1 \end{bmatrix} \\ A^4 &= \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -3 & 1 \end{bmatrix} \\ &= \begin{bmatrix} 1 & 0 \\ -4 & 1 \end{bmatrix} \end{align*} For $n = 1, 2, 3, 4,$ we have the following general equation: \begin{align*} A^n &= \begin{bmatrix} 1 & 0 \\ -n & 1 \end{bmatrix} \end{align*} Now, we will prove by induction that it holds for all integers $n \geq 1.$ Assume that this equation holds for $n \geq 4.$ We then have: \begin{align*} A^{n + 1} &= \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -n & 1 \end{bmatrix} \\ &= \begin{bmatrix} 1 & 0 \\ -(n + 1) & 1 \end{bmatrix} \end{align*} But since this holds for $n + 1,$ by induction, it is true for any integer $n \geq 1,$ which gives us \begin{align*} A^{100} &= \begin{bmatrix} 1 & 0 \\ -100 & 1 \end{bmatrix} \quad \blacksquare \end{align*}