Question 7, Exercise 2.2

Solutions of Question 7 of Exercise 2.2 of Unit 02: Matrices and Determinants. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

If $A=\left[\begin{array}{ll}x & 0 \\ y & 1\end{array}\right]$ then prove that for all positive integers $n, A^{n}=\left[\begin{array}{cc}x^{n} & 0 \\ \dfrac{y\left(x^{n}-1\right)}{x-1} & 1\end{array}\right]$.

Solution.

Given: $$A = \begin{bmatrix} x & 0 \\ y & 1 \end{bmatrix}.$$ We use mathematical induction to prove the given fact. For C-1, put $n = 1$ \begin{align}A^1 =\begin{bmatrix} x^1 & 0 \\ \dfrac{y(x^1 - 1)}{x - 1} & 1 \end{bmatrix} \\ = \begin{bmatrix} x & 0 \\ \dfrac{y(x - 1)}{x - 1} & 1 \end{bmatrix}\\ = \begin{bmatrix} x & 0 \\ y & 1 \end{bmatrix}\end{align} C-1 is satisfied.
For C-2, suppose given statement is true for $n=k$. \begin{align} A^k = \begin{bmatrix} x^k & 0 \\ \frac{y(x^k - 1)}{x - 1} & 1 \end{bmatrix}\end{align} We need to show that the formula holds for $ k + 1 $: \begin{align*} A^{k+1} &= A^k \cdot A\\ &= \begin{bmatrix} x^k & 0 \\ \dfrac{y(x^k - 1)}{x - 1} & 1 \end{bmatrix}\begin{bmatrix} x & 0 \\ y & 1 \end{bmatrix}\\ &= \begin{bmatrix} x^k \cdot x + 0 \cdot y & x^k \cdot 0 + 0 \cdot 1 \\ \frac{y(x^k - 1)}{x - 1} \cdot x + 1 \cdot y & \dfrac{y(x^k - 1)}{x - 1} \cdot 0 + 1 \cdot 1 \end{bmatrix}\\ &=\begin{bmatrix} x^{k+1} & 0 \\ \dfrac{y(x^{k+1} - x) + y(x-1)}{x - 1} & 1 \end{bmatrix}\\ &=\begin{bmatrix} x^{k+1} & 0 \\ \dfrac{y(x^{k+1} - x+x-1)}{x - 1} & 1 \end{bmatrix}\\ &=\begin{bmatrix} x^{k+1} & 0 \\ \dfrac{y(x^{k+1} -1)}{x - 1} & 1 \end{bmatrix} \end{align*} This is true for $n=k+1$.
Hence C-2 is satisfied, and the proof is complete.

If $A=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right]$ then prove that for all positive integers $n$, $ A^{n}=\begin{bmatrix} 1+2 n & -4 n
n & 1-2 n \end{bmatrix} $

Solution. we use mathematical induction. Put $n = 1$ \begin{align*} A^1 = \left[\begin{array}{cc} 1 + 2(1) & -4(1) \\ 1(1) & 1 - 2(1) \end{array}\right]\\ = \left[\begin{array}{cc} 1 + 2 & -4 \\ 1 & 1 - 2 \end{array}\right] \\ = \left[\begin{array}{cc} 3 & -4 \\ 1 & -1 \end{array}\right] \end{align*} Assume that the formula holds for some positive integer $k$, i.e., \begin{align*}A^k = \left[\begin{array}{cc} 1 + 2k & -4k \\ k & 1 - 2k \end{array}\right] \end{align*} For $n=k+1$ \begin{align*} A^{k+1} &= A^k A \\ &= \left[\begin{array}{cc} 1 + 2k & -4k \\ k & 1 - 2k \end{array}\right] \left[\begin{array}{cc} 3 & -4 \\ 1 & -1 \end{array}\right]\\ &= \left[\begin{array}{cc} (1 + 2k) \cdot 3 + (-4k) \cdot 1 & (1 + 2k) \cdot (-4) + (-4k) \cdot (-1) \\ k \cdot 3 + (1 - 2k) \cdot 1 & k \cdot (-4) + (1 - 2k) \cdot (-1) \end{array}\right] \\ &= \left[\begin{array}{cc} 3 + 6k - 4k & -4 - 8k + 4k \\ 3k + 1 - 2k & -4k - 1 + 2k\end{array}\right] \\ &= \left[\begin{array}{cc} 3 + 2k & -4k - 4 \\ k + 1 & -2k - 1\end{array}\right]\\ &= \left[\begin{array}{cc} 1 + 2k + 2 & -4k - 4 \\ k + 1 & 1 - 2k - 2 \end{array}\right] \\ &=\left[\begin{array}{cc} 1 + 2(k + 1) & -4(k + 1) \\ k + 1 & 1 - 2(k + 1) \end{array}\right] \end{align*} By mathematical induction, the formula \[ A^n = \left[\begin{array}{cc} 1 + 2n & -4n \\ n & 1 - 2n \end{array}\right] \] holds for all positive integers $n$.

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