Chapter 9.3, Problem 56E

Solution

To identify: The first seven terms of the “geometric sequence” which has for its first term the matrix $\left[1a\right]$ and for its common ratio the matrix $r=\left[\begin{array}{l}1d\\ 01\end{array}\right]$ . How is this sequence of matrices related to the arithmetic sequence?

Each of the matrices in the sequence $a_n$ has 1 on the first position and the term $a_n$ of the arithmetic sequence which has the first term $a$ and the common difference $d$ .

Given information:

The given term is

  $\begin{array}{c}{a}_1=\left[1a\right]\\ r=\left[\begin{array}{l}1d\\ 01\end{array}\right]\end{array}$

Explanation:

Consider the given information.

Determine the value of $a_2$ by $a_{1}r$ .

  $\begin{array}{c}{a}_2=a_{1}r\\ a_2=\left[1a\right]\cdot \left[\begin{array}{l}1d\\ 01\end{array}\right]\\ a_2=\left[1\left(1\right)+a\left(0\right)\cdot 1\left(d\right)+a\left(1\right)\right]\\ a_2=\left[1d+a\right]\end{array}$

Determine the value of $a_3$ by $a_{2}r$ .

  $\begin{array}{c}{a}_3=a_{2}r\\ a_3=\left[1d+a\right]\cdot \left[\begin{array}{l}0d\\ 11\end{array}\right]\\ a_3=\left[1\left(0\right)+\left(d+a\right)\left(0\right)\cdot 1\left(d\right)+\left(d+a\right)\left(1\right)\right]\\ a_3=\left[12d+a\right]\end{array}$

Determine the value of $a_4$ by $a_{3}r$.

  $\begin{array}{c}{a}_4=a_{3}r\\ a_4=\left[12d+a\right]\cdot \left[\begin{array}{l}0d\\ 11\end{array}\right]\\ a_4=\left[1\left(1\right)+\left(2d+a\right)\left(0\right)\cdot 1\left(d\right)+\left(2d+a\right)\left(1\right)\right]\\ a_4=\left[13d+a\right]\end{array}$

Determine the value of $a_5$ by $a_{4}r$

  $\begin{array}{c}{a}_5=a_{4}r\\ a_5=\left[13d+a\right]\cdot \left[\begin{array}{l}0d\\ 11\end{array}\right]\\ a_5=\left[1\left(0\right)+\left(3d+a\right)\left(0\right)\cdot 1\left(d\right)+\left(3d+a\right)\left(1\right)\right]\\ a_5=\left[14d+a\right]\end{array}$

Determine the value of $a_6$ by $a_{5}r$

  $\begin{array}{c}{a}_6=a_{5}r\\ a_6=\left[14d+a\right]\cdot \left[\begin{array}{l}0d\\ 11\end{array}\right]\\ a_6=\left[1\left(0\right)+\left(4d+a\right)\left(0\right)\cdot 1\left(d\right)+\left(4d+a\right)\left(1\right)\right]\\ a_6=\left[15d+a\right]\end{array}$

Determine the value of $a_7$ by $a_{6}r$

  $\begin{array}{c}{a}_7=a_{6}r\\ a_7=\left[15d+a\right]\cdot \left[\begin{array}{l}0d\\ 11\end{array}\right]\\ a_7=\left[1\left(0\right)+\left(5d+a\right)\left(0\right)\cdot 1\left(d\right)+\left(5d+a\right)\left(1\right)\right]\\ a_7=\left[16d+a\right]\end{array}$

Each of the matrices in the sequence $a_n$ has 1 on the first position and the term $a_n$ of the arithmetic sequence which has the first term $a$ and the common difference $d$ .

  $a_n=\left[1a_n\right]$