Answer
See explanation
Work Step by Step
We will write equation (6) for each iteration as $x_{k}=A_{k} \cdot d,$ where in each iteration we will compute $A$.
By putting matrix $C$ and vector $d$ in Matlab $(7 \times 7 \text { matrix and } 7 \times 1$ vector from exercise 13 ).
Write this code in Matlab to compute $A$ for $k$ -th iteration of $x,$ where $k \geq 2$ :
$I=\operatorname{eye}(7)$
$A=I$
for $i=1: k$ $B=C$
$A=A+B$
$B=B \cdot C ;$
$e n d$
Change $k$ as needed to get matrix $A$ for each iteration, then compute $x_{k}=A_{k} \cdot d$
Like, when $k=5$,
$I=\operatorname{eye}(7)$
$A=I$
for $i=1: 5$ $B=C$
$A=A+B$
$B=B \cdot C ;$
$e n d$
$x=A \cdot d$
Gives us $x_{5}$
D
Results:
$x_{0}=d$
$x_{1}=(89344.2,77730.5,26708.1,72334.7,30325.6,265158.2,9327.8)$
$x_{2}=(94681.2,87714.5,37577.3,100520.5,38598.0,296563.8,11480.0)$
$x_{3}=(97091.9,92573.1,43867.8,115457.0,43491.0,312319.0,12598.8)$
$x_{4}=(98291.6,95033.2,47314.5,123202.5,46247.0,320502.4,13185.5)$
$x_{5}=(98907.2,96305.3,49160.6,127213.7,47756.4,324796.1,13493.8)$
$x_{6}=(99226.6,96969.6,50139.6,129296.7,48569.3,327053.8,13655.9)$
$x_{7}=(99393.1,97317.8,50656.4,130381.6,49002.8,328240.9,13741.1)$
$x_{8}=(99480.0,97500.7,50928.7,130948.0,49232.5,328864.7,13785.9)$
$x_{9}=(99525.5,97596.8,51071.9,131244.1,49353.8,329192.3,13809.4)$
$x_{10}=(99549.4,97647.2,51147.2,131399.2,49417.7,329364.4,13821.7)$
$x_{11}=(99561.9,97673.7,51186.8,131480.4,49451.3,329454.7,13828.2)$
$x_{12}=(99568.4,97687.6,51207.5,131523.0,49469.0,329502.1,13831.6)$
$x_{12}$ is the first vector whose entries are accurate to the nearest thousand.
Because we started with $x_{0},$ it takes $12+1=13$ steps to obtain the answer.
