Chapter 8.3, Problem 66E

(a)

To determine

To Find: The system of linear equations for the data.

Answer to Problem 66E

The required system of linear equation is $\{\begin{array}{l}144a+12b+c=1474\\ 169a+13b+c=1807\\ 196a+14b+c=2188\end{array}$ .

Explanation of Solution

Given:

The given data is shown in Table 1

Table 1

Year	Visitor, y in thousand
2012	1474
2013	1807
2014	2188

The given quadratic equation is,

  $y=a{t}^2+bt+c$

The year is $t=12$ corresponds to 2012.

Calculation:

Consider the given equation is,

  $y=a{t}^2+bt+c$

Then, from the above equation and from the given Table, the system of linear equation is,

  $\{\begin{array}{l}144a+12b+c=1474\\ 169a+13b+c=1807\\ 196a+14b+c=2188\end{array}$

(b)

To determine

To Find: The inverse of the coefficient matrix of the system from part (a).

Answer to Problem 66E

The inverse of the coefficient matrix is $\left[\begin{array}{ccc}\frac{1}{2}& -1& \frac{1}{2}\\ \frac{-27}{2}& 26& \frac{-25}{2}\\ 91& -168& 78\end{array}\right]$ .

Explanation of Solution

Consider the given system of linear equations is,

  $\{\begin{array}{l}144a+12b+c=1474\\ 169a+13b+c=1807\\ 196a+14b+c=2188\end{array}$

From above set of equations, the required matrix is,

  $\left[\begin{array}{ccc}144& 12& 1\\ 169& 13& 1\\ 196& 14& 1\end{array}\right]$

The inverse of the above matrix is,

  $\left[\begin{array}{ccc}\frac{1}{2}& -1& \frac{1}{2}\\ \frac{-27}{2}& 26& \frac{-25}{2}\\ 91& -168& 78\end{array}\right]$

(c)

To determine

To Find: The solution for the system and then write the model in the form of $a{t}^2+bt+c$ .

Answer to Problem 66E

The solution for the system of equation is $\left[\begin{array}{c}a\\ b\\ c\end{array}\right]=\left[\begin{array}{c}24\\ -267\\ 1222\end{array}\right]$ and the required equation of the parabola is $y=24t^2-267t+1222$ .

Explanation of Solution

Consider the inverse of the coefficient matrix written in the form $X=A^{-1}B$ as,

  $\left[\begin{array}{c}a\\ b\\ c\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{2}& -1& \frac{1}{2}\\ \frac{-27}{2}& 26& \frac{-25}{2}\\ 91& -168& 78\end{array}\right]\left[\begin{array}{c}1474\\ 1807\\ 2188\end{array}\right]$

By the use of graphing utility the results are,

  $\left[\begin{array}{c}a\\ b\\ c\end{array}\right]=\left[\begin{array}{c}24\\ -267\\ 1222\end{array}\right]$

Thus, the required equation of the parabola is $y=24t^2-267t+1222$ .

(d)

To determine

To Find: The graph for the model with the given data.

Answer to Problem 66E

The required graph for the parabola is shown in Figure 1

Explanation of Solution

Consider the required equation of the parabola is $y=24t^2-267t+1222$ .

Then, from the given model the graph for the parabola is shown in Figure 1

  

Figure 1