contestada

Matt is a software engineer writing a script involving 6 tasks. Each must be done one after the other. Let ti be the time for the ith task. These times have a certain structure:


•Any 3 adjacent tasks will take half as long as the next two tasks.


•The second task takes 1 second.


•The fourth task takes 10 seconds.


a) Write an augmented matrix for the system of equations describing the length of each task.


b) Reduce this augmented matrix to reduced echelon form.


c) Suppose he knows additionally that the sixth task takes 20 seconds and the first three tasks will run in 50 seconds. Write the extra rows that you would add to your answer in b) to take account of this new information.


d) Solve the system of equations in c).

Respuesta :

Answer:

Let [tex]t_i[/tex] be the time for the [tex]i[/tex]th task.

We know these times have a certain structure:

  • Any 3 adjacent tasks will take half as long as the next two tasks.

In the form of an equations we have

[tex]t_1+t_2+t_3=\frac{1}{2}t_4+\frac{1}{2}t_5 \\\\t_2+t_3+t_4=\frac{1}{2}t_5+\frac{1}{2}t_6[/tex]

  • The second task takes 1 second [tex]t_2=1[/tex]
  • The fourth task takes 10 seconds [tex]t_4=10[/tex]

So, we have the following system of equations:

[tex]t_1+t_2+t_3-\frac{1}{2}t_4-\frac{1}{2}t_5=0 \\\\t_2+t_3+t_4-\frac{1}{2}t_5-\frac{1}{2}t_6=0\\\\t_2=1\\\\t_4=10[/tex]

a) An augmented matrix for a system of equations is a matrix of numbers in which each row represents the constants from one equation (both the coefficients and the constant on the other side of the equal sign) and each column represents all the coefficients for a single variable.

Here is the augmented matrix for this system.

[tex]\left[ \begin{array}{cccccc|c} 1 & 1 & 1 & - \frac{1}{2} & - \frac{1}{2} & 0 & 0 \\\\ 0 & 1 & 1 & 1 & - \frac{1}{2} & - \frac{1}{2} & 0 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \end{array} \right][/tex]

b) To reduce this augmented matrix to reduced echelon form, you must use these row operations.

  • Subtract row 2 from row 1 [tex]\left(R_1=R_1-R_2\right)[/tex].
  • Subtract row 2 from row 3 [tex]\left(R_3=R_3-R_2\right)[/tex].
  • Add row 3 to row 2 [tex]\left(R_2=R_2+R_3\right)[/tex].
  • Multiply row 3 by −1 [tex]\left({R}_{{3}}=-{1}\cdot{R}_{{3}}\right)[/tex].
  • Add row 4 multiplied by [tex]\frac{3}{2}[/tex] to row 1 [tex]\left(R_1=R_1+\left(\frac{3}{2}\right)R_4\right)[/tex].
  • Subtract row 4 from row 3 [tex]\left(R_3=R_3-R_4\right)[/tex].

Here is the reduced echelon form for the augmented matrix.

[tex]\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & \frac{1}{2} & 15 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & - \frac{1}{2} & - \frac{1}{2} & -11 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \end{array} \right][/tex]

c) The additional rows are

[tex]\begin{array}{ccccccc} 0 & 0 & 0 & 0 & 0 & 1 & 20 \\\\ 1 & 1 & 1 & 0 & 0 & 0 & 50 \end{array} \right[/tex]

and the augmented matrix is

[tex]\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & \frac{1}{2} & 15 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & - \frac{1}{2} & - \frac{1}{2} & -11 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \\\\ 0 & 0 & 0 & 0 & 0 & 1 & 20 \\\\ 1 & 1 & 1 & 0 & 0 & 0 & 50 \end{array} \right][/tex]

d) To solve the system you must use these row operations.

  • Subtract row 1 from row 6 [tex]\left(R_6=R_6-R_1\right)[/tex].
  • Subtract row 2 from row 6 [tex]\left(R_6=R_6-R_2\right)[/tex].
  • Subtract row 3 from row 6 [tex]\left(R_6=R_6-R_3\right)[/tex].
  • Swap rows 5 and 6.
  • Add row 5 to row 3 [tex]\left(R_3=R_3+R_5\right)[/tex].
  • Multiply row 5 by 2 [tex]\left(R_5=\left(2\right)R_5\right)[/tex].
  • Subtract row 6 multiplied by 1/2 from row 1 [tex]\left(R_1=R_1-\left(\frac{1}{2}\right)R_6\right)[/tex].
  • Add row 6 multiplied by 1/2 to row 3 [tex]\left(R_3=R_3+\left(\frac{1}{2}\right)R_6\right)[/tex].

[tex]\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & 0 & 0 & 44 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \\\\ 0 & 0 & 0 & 0 & 1 & 0 & 90 \\\\ 0 & 0 & 0 & 0 & 0 & 1 & 20 \end{array} \right][/tex]

The solutions are: [tex](t_1,...,t_6)=(5,1,44,10,90,20)[/tex].

Otras preguntas