Question: User: (1) (1.1: Modeling) Matt is a software engineer writing a script involving 6 tasks. Each must be done one after the other. Let $t_i$ be the time needed to complete the $i$th task. These times have a certain structure:
•The total time needed to complete any 3 adjacent tasks is half the total time required to complete 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 will take 20 seconds and the first three tasks together will take 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).
1. **State the problem:** We have 6 tasks with times $t_1, t_2, t_3, t_4, t_5, t_6$. The conditions are:
- The total time for any 3 adjacent tasks is half the total time for the next 2 tasks.
- $t_2 = 1$
- $t_4 = 10$
- Additional info later: $t_6 = 20$ and $t_1 + t_2 + t_3 = 50$
2. **Write equations from the first condition:** For $i=1$ to $4$ (since 3 adjacent tasks and next 2 tasks must fit in range):
$$t_i + t_{i+1} + t_{i+2} = \frac{1}{2}(t_{i+3} + t_{i+4})$$
Specifically:
- For $i=1$: $t_1 + t_2 + t_3 = \frac{1}{2}(t_4 + t_5)$
- For $i=2$: $t_2 + t_3 + t_4 = \frac{1}{2}(t_5 + t_6)$
- For $i=3$: $t_3 + t_4 + t_5 = \frac{1}{2}(t_6 + t_7)$ but $t_7$ does not exist, so only up to $i=2$ is valid.
So only two equations from this condition:
$$t_1 + t_2 + t_3 = \frac{1}{2}(t_4 + t_5)$$
$$t_2 + t_3 + t_4 = \frac{1}{2}(t_5 + t_6)$$
3. **Rewrite equations to standard form:** Multiply both sides by 2 to clear fractions:
$$2t_1 + 2t_2 + 2t_3 = t_4 + t_5$$
$$2t_2 + 2t_3 + 2t_4 = t_5 + t_6$$
Bring all terms to one side:
$$2t_1 + 2t_2 + 2t_3 - t_4 - t_5 = 0$$
$$0t_1 + 2t_2 + 2t_3 + 2t_4 - t_5 - t_6 = 0$$
4. **Known values:**
$$t_2 = 1$$
$$t_4 = 10$$
5. **Augmented matrix for (a):** Variables in order $t_1, t_2, t_3, t_4, t_5, t_6$
\[
\begin{bmatrix}
2 & 2 & 2 & -1 & -1 & 0 & | & 0 \\
0 & 2 & 2 & 2 & -1 & -1 & | & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & | & 1 \\
0 & 0 & 0 & 1 & 0 & 0 & | & 10
\end{bmatrix}
\]
6. **Reduce to reduced echelon form (b):**
- From row 3: $t_2 = 1$
- From row 4: $t_4 = 10$
Substitute $t_2=1$, $t_4=10$ into first two equations:
- Eq1: $2t_1 + 2(1) + 2t_3 - 10 - t_5 = 0 \Rightarrow 2t_1 + 2t_3 - t_5 = 8$
- Eq2: $2(1) + 2t_3 + 2(10) - t_5 - t_6 = 0 \Rightarrow 2t_3 - t_5 - t_6 = -22$
Rewrite system:
$$2t_1 + 2t_3 - t_5 = 8$$
$$2t_3 - t_5 - t_6 = -22$$
7. **Add new info (c):**
$$t_6 = 20$$
$$t_1 + t_2 + t_3 = 50$$
Add rows:
$$0,0,0,0,0,1 | 20$$
$$1,1,1,0,0,0 | 50$$
8. **Solve system (d):**
From $t_6=20$ and $t_2=1$, $t_4=10$ known.
From second reduced equation:
$$2t_3 - t_5 - 20 = -22 \Rightarrow 2t_3 - t_5 = -2$$
From first reduced equation:
$$2t_1 + 2t_3 - t_5 = 8$$
From sum equation:
$$t_1 + 1 + t_3 = 50 \Rightarrow t_1 + t_3 = 49$$
Express $t_1 = 49 - t_3$
Substitute $t_1$ into first equation:
$$2(49 - t_3) + 2t_3 - t_5 = 8 \Rightarrow 98 - 2t_3 + 2t_3 - t_5 = 8 \Rightarrow 98 - t_5 = 8 \Rightarrow t_5 = 90$$
Use $t_5=90$ in second equation:
$$2t_3 - 90 = -2 \Rightarrow 2t_3 = 88 \Rightarrow t_3 = 44$$
Then $t_1 = 49 - 44 = 5$
**Final solution:**
$$t_1 = 5, t_2 = 1, t_3 = 44, t_4 = 10, t_5 = 90, t_6 = 20$$