US2563
### Linear Algebra - Homework 1
**Due: Saturday, July 13, by 11:59 pm, via Gradescope**
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### Instructions:
1. Assign a page to each problem you submit. Refer to Gradescope's YouTube videos for submission guidelines.
2. Late homework is not accepted. Don't ask.
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### Problem 1: Solving Systems of Equations using Augmented Matrices and EROs
#### (a) Solve the system:
\[
\begin{cases}
4x - 9y - 12z = 1 \\
3x - 4y - 11z = 2 \\
-x + 9y - 2z = 3
\end{cases}
\]
**Step 1: Construct the augmented matrix:**
\[
\left[\begin{array}{ccc|c}
4 & -9 & -12 & 1 \\
3 & -4 & -11 & 2 \\
-1 & 9 & -2 & 3
\end{array}\right]
\]
**Step 2: Use elementary row operations (EROs) to get the matrix in Row Echelon Form (REF):**
1. \( R_2 \leftarrow R_2 - \frac{3}{4}R_1 \)
2. \( R_3 \leftarrow R_3 + \frac{1}{4}R_1 \)
\[
\left[\begin{array}{ccc|c}
4 & -9 & -12 & 1 \\
0 & \frac{11}{4} & -\frac{17}{4} & \frac{5}{4} \\
0 & \frac{27}{4} & -\frac{41}{4} & \frac{13}{4}
\end{array}\right]
\]
3. \( R_3 \leftarrow R_3 - \frac{27}{11}R_2 \)
\[
\left[\begin{array}{ccc|c}
4 & -9 & -12 & 1 \\
0 & \frac{11}{4} & -\frac{17}{4} & \frac{5}{4} \\
0 & 0 & -\frac{10}{11} & \frac{2}{11}
\end{array}\right]
\]
4. \( R_3 \leftarrow -\frac{11}{10}R_3 \)
\[
\left[\begin{array}{ccc|c}
4 & -9 & -12 & 1 \\
0 & \frac{11}{4} & -\frac{17}{4} & \frac{5}{4} \\
0 & 0 & 1 & -\frac{1}{5}
\end{array}\right]
\]
5. \( R_2 \leftarrow R_2 + \frac{17}{4}R_3 \)
\[
\left[\begin{array}{ccc|c}
4 & -9 & -12 & 1 \\
0 & \frac{11}{4} & 0 & \frac{3}{2} \\
0 & 0 & 1 & -\frac{1}{5}
\end{array}\right]
\]
6. \( R_2 \leftarrow \frac{4}{11}R_2 \)
\[
\left[\begin{array}{ccc|c}
4 & -9 & -12 & 1 \\
0 & 1 & 0 & \frac{6}{11} \\
0 & 0 & 1 & -\frac{1}{5}
\end{array}\right]
\]
7. \( R_1 \leftarrow R_1 + 9R_2 \)
8. \( R_1 \leftarrow R_1 + 12R_3 \)
\[
\left[\begin{array}{ccc|c}
4 & 0 & 0 & \frac{16}{11} \\
0 & 1 & 0 & \frac{6}{11} \\
0 & 0 & 1 & -\frac{1}{5}
\end{array}\right]
\]
9. \( R_1 \leftarrow \frac{1}{4}R_1 \)
\[
\left[\begin{array}{ccc|c}
1 & 0 & 0 & \frac{4}{11} \\
0 & 1 & 0 & \frac{6}{11} \\
0 & 0 & 1 & -\frac{1}{5}
\end{array}\right]
\]
**Solution:**
\[
\boxed{x = \frac{4}{11}, \; y = \frac{6}{11}, \; z = -\frac{1}{5}}
\]
---
#### (b) Solve the system:
\[
\begin{cases}
-7x + 7y - z = 4 \\
x - 11y + 6z = 5 \\
-2x - 7y + 5z = 6
\end{cases}
\]
**Step 1: Construct the augmented matrix:**
\[
\left[\begin{array}{ccc|c}
-7 & 7 & -1 & 4 \\
1 & -11 & 6 & 5 \\
-2 & -7 & 5 & 6
\end{array}\right]
\]
**Step 2: Use elementary row operations (EROs) to get the matrix in Row Echelon Form (REF):**
1. \( R_1 \leftarrow -\frac{1}{7}R_1 \)
\[
\left[\begin{array}{ccc|c}
1 & -1 & \frac{1}{7} & -\frac{4}{7} \\
1 & -11 & 6 & 5 \\
-2 & -7 & 5 & 6
\end{array}\right]
\]
2. \( R_2 \leftarrow R_2 - R_1 \)
3. \( R_3 \leftarrow R_3 + 2R_1 \)
\[
\left[\begin{array}{ccc|c}
1 & -1 & \frac{1}{7} & -\frac{4}{7} \\
0 & -10 & \frac{41}{7} & \frac{39}{7} \\
0 & -9 & \frac{39}{7} & \frac{38}{7}
\end{array}\right]
\]
4. \( R_2 \leftarrow -\frac{1}{10}R_2 \)
\[
\left[\begin{array}{ccc|c}
1 & -1 & \frac{1}{7} & -\frac{4}{7} \\
0 & 1 & -\frac{41}{70} & -\frac{39}{70} \\
0 & -9 & \frac{39}{7} & \frac{38}{7}
\end{array}\right]
\]
5. \( R_3 \leftarrow R_3 + 9R_2 \)
\[
\left[\begin{array}{ccc|c}
1 & -1 & \frac{1}{7} & -\frac{4}{7} \\
0 & 1 & -\frac{41}{70} & -\frac{39}{70} \\
0 & 0 & \frac{90}{7} & \frac{71}{7}
\end{array}\right]
\]
6. \( R_3 \leftarrow \frac{7}{90}R_3 \)
\[
\left[\begin{array}{ccc|c}
1 & -1 & \frac{1}{7} & -\frac{4}{7} \\
0 & 1 & -\frac{41}{70} & -\frac{39}{70} \\
0 & 0 & 1 & \frac{71}{90}
\end{array}\right]
\]
7. \( R_2 \leftarrow R_2 + \frac{41}{70}R_3 \)
\[
\left[\begin{array}{ccc|c}
1 & -1 & \frac{1}{7} & -\frac{4}{7} \\
0 & 1 & 0 & -\frac{1}{2} \\
0 & 0 & 1 & \frac{71}{90}
\end{array}\right]
\]
8. \( R_1 \leftarrow R_1 - \frac{1}{7}R_3 \)
9. \( R_1 \leftarrow R_1 + R_2 \)
\[
\left[\begin{array}{ccc|c}
1 & 0 & 0 & \frac{1}{18} \\
0 & 1 & 0 & -\frac{1}{2} \\
0 & 0 & 1 & \frac{71}{90}
\end{array}\right]
\]
**Solution:**
\[
\boxed{x = \frac{1}{18}, \; y = -\frac{1}{2}, \; z = \frac{71}{90}}
\]
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### Problem 2: Proof of Unique Solution
**Prove that it is impossible for the equation \( A\vec{x} = \vec{b} \) to have exactly two distinct solutions.**
**Proof:**
Assume \( \vec{x