To solve the problem, we need to find the coordinates of points L and M from the equations \( AX = C \) and \( BX = D \), and then find their reflections \( L' \) and \( M' \) in the plane defined by \( x + y + z = 9 \).
### Step 1: Solve \( AX = C \)
Given:
\[
A = \begin{pmatrix}
1 & 2 & 3 \\
4 & 1 & 2 \\
1 & -1 & 1
\end{pmatrix}, \quad C = \begin{pmatrix}
14 \\
12 \\
2
\end{pmatrix}
\]
We need to solve the system of equations represented by \( AX = C \).
1. Write the augmented matrix:
\[
\begin{pmatrix}
1 & 2 & 3 & | & 14 \\
4 & 1 & 2 & | & 12 \\
1 & -1 & 1 & | & 2
\end{pmatrix}
\]
2. Perform row operations to reduce this matrix to row echelon form.
- Replace \( R_2 \) with \( R_2 - 4R_1 \):
\[
R_2 = (4 - 4 \cdot 1, 1 - 4 \cdot 2, 2 - 4 \cdot 3 | 12 - 4 \cdot 14) = (0, -7, -10 | -40)
\]
- Replace \( R_3 \) with \( R_3 - R_1 \):
\[
R_3 = (1 - 1, -1 - 2, 1 - 3 | 2 - 14) = (0, -3, -2 | -12)
\]
The matrix becomes:
\[
\begin{pmatrix}
1 & 2 & 3 & | & 14 \\
0 & -7 & -10 & | & -40 \\
0 & -3 & -2 & | & -12
\end{pmatrix}
\]
3. Now, simplify \( R_2 \) and \( R_3 \):
- Divide \( R_2 \) by -7:
\[
R_2 = (0, 1, \frac{10}{7} | \frac{40}{7})
\]
- Replace \( R_3 \) with \( R_3 + 3R_2 \):
\[
R_3 = (0, 0, \frac{16}{7} | \frac{12 + 120/7}{7}) = (0, 0, \frac{16}{7} | \frac{84}{7}) = (0, 0, 1 | 5)
\]
4. The final augmented matrix is:
\[
\begin{pmatrix}
1 & 2 & 3 & | & 14 \\
0 & 1 & \frac{10}{7} & | & \frac{40}{7} \\
0 & 0 & 1 & | & 5
\end{pmatrix}
\]
5. Back-substituting to find \( x, y, z \):
- From \( z = 5 \)
- Substitute \( z \) into \( R_2 \):
\[
y + \frac{10}{7} \cdot 5 = \frac{40}{7} \implies y + \frac{50}{7} = \frac{40}{7} \implies y = -\frac{10}{7}
\]
- Substitute \( y \) and \( z \) into \( R_1 \):
\[
x + 2(-\frac{10}{7}) + 3(5) = 14 \implies x - \frac{20}{7} + 15 = 14 \implies x = 14 - 15 + \frac{20}{7} = \frac{20}{7} - 1 = \frac{13}{7}
\]
Thus, the coordinates of point \( L \) are:
\[
L = \left( \frac{13}{7}, -\frac{10}{7}, 5 \right)
\]
### Step 2: Solve \( BX = D \)
Given:
\[
B = \begin{pmatrix}
2 & 1 & 3 \\
4 & 1 & -1 \\
2 & 2 & 3
\end{pmatrix}, \quad D = \begin{pmatrix}
13 \\
11 \\
14
\end{pmatrix}
\]
1. Write the augmented matrix:
\[
\begin{pmatrix}
2 & 1 & 3 & | & 13 \\
4 & 1 & -1 & | & 11 \\
2 & 2 & 3 & | & 14
\end{pmatrix}
\]
2. Perform row operations to reduce this matrix to row echelon form.
- Replace \( R_2 \) with \( R_2 - 2R_1 \):
\[
R_2 = (0, -1, -7 | -15)
\]
- Replace \( R_3 \) with \( R_3 - R_1 \):
\[
R_3 = (0, 1, 0 | 1)
\]
The matrix becomes:
\[
\begin{pmatrix}
2 & 1 & 3 & | & 13 \\
0 & -1 & -7 & | & -15 \\
0 & 1 & 0 & | & 1
\end{pmatrix}
\]
3. Now, simplify \( R_2 \) and \( R_3 \):
- Multiply \( R_2 \) by -1:
\[
R_2 = (0, 1, 7 | 15)
\]
4. The final augmented matrix is:
\[
\begin{pmatrix}
2 & 1 & 3 & | & 13 \\
0 & 1 & 7 & | & 15 \\
0 & 0 & 1 & | & 1
\end{pmatrix}
\]
5. Back-substituting to find \( x, y, z \):
- From \( z = 1 \)
- Substitute \( z \) into \( R_2 \):
\[
y + 7(1) = 15 \implies y = 8
\]
- Substitute \( y \) and \( z \) into \( R_1 \):
\[
2x + 1(8) + 3(1) = 13 \implies 2x + 8 + 3 = 13 \implies 2x = 2 \implies x = 1
\]
Thus, the coordinates of point \( M \) are:
\[
M = (1, 8, 1)
\]
### Step 3: Find Reflections \( L' \) and \( M' \)
The plane equation is \( x + y + z = 9 \).
1. For point \( L \):
- Coordinates of \( L \) are \( \left( \frac{13}{7}, -\frac{10}{7}, 5 \right) \).
- Substitute into the plane equation:
\[
d = \frac{L_x + L_y + L_z - 9}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{\frac{13}{7} - \frac{10}{7} + 5 - 9}{\sqrt{3}} = \frac{\frac{3}{7} - 4}{\sqrt{3}} = \frac{\frac{3 - 28}{7}}{\sqrt{3}} = \frac{-25/7}{\sqrt{3}} = -\frac{25}{7\sqrt{3}}
\]
- The reflection point \( L' \) is given by:
\[
L' = L + 2d \cdot \text{normal vector}
\]
- Normal vector is \( (1, 1, 1) \), so:
\[
L' = \left( \frac{13}{7}, -\frac{10}{7}, 5 \right) + 2 \left(-\frac{25}{7\sqrt{3}}\right)(1, 1, 1)
\]
2. For point \( M \):
- Coordinates of \( M \) are \( (1, 8, 1) \).
- Substitute into the plane equation:
\[
d = \frac{1 + 8 + 1 - 9}{\sqrt{3}} = \frac{1}{\sqrt{3}}
\]
- The reflection point \( M' \) is given by:
\[
M' = M + 2d \cdot \text{normal vector}
\]
- So:
\[
M' = (1, 8, 1) + 2 \left(\frac{1}{\sqrt{3}}\right)(1, 1, 1)
\]
### Final Coordinates
Thus, the coordinates of points are:
- \( L = \left( \frac{13}{7}, -\frac{10}{7}, 5 \right) \)
- \( M = (1, 8, 1) \)
- \( L' \) and \( M' \) can be calculated using the reflection formulas as shown.
