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In Exercises 16-18, determine whether the given [tex]\(2 \times 3\)[/tex] system of linear equations represents coincident planes (that is, the same plane), two parallel planes, or two planes whose intersection is a line. In the latter case, give the parametric equations for the line; that is, give equations of the form [tex]\(x = at + b\)[/tex], [tex]\(y = ct + d\)[/tex], [tex]\(z = et + f\)[/tex].

16.
[tex]\[ 2x_1 + x_2 + x_3 = 3 \][/tex]

17.
[tex]\[
\begin{aligned}
x_1 + 2x_2 - x_3 &= 2 \\
-2x_1 + x_2 - x_3 &= 1 \\
x_1 + x_2 + x_3 &= 3
\end{aligned}
\][/tex]

18.
[tex]\[
\begin{aligned}
x_1 + 3x_2 - 2x_3 &= -1 \\
2x_1 + 6x_2 - 4x_3 &= -2
\end{aligned}
\][/tex]



Answer :

GinnyAnswer
Let's examine each system of linear equations and determine if they represent coincident planes, parallel planes, or if they intersect along a line. We'll also provide parametric equations for the line of intersection if applicable.

### 16. [tex]\( 2x_1 + x_2 + x_3 = 3 \)[/tex]

This single equation represents a plane in 3-dimensional space. Since there is no other plane given to compare with, we cannot classify this system as coincident planes, parallel planes, or intersecting planes. It's just one plane by itself.

### 17.
[tex]\( x_1 + 2x_2 - x_3 = 2 \)[/tex]

[tex]\( -2x_1 + x_2 - x_3 = 1 \)[/tex]

[tex]\( x_1 + x_2 + x_3 = 3 \)[/tex]

Let's solve this system of equations step by step.

1. Write the system as an augmented matrix:

[tex]\[ \begin{pmatrix} 1 & 2 & -1 & | & 2 \\ -2 & 1 & -1 & | & 1 \\ 1 & 1 & 1 & | & 3 \end{pmatrix} \][/tex]

2. Row reduce the matrix to its row echelon form:
- [tex]\( R_2 + 2R_1 \rightarrow R_2 \)[/tex]
- [tex]\( R_3 - R_1 \rightarrow R_3 \)[/tex]

[tex]\[ \begin{pmatrix} 1 & 2 & -1 & | & 2 \\ 0 & 5 & -3 & | & 5 \\ 0 & -1 & 2 & | & 1 \end{pmatrix} \][/tex]

3. Further row reduction:
- [tex]\( R_3 + \frac{1}{5} R_2 \rightarrow R_3 \)[/tex]

[tex]\[ \begin{pmatrix} 1 & 2 & -1 & | & 2 \\ 0 & 5 & -3 & | & 5 \\ 0 & 0 & \frac{7}{5} & | & 2 \end{pmatrix} \][/tex]

4. Normalize the third row:
- Multiply [tex]\( R_3 \)[/tex] by [tex]\( \frac{5}{7} \)[/tex]

[tex]\[ \begin{pmatrix} 1 & 2 & -1 & | & 2 \\ 0 & 5 & -3 & | & 5 \\ 0 & 0 & 1 & | & \frac{10}{7} \end{pmatrix} \][/tex]

5. Back substitution to achieve row-reduction echelon form:
- Update [tex]\( R_2 \)[/tex]
- Update [tex]\( R_1 \)[/tex]

[tex]\[ \begin{pmatrix} 1 & 0 & \frac{1}{7} & | & \frac{4}{7} \\ 0 & 1 & -\frac{3}{5} & | & 1 \\ 0 & 0 & 1 & | & \frac{10}{7} \end{pmatrix} \][/tex]

This gives us the simplified system of linear equations:

[tex]\[ x_1 = \frac{4}{7} - \frac{1}{7}x_3 \][/tex]
[tex]\[ x_2 = 1 + \frac{3}{5}x_3 \][/tex]
[tex]\[ x_3 = \frac{10}{7} \][/tex]

From this system, we can see that we have parametric equations for the system where [tex]\( t = x_3 \)[/tex]:

[tex]\[ x_1 = -\frac{4}{7} + \frac{1}{7} x_3 \][/tex]
[tex]\[ x_2 = 1 - \frac{3}{5} x_3 \][/tex]
[tex]\[ x_3 = \frac{10}{7} t \][/tex]

So, the parametric equations for the line of intersection are:

[tex]\( x_1 = \frac{4}{7} - \frac{1}{7} t \)[/tex]

[tex]\( x_2 = 1 + \frac{3}{5} t \)[/tex]

[tex]\( x_3 = t \)[/tex]

Where [tex]\( t \)[/tex] is a parameter.

### 18.
[tex]\( x_1 + 3x_2 - 2x_3 = -1 \)[/tex]

[tex]\( 2x_1 + 6x_2 - 4x_3 = -2 \)[/tex]

Let's solve this system of equations step by step:

1. Write the system as an augmented matrix:

[tex]\[ \begin{pmatrix} 1 & 3 & -2 & | & -1 \\ 2 & 6 & -4 & | & -2 \end{pmatrix} \][/tex]

2. Row reduce the matrix:
- [tex]\( R_2 - 2R_1 \rightarrow R_2 \)[/tex]

[tex]\[ \begin{pmatrix} 1 & 3 & -2 & | & -1 \\ 0 & 0 & 0 & | & 0 \end{pmatrix} \][/tex]

The second row is all zeros, implying the second equation does not provide additional information beyond the first equation. Thus, this system represents coincident planes (the same plane), which is a redundant representation of a single plane.

### Summary

- 16: One plane only.
- 17: Intersect in a line. Parametric equations for the line:
[tex]\[ x_1 = \frac{4}{7} + \frac{1}{7}t, \quad x_2 = 1 - \frac{3}{5}t, \quad x_3 = t \][/tex]
- 18: Coincident planes (the same plane).

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