Describe the solutions of the first system of equations below in parametric vector form. Provide a geometric comparison with the solution set of the second system of equations below. 2x, +2x2 + 4x3 = 8 - 4x, - 4x2 - 8x3 = - 16 - 6x2 + 6x3 = 12 2x, +2x2 + 4x3 = 0 - 4x, - 4x2 - 8x3 = 0 - 6x2 + 6x3 = 0 X1 Describe the solution set, x= X2 of the first system of equations in parametric vector form. Select the correct choice below and fill in the answer box(es) within your X3 choice. (Type an integer or fraction for each matrix element.) O A. X= O B. X=X2 Oc. x= + X2 O D. X=X2 + X3

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Chapter1: Equations And Graphs
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Linear Algebra

Describe the solutions of the first system of equations below in parametric vector form. Provide a geometric comparison with the solution set of the second system of equations below.

\[
\begin{align*}
2x_1 + 2x_2 + 4x_3 &= 8 \\
-4x_1 - 4x_2 - 8x_3 &= -16 \\
-6x_2 + 6x_3 &= 12 \\
\end{align*}
\]

\[
\begin{align*}
2x_1 + 2x_2 + 4x_3 &= 0 \\
-4x_1 - 4x_2 - 8x_3 &= 0 \\
6x_2 + 6x_3 &= 0 \\
\end{align*}
\]

Describe the solution set, \( \mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \), of the first system of equations in parametric vector form. Select the correct choice below and fill in the answer box(es) within your choice. 
(Type an integer or fraction for each matrix element.)

A. \( \mathbf{x} = \begin{bmatrix} \boxed{} \\ \boxed{} \\ \boxed{} \end{bmatrix} \)

B. \( \mathbf{x} = \begin{bmatrix} \boxed{} \\ x_2 \\\boxed{} \end{bmatrix} \)

C. \( \mathbf{x} = \begin{bmatrix} \boxed{} \\ \boxed{} \\ x_3 \end{bmatrix} \)

D. \( \mathbf{x} = \begin{bmatrix} \boxed{} \\ x_2 \\ \boxed{} + x_3 \end{bmatrix} \)
Transcribed Image Text:Describe the solutions of the first system of equations below in parametric vector form. Provide a geometric comparison with the solution set of the second system of equations below. \[ \begin{align*} 2x_1 + 2x_2 + 4x_3 &= 8 \\ -4x_1 - 4x_2 - 8x_3 &= -16 \\ -6x_2 + 6x_3 &= 12 \\ \end{align*} \] \[ \begin{align*} 2x_1 + 2x_2 + 4x_3 &= 0 \\ -4x_1 - 4x_2 - 8x_3 &= 0 \\ 6x_2 + 6x_3 &= 0 \\ \end{align*} \] Describe the solution set, \( \mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \), of the first system of equations in parametric vector form. Select the correct choice below and fill in the answer box(es) within your choice. (Type an integer or fraction for each matrix element.) A. \( \mathbf{x} = \begin{bmatrix} \boxed{} \\ \boxed{} \\ \boxed{} \end{bmatrix} \) B. \( \mathbf{x} = \begin{bmatrix} \boxed{} \\ x_2 \\\boxed{} \end{bmatrix} \) C. \( \mathbf{x} = \begin{bmatrix} \boxed{} \\ \boxed{} \\ x_3 \end{bmatrix} \) D. \( \mathbf{x} = \begin{bmatrix} \boxed{} \\ x_2 \\ \boxed{} + x_3 \end{bmatrix} \)
**Question 1:**

Construct a nonzero \(2 \times 2\) matrix \(A\) such that the solution set of the equation \(Ax = 0\) is the line in \(\mathbb{R}^2\) through (6,1) and the origin. Then, find a vector \(b\) in \(\mathbb{R}^2\) such that the solution set of \(Ax = b\) is not a line in \(\mathbb{R}^2\) parallel to the solution set of \(Ax = 0\). Why does this not contradict the theorem that states that if the equation \(Ax = b\) is consistent for some given \(b\) and \(p\) is a solution, then the solution set of \(Ax = b\) is the set of all vectors of the form \(w = p + v_h\), where \(v_h\) is any solution of the homogeneous equation \(Ax = 0\)?

---

**Question 2:**

Construct a nonzero \(2 \times 2\) matrix \(A\) such that the solution set of the equation \(Ax = 0\) is the line in \(\mathbb{R}^2\) through (6,1) and the origin. Choose the correct answer below:

- \( \text{A.} \quad \begin{bmatrix} 1 & 6 \\ 1 & 6 \end{bmatrix} \)

- \( \text{B.} \quad \begin{bmatrix} 1 & 1 \\ 6 & 6 \end{bmatrix} \)

- \( \text{C.} \quad \begin{bmatrix} 1 & -6 \\ 1 & -6 \end{bmatrix} \)

- \( \text{D.} \quad \begin{bmatrix} 1 & 1 \\ -6 & -6 \end{bmatrix} \)
Transcribed Image Text:**Question 1:** Construct a nonzero \(2 \times 2\) matrix \(A\) such that the solution set of the equation \(Ax = 0\) is the line in \(\mathbb{R}^2\) through (6,1) and the origin. Then, find a vector \(b\) in \(\mathbb{R}^2\) such that the solution set of \(Ax = b\) is not a line in \(\mathbb{R}^2\) parallel to the solution set of \(Ax = 0\). Why does this not contradict the theorem that states that if the equation \(Ax = b\) is consistent for some given \(b\) and \(p\) is a solution, then the solution set of \(Ax = b\) is the set of all vectors of the form \(w = p + v_h\), where \(v_h\) is any solution of the homogeneous equation \(Ax = 0\)? --- **Question 2:** Construct a nonzero \(2 \times 2\) matrix \(A\) such that the solution set of the equation \(Ax = 0\) is the line in \(\mathbb{R}^2\) through (6,1) and the origin. Choose the correct answer below: - \( \text{A.} \quad \begin{bmatrix} 1 & 6 \\ 1 & 6 \end{bmatrix} \) - \( \text{B.} \quad \begin{bmatrix} 1 & 1 \\ 6 & 6 \end{bmatrix} \) - \( \text{C.} \quad \begin{bmatrix} 1 & -6 \\ 1 & -6 \end{bmatrix} \) - \( \text{D.} \quad \begin{bmatrix} 1 & 1 \\ -6 & -6 \end{bmatrix} \)
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