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. First system: $3x_1 + 3x_2 + 6x_3 = 12$ $-9x_1 - 9x_2 - 18x_3 = -36$ $-6x_2 + 18x_3 = 12$ Second system: $3x_1 + 3x_2 + 6x_3 = 0$ $-9x_1 - 9x_2 - 18x_3 = 0$ $-6x_2 + 18x_3 = 0$ Describe the solution set $x = \begin{bmatrix}x_1 \\ x_2 \\ x_3 \end{bmatrix}$ of the first system of equations in parametric vector form. Choose the correct answer: A. $x =$ B. $x = x_2$ C. $x = x_2 \begin{bmatrix} 6 \\ -2 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} -5 \\ 3 \\ 1 \end{bmatrix}$ D. $x = x_2 + x_3$ Which option best compares the two systems? A. The solution set of the first system is a plane parallel to the plane that is the solution set of the second system. B. The solution set of the first system is a line perpendicular to the line that is the solution set of the second system. C. The solution set of the first system is a plane parallel to the line that is the solution set of the second system. D. The solution set of the first system is a line parallel to the line that is the solution set of the second system.

Neetesh Kumar | September 23, 2024

Math2B Course: Linear Algebra

This is the solution to Math2B Course: Linear Algebra
Assignment: Ch1 Section 05 Question Number 5
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Step-by-Step Solution:

Step 1: Solve the first system of equations.

The system is:

$3x_1 + 3x_2 + 6x_3 = 12$

$-9x_1 - 9x_2 - 18x_3 = -36$

$-6x_2 + 18x_3 = 12$

Notice that the second equation is just a multiple of the first equation, so we can ignore it.

Now, we have: $3x_1 + 3x_2 + 6x_3 = 12$ $-6x_2 + 18x_3 = 12$

Dividing the first equation by 3: $x_1 + x_2 + 2x_3 = 4$

Dividing the second equation by 6: $-x_2 + 3x_3 = 2$

Now, solve for $x_2$ in terms of $x_3$: $x_2 = 3x_3 - 2$

Substitute this into the first equation: $x_1 + (3x_3 - 2) + 2x_3 = 4$ $x_1 + 5x_3 - 2 = 4$ $x_1 = -5x_3 + 6$

Thus, the solution is: $x_1 = -5x_3 + 6$ $x_2 = 3x_3 - 2$ $x_3 = x_3$

Step 2: Parametric vector form.

We can express the solution in parametric vector form: $x = x_3 \begin{bmatrix} -5 \\ 3 \\ 1 \end{bmatrix} + \begin{bmatrix} 6 \\ -2 \\ 0 \end{bmatrix}$

So the solution is: $x = x_3 \begin{bmatrix} -5 \\ 3 \\ 1 \end{bmatrix} + x_2 \begin{bmatrix} 6 \\ -2 \\ 0 \end{bmatrix}$

Step 3: Comparison with the second system.

The second system is the homogeneous form of the first system, which means that the solution of the second system will be a line, and the solution of the first system is a plane parallel to that line.

Final Answer:

Part (i) - Option C $\checkmark$

The correct choice for parametric vector form is:

$x = x_2 \begin{bmatrix} 6 \\ -2 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} -5 \\ 3 \\ 1 \end{bmatrix}$

The geometric comparison is:

Part (ii) - Option D $\checkmark$

The solution set of the first system is a plane parallel to the line that is the solution set of the second system.

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