In order to solve the following system of equations by using row operations to convert its augmented matrix to row echelon form, which matrix can be used in the first step?

Question

3y+4z=−20
3
𝑦
+
4
𝑧
=
−
20

6x−5y=3
6
𝑥
−
5
𝑦
=
3

−8x−z=10
−
8
𝑥
−
𝑧
=
10

(1 point)
Responses

⎡⎣⎢0−8630−54−10−20103⎤⎦⎥
[
0
3
4
−
20
−
8
0
−
1
10
6
−
5
0
3
]
the 3 by 4 matrix, row 1, column 1, 0 column 2, 3 column 3, 4 column 4, negative 20 row 2, column 1, negative 8 column 2, 0 column 3, negative 1 column 4, 10 row 3, column 1, 6 column 2, negative 5 column 3, 0 column 4, 3

⎡⎣⎢608−5300413−20−10⎤⎦⎥
[
6
−
5
0
3
0
3
4
−
20
8
0
1
−
10
]
the 3 by 4 matrix, row 1, column 1, 6 column 2, negative 5 column 3, 0 column 4, 3 row 2, column 1, 0 column 2, 3 column 3, 4 column 4, negative 20 row 3, column 1, 8 column 2, 0 column 3, 1 column 4, negative 10

⎡⎣⎢06−8−3−50−40−1−20310⎤⎦⎥
[
0
−
3
−
4
−
20
6
−
5
0
3
−
8
0
−
1
10
]

Answer 1

To convert the system of equations into an augmented matrix, we first need to write down the coefficients of the variables and the constants from the equations.

Given the system of equations:

1. \( 3y + 4z = -20 \)
2. \( 6x - 5y = 3 \)
3. \( -8x - z = 10 \)

We can express this in matrix form (augmented matrix). The augmented matrix will be structured as follows:

\[ \begin{bmatrix} \text{Coefficient of } x & \text{Coefficient of } y & \text{Coefficient of } z & \text{Constant} \ \end{bmatrix} \]

Writing the equations out, we have:

• For the first equation \( (0, 3, 4 | -20) \)
• For the second equation \( (6, -5, 0 | 3) \)
• For the third equation \( (-8, 0, -1 | 10) \)

So, the augmented matrix becomes:

\[ \begin{bmatrix} 0 & 3 & 4 & -20 \ 6 & -5 & 0 & 3 \ -8 & 0 & -1 & 10 \end{bmatrix} \]

Thus, the correct augmented matrix to begin with row operations that is consistent with this system is:

\[ \begin{bmatrix} 0 & 3 & 4 & -20 \ 6 & -5 & 0 & 3 \ -8 & 0 & -1 & 10 \end{bmatrix} \]

Looking at the provided options, the matrix that matches this establishment in terms of its layout (keeping the correct order) would be:

\[ \begin{bmatrix} 0 & 3 & 4 & -20 \ 6 & -5 & 0 & 3 \ -8 & 0 & -1 & 10 \end{bmatrix} \]

Thus, the first step would be to use the first option, which correctly represents the augmented matrix for the given system of equations.