Question

Question

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?

3y+4z=−20

6x−5y=3

−8x−z=10

(1 point)
Responses

⎡⎣⎢60−8−53004−13−2010⎤⎦⎥
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, negative 8 column 2, 0 column 3, negative 1 column 4, 10

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

⎡⎣⎢0−8630−54−10−20103⎤⎦⎥
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
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Answer 1

To convert the given system of equations into an augmented matrix, we first need to organize the coefficients of \(x\), \(y\), and \(z\) along with the constants from the right-hand side of the equations into a 3x4 matrix.

The given equations are:

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

Now, we can rewrite these equations to fit the standard form:

• For the first equation, we note that \(x\) has a coefficient of 0, \(y\) has a coefficient of 3, \(z\) has a coefficient of 4, and the constant on the right side is -20. Therefore, it corresponds to the row: \([0, 3, 4 | -20]\).

• For the second equation: \(6x - 5y + 0z = 3\). This corresponds to the row: \([6, -5, 0 | 3]\).

• For the third equation: \(-8x + 0y - z = 10\). This corresponds to the row: \([-8, 0, -1 | 10]\).

Putting this together, the augmented matrix becomes:

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

Now, let's analyze the options given:

1. \(\begin{bmatrix} 60 & -8 & -5 & 3 \ 0 & 4 & -1 & -20 \ 10 \end{bmatrix}\)
2. \(\begin{bmatrix} 6 & 0 & 8 & -5 \ 0 & 3 & 4 & -20 \ -10 \end{bmatrix}\)
3. \(\begin{bmatrix} 0 & -8 & 6 & -3 \ -5 & 0 & -4 & -1 \ -20 & 3 & 10 \end{bmatrix}\)
4. \(\begin{bmatrix} 0 & 3 & 4 & -20 \ 6 & -5 & 0 & 3 \ -8 & 0 & -1 & 10 \end{bmatrix}\)

Among the options, the correct augmented matrix that corresponds to the equations after rewriting them is:

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

This corresponds to option 4.

Therefore, option 4 is the correct matrix that can be used in the first step to solve the system of equations.