Question
Which of the following matrices correctly converts the system to reduced row echelon form?
⎧⎩⎨⎪⎪2x+y−4z=−28−2x+y+3z=21x+y−z=−12
{
2
𝑥
+
𝑦
−
4
𝑧
=
−
28
−
2
𝑥
+
𝑦
+
3
𝑧
=
21
𝑥
+
𝑦
−
𝑧
=
−
12
(1 point)
Responses
⎡⎣⎢1000100−121−7−723⎤⎦⎥
[
1
0
0
−
7
0
1
−
1
2
−
7
2
0
0
1
3
]
the 3 by 4 matrix, row 1, column 1, 1 column 2, 0 column 3, 0 column 4, negative 7 row 2, column 1, 0 column 2, 1 column 3, negative 1 over 2 column 4, negative 7 over 2 row 3, column 1, 0 column 2, 0 column 3, 1 column 4, 3
⎡⎣⎢100010001−7−23⎤⎦⎥
[
1
0
0
−
7
0
1
0
−
2
0
0
1
3
]
the 3 by 4 matrix, row 1, column 1, 1 column 2, 0 column 3, 0 column 4, negative 7 row 2, column 1, 0 column 2, 1 column 3, 0 column 4, negative 2 row 3, column 1, 0 column 2, 0 column 3, 1 column 4, 3
⎡⎣⎢⎢100010−74−121−494−723⎤⎦⎥⎥
[
1
0
−
7
4
−
49
4
0
1
−
1
2
−
7
2
0
0
1
3
]
the 3 by 4 matrix, row 1, column 1, 1 column 2, 0 column 3,
⎧⎩⎨⎪⎪2x+y−4z=−28−2x+y+3z=21x+y−z=−12
{
2
𝑥
+
𝑦
−
4
𝑧
=
−
28
−
2
𝑥
+
𝑦
+
3
𝑧
=
21
𝑥
+
𝑦
−
𝑧
=
−
12
(1 point)
Responses
⎡⎣⎢1000100−121−7−723⎤⎦⎥
[
1
0
0
−
7
0
1
−
1
2
−
7
2
0
0
1
3
]
the 3 by 4 matrix, row 1, column 1, 1 column 2, 0 column 3, 0 column 4, negative 7 row 2, column 1, 0 column 2, 1 column 3, negative 1 over 2 column 4, negative 7 over 2 row 3, column 1, 0 column 2, 0 column 3, 1 column 4, 3
⎡⎣⎢100010001−7−23⎤⎦⎥
[
1
0
0
−
7
0
1
0
−
2
0
0
1
3
]
the 3 by 4 matrix, row 1, column 1, 1 column 2, 0 column 3, 0 column 4, negative 7 row 2, column 1, 0 column 2, 1 column 3, 0 column 4, negative 2 row 3, column 1, 0 column 2, 0 column 3, 1 column 4, 3
⎡⎣⎢⎢100010−74−121−494−723⎤⎦⎥⎥
[
1
0
−
7
4
−
49
4
0
1
−
1
2
−
7
2
0
0
1
3
]
the 3 by 4 matrix, row 1, column 1, 1 column 2, 0 column 3,
Answers
GPT-4o mini
To convert the given system of equations into reduced row echelon form (RREF), we must examine the provided matrices step by step.
The system of equations is:
1. \( 2x + y - 4z = -28 \)
2. \( -2x + y + 3z = 21 \)
3. \( x + y - z = -12 \)
We can represent this system as an augmented matrix:
\[
\begin{bmatrix}
2 & 1 & -4 & | & -28 \\
-2 & 1 & 3 & | & 21 \\
1 & 1 & -1 & | & -12 \\
\end{bmatrix}
\]
We aim to use row operations to reduce it to RREF.
Let's analyze each provided matrix to check if they correctly represent the RREF of the given system:
1. **Matrix A:**
\[
\begin{bmatrix}
1 & 0 & 0 & | & -7 \\
0 & 1 & -1/2 & | & -7/2 \\
0 & 0 & 1 & | & 3 \\
\end{bmatrix}
\]
In RREF, there should be a "1" in each leading position, and every leading "1" should be the only nonzero entry in its column. This matrix meets the RREF criteria.
2. **Matrix B:**
\[
\begin{bmatrix}
1 & 0 & 0 & | & -7 \\
0 & 1 & 0 & | & -2 \\
0 & 0 & 1 & | & 3 \\
\end{bmatrix}
\]
This matrix also fulfills the RREF conditions. Each leading entry is "1" and is the only non-zero number in its respective column.
3. **Matrix C:**
\[
\begin{bmatrix}
1 & 0 & -7/4 & | & -49/4 \\
0 & 1 & -1/2 & | & -7/2 \\
0 & 0 & 1 & | & 3 \\
\end{bmatrix}
\]
This matrix is not in RREF since the first column should not have any other coefficients apart from "1" in the leading position.
Given this analysis, Matrices A and B correspond to valid RREF forms of the original system. However, Matrix B is more straightforward as it maintains the values directly applicable from the original equations without introducing fractions.
**Conclusion:**
Both Matrix A and Matrix B are valid as they produce RREF. If you are specifically looking for the reduced form without fractions, Matrix B is preferable. Thus, Matrix B is more appropriate for the context given.
The system of equations is:
1. \( 2x + y - 4z = -28 \)
2. \( -2x + y + 3z = 21 \)
3. \( x + y - z = -12 \)
We can represent this system as an augmented matrix:
\[
\begin{bmatrix}
2 & 1 & -4 & | & -28 \\
-2 & 1 & 3 & | & 21 \\
1 & 1 & -1 & | & -12 \\
\end{bmatrix}
\]
We aim to use row operations to reduce it to RREF.
Let's analyze each provided matrix to check if they correctly represent the RREF of the given system:
1. **Matrix A:**
\[
\begin{bmatrix}
1 & 0 & 0 & | & -7 \\
0 & 1 & -1/2 & | & -7/2 \\
0 & 0 & 1 & | & 3 \\
\end{bmatrix}
\]
In RREF, there should be a "1" in each leading position, and every leading "1" should be the only nonzero entry in its column. This matrix meets the RREF criteria.
2. **Matrix B:**
\[
\begin{bmatrix}
1 & 0 & 0 & | & -7 \\
0 & 1 & 0 & | & -2 \\
0 & 0 & 1 & | & 3 \\
\end{bmatrix}
\]
This matrix also fulfills the RREF conditions. Each leading entry is "1" and is the only non-zero number in its respective column.
3. **Matrix C:**
\[
\begin{bmatrix}
1 & 0 & -7/4 & | & -49/4 \\
0 & 1 & -1/2 & | & -7/2 \\
0 & 0 & 1 & | & 3 \\
\end{bmatrix}
\]
This matrix is not in RREF since the first column should not have any other coefficients apart from "1" in the leading position.
Given this analysis, Matrices A and B correspond to valid RREF forms of the original system. However, Matrix B is more straightforward as it maintains the values directly applicable from the original equations without introducing fractions.
**Conclusion:**
Both Matrix A and Matrix B are valid as they produce RREF. If you are specifically looking for the reduced form without fractions, Matrix B is preferable. Thus, Matrix B is more appropriate for the context given.