Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![**Finding Bases for Null Space, Column Space, and Row Space**
Given that matrices \( A \) and \( B \) are row equivalent, we need to find the bases for the Null space (Nul \( A \)), Column space (Col \( A \)), and Row space (Row \( A \)) of matrix \( A \).
The matrices are given as:
\[
A = \begin{bmatrix}
-2 & 8 & -2 & -8 \\
2 & -12 & -4 & 2 \\
-3 & 16 & 3 & -6
\end{bmatrix}, \quad
B = \begin{bmatrix}
1 & 0 & 7 & 10 \\
0 & 4 & 6 & 6 \\
0 & 0 & 0 & 0
\end{bmatrix}
\]
### Null Space (Nul \( A \))
To find the basis for the null space, we need to solve \( B \mathbf{x} = \mathbf{0} \).
Given matrix \( B \):
\[
B = \begin{bmatrix}
1 & 0 & 7 & 10 \\
0 & 4 & 6 & 6 \\
0 & 0 & 0 & 0
\end{bmatrix}
\]
Transform \( B \) to its row-reduced echelon form (if it's not already). In this case:
\[
B = \begin{bmatrix}
1 & 0 & 7 & 10 \\
0 & 4 & 6 & 6 \\
0 & 0 & 0 & 0
\end{bmatrix}
\]
From this, we can see that the null space is spanned by non-pivot columns.
Let \( \mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} \):
\[ x_1 + 7x_3 + 10x_4 = 0 \]
\[ 4x_2 + 6x_3 + 6x_4 = 0 \]
Solve for \( x_1 \) and \( x_2 \). Set \( x_3 = t \) and \( x_4 = s \):
\[ x_1 = -7t - 10s \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff893ad80-e3fa-4d69-87b2-5d427e54a923%2F2002894d-7373-4576-adf9-dddfcc0bc65b%2F194b5g_processed.png&w=3840&q=75)
Transcribed Image Text:**Finding Bases for Null Space, Column Space, and Row Space**
Given that matrices \( A \) and \( B \) are row equivalent, we need to find the bases for the Null space (Nul \( A \)), Column space (Col \( A \)), and Row space (Row \( A \)) of matrix \( A \).
The matrices are given as:
\[
A = \begin{bmatrix}
-2 & 8 & -2 & -8 \\
2 & -12 & -4 & 2 \\
-3 & 16 & 3 & -6
\end{bmatrix}, \quad
B = \begin{bmatrix}
1 & 0 & 7 & 10 \\
0 & 4 & 6 & 6 \\
0 & 0 & 0 & 0
\end{bmatrix}
\]
### Null Space (Nul \( A \))
To find the basis for the null space, we need to solve \( B \mathbf{x} = \mathbf{0} \).
Given matrix \( B \):
\[
B = \begin{bmatrix}
1 & 0 & 7 & 10 \\
0 & 4 & 6 & 6 \\
0 & 0 & 0 & 0
\end{bmatrix}
\]
Transform \( B \) to its row-reduced echelon form (if it's not already). In this case:
\[
B = \begin{bmatrix}
1 & 0 & 7 & 10 \\
0 & 4 & 6 & 6 \\
0 & 0 & 0 & 0
\end{bmatrix}
\]
From this, we can see that the null space is spanned by non-pivot columns.
Let \( \mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} \):
\[ x_1 + 7x_3 + 10x_4 = 0 \]
\[ 4x_2 + 6x_3 + 6x_4 = 0 \]
Solve for \( x_1 \) and \( x_2 \). Set \( x_3 = t \) and \( x_4 = s \):
\[ x_1 = -7t - 10s \
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