If the subspace of all solutions of Ax = 0 has a basis consisting of four vectors and if A is a 7x9 matrix, what is the rank of A? rank A = (Type a whole number.)
If the subspace of all solutions of Ax = 0 has a basis consisting of four vectors and if A is a 7x9 matrix, what is the rank of A? rank A = (Type a whole number.)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Problem:
If the subspace of all solutions of \( Ax = 0 \) has a basis consisting of four vectors and if \( A \) is a \( 7 \times 9 \) matrix, what is the rank of \( A \)?
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### Solution:
\[ \text{Rank}(A) = \_\_\_ \] (Type a whole number.)
**Explanation:**
In this problem, you are given:
- The subspace of all solutions of \( A x = 0 \) (the null space of \( A \)) has a basis consisting of 4 vectors.
- \( A \) is a \( 7 \times 9 \) matrix.
To find the rank of \( A \):
1. Remember the Rank-Nullity Theorem:
\[ \text{Rank}(A) + \text{Nullity}(A) = \text{Number of columns of } A \]
2. The number of columns of \( A \) is 9.
3. The nullity of \( A \) is the dimension of the null space, which is given as 4 (the number of vectors in the basis of the subspace of solutions to \( A x = 0 \)).
Thus:
\[ \text{Rank}(A) + 4 = 9 \]
Solving the equation for Rank(A):
\[ \text{Rank}(A) = 9 - 4 \]
\[ \text{Rank}(A) = 5 \]
So,
\[ \text{Rank}(A) = 5 \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F70fe4f99-9e69-4bef-afe6-80d54e8903ee%2Fe4685dba-43ec-4398-9d53-221eb814bf6f%2Fp9am05r_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem:
If the subspace of all solutions of \( Ax = 0 \) has a basis consisting of four vectors and if \( A \) is a \( 7 \times 9 \) matrix, what is the rank of \( A \)?
---
### Solution:
\[ \text{Rank}(A) = \_\_\_ \] (Type a whole number.)
**Explanation:**
In this problem, you are given:
- The subspace of all solutions of \( A x = 0 \) (the null space of \( A \)) has a basis consisting of 4 vectors.
- \( A \) is a \( 7 \times 9 \) matrix.
To find the rank of \( A \):
1. Remember the Rank-Nullity Theorem:
\[ \text{Rank}(A) + \text{Nullity}(A) = \text{Number of columns of } A \]
2. The number of columns of \( A \) is 9.
3. The nullity of \( A \) is the dimension of the null space, which is given as 4 (the number of vectors in the basis of the subspace of solutions to \( A x = 0 \)).
Thus:
\[ \text{Rank}(A) + 4 = 9 \]
Solving the equation for Rank(A):
\[ \text{Rank}(A) = 9 - 4 \]
\[ \text{Rank}(A) = 5 \]
So,
\[ \text{Rank}(A) = 5 \]
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