|
Question 1181408: Given the following matrices:
b1 b4 b7
b2 b5 b8 = B
b3 b6 b9
r(B)= 3
For easier understanding let’s say that bi is the column i in matrix B.
A is a matrix of dimensions 3x3 in which the following exists:
K = { (b1+b3) , (2b2-b3) } is a basis for the equation: Ax=0.
r(A)=1
Also:
1 2 4
3 1 2 =A*B
1 2 4
r(AB)=2.
1) find a vector basis for the vector space that is A’s columns.
2) given the following:
K = { (1,2,0) , (0,1,1) }
Find a reduced echelon form (C) for A.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! There's a fundamental contradiction in the given information. If r(A) = 1, then the rank of any product involving A, including AB, *cannot* be greater than 1. You've stated that r(AB) = 2, which is impossible if A has rank 1. Therefore, the problem as stated is inconsistent.
However, let's address the *method* of solving this type of problem, assuming we correct the ranks. It's likely that r(A) should be 2, not 1. This would make r(AB) = 2 possible. Here's how we'd proceed *if* the ranks were consistent:
**1) Finding a vector basis for the column space of A (assuming r(A) = 2):**
* **The key idea:** The column space of AB is a *subspace* of the column space of A. If r(AB) = 2 and we assume r(A) = 2, then the column space of AB *is* the column space of A.
* **Basis for the column space of A:** Since r(AB) = 2, any two linearly independent columns of AB will form a basis for the column space of AB, and thus also for the column space of A. Looking at AB:
```
1 2 4
3 1 2
1 2 4
```
We can see that the first two columns, (1, 3, 1) and (2, 1, 2), are linearly independent. Therefore, {(1, 3, 1), (2, 1, 2)} is a basis for the column space of A.
**2) Finding the reduced echelon form (C) for A (assuming r(A) = 2 - This part is also problematic):**
* **Understanding the null space (kernel):** K = {(1, 2, 0), (0, 1, 1)} being a basis for the null space of A (Ax = 0) tells us about the relationships between the *columns* of A. Let the columns of A be denoted as a1, a2, and a3. Then:
* 1*a1 + 2*a2 + 0*a3 = 0 => a1 = -2a2
* 0*a1 + 1*a2 + 1*a3 = 0 => a3 = -a2
* **The problem:** These relationships imply that all columns of A are scalar multiples of a single vector (a2). This means the rank of A *must* be 1, not 2. The given null space basis is inconsistent with r(A) = 2.
* **What if r(A) was 1 (and K was a correct null space basis):** If r(A) = 1, then a1 = -2a2 and a3 = -a2. We can choose a2 = (1, 0, 0) (or any other non-zero vector). Then a1 = (-2, 0, 0) and a3 = (-1, 0, 0). So, A would be:
```
-2 1 -1
0 0 0
0 0 0
```
The reduced row echelon form of this A would be:
```
1 -1/2 1/2
0 0 0
0 0 0
```
**In summary:** The problem as stated is inconsistent due to the conflicting ranks. If you correct the rank of A to 2, then the column space basis is found as described above. However, the given null space basis K is incompatible with *any* rank of A other than 1. If r(A) = 1, then the reduced echelon form can be found, but it will be different. You need to double-check the original problem statement for errors.
|
|
|
| |
