when i reduce the augmented matrix in #11, i get:
the last row implies
doesn't look consistent to me.
Hello,
Could someone please tell me how to solve exercises 11, 22 and 26?
Exercise 11
The answer of exercise 11 should be that b is not a linear combination of a1, a2 and a3. I computed the echelon form of the augmented matrix where a1, a2, a3 and b are the columns of the augmented matrix, but I end up with a consistent system, so I don't understand why b is not a linear combination of a1, a2 and a3.
Exercise 22
Not even a clue where to start...???
Exercise 26
In 26(a) I found that the system is consistent, so b is in W. How do I solve 26(b)?
Thanks for your help!
Lotte
I used MATLAB to get the row reduced echelon form of this matrix. MATLAB gives the same row reduced echelon form I found.
Does anyone know how Deveno ended up with an inconsistent row? The answer of exercise 11 should be that b is not a linear combination of a1, a2 and a3.
I don't understand why, because the row reduced echelon form of the matrix shows a consistent system...
Please help!
hmm...i must have made an error, because i re-checked and got the same matrix as you.
so, let's see what that means:
it means all vectors in col(A), are linear combinations of the first two columns of A (those are the pivot columns).
suppose that (2,-1,6) = a(1,-2,0) + b(0,1,2) = (a,b-2a,b). then a = 2, and b = 6, but then b-2a = 6 - 4 - 2 ≠ -1.