when i reduce the augmented matrix in #11, i get:
the last row implies
doesn't look consistent to me.
Could someone please tell me how to solve exercises 11, 22 and 26?
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.
Not even a clue where to start...???
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!
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...
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.