determine whether the given vector A in M22 belongs to the span (a1,a2,a3) where a1=[1 -1; 0 3]
a2=[1 1; 0 2]
a3=[2 2; -1 1]
and A=[5 1; -1 9]
so I set up an augmented matrix , but im not sure if I set it up correctly. I made a1 a2 and a3 into columns.
1 1 2 | 5
0 0 -1 | -1
-1 1 2 | 1
3 2 1 | 9
then I did the ref and found that it was inconsistent so I concluded A is not in the span.
is this correct?
Well, it seems like you made a small mistake while setting up the augmented matrix. Remember, when forming the augmented matrix, you need to consider the coefficients of the variables, not the actual vectors.
So, the correct augmented matrix for determining whether vector A is in the span of a1, a2, and a3 would be:
1 1 2 | 5
0 0 -1 | 1
-1 1 2 | -1
3 2 1 | 9
Now, let's go through the steps to determine if A is in the span. Starting with the row echelon form:
1 1 2 | 5
0 0 -1 | 1
0 2 4 | 4
0 0 -5 | -2
Now, if you continue with the reduced row echelon form, you will find that there are both non-zero coefficients on the right side but no possible solution for the variables. Therefore, the system is inconsistent, and as a result, A does not belong to the span of a1, a2, and a3.
So, your conclusion is correct! However, let's bring in the humor:
It looks like A just couldn't find its place in the span party. It tried to squeeze in, but the system said, "You're not on the guest list!" So, A had to leave, feeling a bit excluded. But hey, there are plenty of other vectors out there to mingle with!
Yes, your approach is correct. You correctly set up an augmented matrix using column vectors for a1, a2, a3, and A. After performing row operations to put the augmented matrix into row-echelon form or reduced row-echelon form, you concluded that the system of equations is inconsistent. This means that vector A does not belong to the span of a1, a2, a3. Therefore, your conclusion is correct – A is not in the span.
Yes, you set up the augmented matrix correctly by making columns out of vectors a1, a2, and a3. The augmented matrix is:
1 1 2 | 5
0 0 -1 | -1
-1 1 2 | 1
3 2 1 | 9
To determine whether vector A belongs to the span of a1, a2, and a3, you can perform row operations on the augmented matrix to determine if it is consistent or inconsistent.
However, I noticed a minor mistake in the augmented matrix you provided. The last row should be [3 2 1 | -1] instead of [3 2 1 | 9].
With this correction, you can perform row reduction or row operations to obtain the row echelon form (REF) or reduced row echelon form (RREF) of the augmented matrix.
Let's perform the row operations to find the RREF:
R2 = R2 + R1, R3 = R3 + R1, and R4 = R4 - 3R1:
1 1 2 | 5
0 1 1 | 4
0 2 4 | 6
0 -1 -5 |-6
R3 = R3 - 2R2 and R4 = R4 + R2:
1 1 2 | 5
0 1 1 | 4
0 0 2 | -2
0 0 -4 | -2
R4 = R4 + R3/2:
1 1 2 | 5
0 1 1 | 4
0 0 2 | -2
0 0 0 | -3
Since the last row of the RREF has a non-zero entry in the augmented column, while the coefficient of the variable is zero, this indicates that the system is inconsistent. Therefore, the vector A=[5 1; -1 9] does not belong to the span of vectors a1, a2, and a3.
So, your conclusion is correct. Vector A is not in the span of a1, a2, and a3.