Homework SourseLet w 1 1 1 0 e3 0 0 1 e1 1 0 0 A 6 1 2 2 3 3 1 1 7 5 3
Let w = [1 -1 1 0] e_3 = [0 0 1] e_1 = [1 0 0] A = [6 1 2 2 3 3 1 1 7 5 3 -2] B = [1 0 -1 -2 2 0 0 1 -1 -2 1 1] Perform the following multiplications if possible. If not possible, explain why not. e_3^T AB Ae_3 Awe_3^T e_3^T e_1 w^T Be_3.
![Let w = [1 -1 1 0] e_3 = [0 0 1] e_1 = [1 0 0] A = [6 1 2 2 3 3 1 1 7 5 3 -2] B = [1 0 -1 -2 2 0 0 1 -1 -2 1 1] Perform the following multiplications if possib](/WebImages/36/let-w-1-1-1-0-e3-0-0-1-e1-1-0-0-a-6-1-2-2-3-3-1-1-7-5-3-1108019-1761586862-0.webp "Let w = [1 -1 1 0] e_3 = [0 0 1] e_1 = [1 0 0] A = [6 1 2 2 3 3 1 1 7 5 3 -2] B = [1 0 -1 -2 2 0 0 1 -1 -2 1 1] Perform the following multiplications if possib")
![Let w = [1 -1 1 0] e_3 = [0 0 1] e_1 = [1 0 0] A = [6 1 2 2 3 3 1 1 7 5 3 -2] B = [1 0 -1 -2 2 0 0 1 -1 -2 1 1] Perform the following multiplications if possib](/WebImages/36/let-w-1-1-1-0-e3-0-0-1-e1-1-0-0-a-6-1-2-2-3-3-1-1-7-5-3-1108019-1761586862-1.webp "Let w = [1 -1 1 0] e_3 = [0 0 1] e_1 = [1 0 0] A = [6 1 2 2 3 3 1 1 7 5 3 -2] B = [1 0 -1 -2 2 0 0 1 -1 -2 1 1] Perform the following multiplications if possib")
Solution
1. A is a 3x4 matrix and B is a 4x3 matrix so that AB is a 3x3 matrix
0
6
-6
-5
8
-3
1
11
-12
e3 is a 3x1 vector so that e3T is a 1x3 vector = (0,0,1) Hence e3TAB =
1
11
-12
2. A is a 3x4 matrix and e3 is a 3x1 vector so that A e3 is not defined/does not exist.
3. A is a 3x4 matrix and w is a 4x1 vector so that Aw will be a 3x1 matrix. Further, e3T is a 1x3 vector = (0,0,1) . Here, Aw =
7
1
5
and Awe3T =
0
0
7
0
0
1
0
0
5
4. e3T is a 1x3 vector = (0,0,1) and e1 = (1,0,0)T is a 3x1 vector so that e3T e1 is a singleton =
0
5. wT is a 1x4 vector, B is a 4x3 matrix so that wT B is a 1x3 matrix. e3 is a 3x1 vector. Therefore, wT B e3 is a singleton =
-2
| 0 | 6 | -6 |
| -5 | 8 | -3 |
| 1 | 11 | -12 |
