Matrix and vector multiplication examples

Example 1

Compute Ax where x = (-2,-1, 0)

A = ⌊ ⌋ 1 2 3 | 4 5 6 | |⌈ |⌉ 7 8 9 10 11 12.

Solution:

Ax = ⌊ 1 2 3 ⌋ | | | 4 5 6 | ⌈ 7 8 9 ⌉ 10 11 12⌊ ⌋ - 2 ⌈ 1 ⌉ 0
= ⌊ ⌋ - 2 ⋅ 1 + 1 ⋅ 2 + 0 ⋅ 3 || - 2 ⋅ 4 + 1 ⋅ 5 + 0 ⋅ 6 || ⌈ - 2 ⋅ 7 + 1 ⋅ 8 + 0 ⋅ 9 ⌉ - 2 ⋅ 10 + 1 ⋅ 11 + 0 ⋅ 12
= ⌊ ⌋ 0 || - 3 || ⌈ - 6 ⌉ - 9 = (0,-3,-6,-6).

Example 2

Compute Ay where y = (-3,-2,-1, 0) and A is as in Example 1.

Solution: The matrix-vector product is not defined. A is 4 × 3 and y is 4 × 1 (viewed as column vector).

Example 3

Compute BC, where

B = [ 1 2 3 ] 4 5 6 and C = ⌊ ⌋ 1 2 ⌈ 3 4 ⌉ 5 6.

Solution:

BC = [ ] 1 2 3 4 5 6⌊ ⌋ 1 2 ⌈ 3 4 ⌉ 5 6
= [ ] 1 ⋅ 1 + 2 ⋅ 3 + 3 ⋅ 5 1 ⋅ 2 + 2 ⋅ 4 + 3 ⋅ 6 4 ⋅ 1 + 5 ⋅ 3 + 6 ⋅ 5 4 ⋅ 2 + 5 ⋅ 4 + 6 ⋅ 6
= [ ] 22 28 49 64

Example 4

Using B and C as defined in Example 3, calculate CB.

Solution:

CB = ⌊ ⌋ 1 2 ⌈ 3 4 ⌉ 5 6[ ] 1 2 3 4 5 6
= ⌊ ⌋ 1 ⋅ 1 + 2 ⋅ 4 1 ⋅ 2 + 2 ⋅ 5 1 ⋅ 3 + 2 ⋅ 6 ⌈ 3 ⋅ 1 + 4 ⋅ 4 3 ⋅ 2 + 4 ⋅ 5 3 ⋅ 3 + 4 ⋅ 6 ⌉ 5 ⋅ 1 + 6 ⋅ 4 5 ⋅ 2 + 6 ⋅ 5 5 ⋅ 3 + 6 ⋅ 6
= ⌊ ⌋ 9 12 15 ⌈ 19 26 33 ⌉ 29 40 51
Clearly, one can see that matrix multiplication is not commutative, i.e., BC≠CB. In the case of examples 3 and 4, BC isn’t even the same size matrix as CB. In some other cases, BC might be defined but CB won’t be defined (for example, when B is a 3 × 2 matrix and C is a 2 × 4 matrix). It is even true that when B and C are square matrices, matrix multiplication is not commutative. You can try yourself and see that BC≠CB if
B = [ ] 1 2 3 4 and C = [ ] 5 6 7 8.