A First Course in Linear Algebra, 2017a

66 Matrices
Solution. First check if the product is possible. It is of the form (3 × 2)(2 × 3) and since the inside
numbers match, it is possible to do the multiplication. The result should be a 3 × 3 matrix. We can first
compute AB:
⎡⎡
⎣⎣
1 2
3 1
2 6
⎤ ⎡
[ ]
⎦ 2
, ⎣
7
1 2
3 1
2 6
⎤ ⎡
[ ]
⎦ 3
, ⎣
6
1 2
3 1
2 6
⎤
[ ] ⎤
⎦ 1
⎦
2
where the commas separate the columns in the resulting product. Thus the above product equals
⎡
16 15
⎤
5
⎣ 13 15 5 ⎦
46 42 14
which is a 3 × 3 matrix as desired. Thus, the (3,2)-entry equals 42.
Now using Definition 2.21, we can find that the (3,2)-entry equals
2
∑ a 3k b k2 = a 31 b 12 + a 32 b 22
k=1
= 2 × 3 + 6 × 6 = 42
Consulting our result for AB above, this is correct!
You may wish to use this method to verify that the rest of the entries in AB are correct.
♠
Here is another example.
Example 2.23: Finding the Entries of a Product
Determine if the product AB is defined. If it is, find the (2,1)-entry of the product.
⎡ ⎤ ⎡ ⎤
2 3 1 1 2
A = ⎣ 7 6 2 ⎦,B = ⎣ 3 1 ⎦
0 0 0 2 6
Solution. This product is of the form (3 × 3)(3 × 2). The middle numbers match so the matrices are
conformable and it is possible to compute the product.
We want to find the (2,1)-entry of AB, that is, the entry in the second row and first column of the
product. We will use Definition 2.21, which states
(AB) ij =
n
∑ a ik b kj
k=1
In this case, n = 3, i = 2and j = 1. Hence the (2,1)-entry is found by computing
⎡ ⎤
3
(AB) 21 = ∑ a 2k b k1 = [ ]
b 11
a 21 a 22 a 23
⎣ b 21
⎦
k=1
b 31