Homepage › Solution manuals › Stephen Friedberg › Linear Algebra › Exercise 3.2.15

Exercise 3.2.15

Answers

Let $P = M (A | B)$ and $Q = (MA | MB)$ . We want to show that $P_{ij} = Q_{ij}$ for all $i$ and for all $j$ . Assume that $A$ and $B$ has $a$ and $b$ columns respectly. For $j = 1, 2, \dots, a$ , we have that

P_{ij} = \sum_{k = 1}^{n} M_{ik} A_{kj} = {(MA)}_{ij} = Q_{ij} .

For $j = a + 1, a + 2, \dots, a + b$ , we have that

P_{ij} = \sum_{k = 1}^{n} M_{ik} B_{kj} = {(MB)}_{ij} = Q_{ij} .

jephianlin

2011-06-27 00:00

Exercise 3.2.15

Answers

Comments

Add answer