Find the adjacency matrices for the directed graphs in (a) and (b) (in attachment).

Two vertices in a graph are adjacent if there exists an edge between them. The entry in row i and column j of the adjacency matrix for a graph is the number of one-edge paths that take you from vertex i to vertex j.

(a)

• None of the vertices have edges that connect them to themselves

• v₁ shares 2 edges with v₂ and 1 edge with v₃

• v₂ shares 2 edges with v₁ and no edges with v₃

• v₃ shares 1 edge with v₁ and none with v₂

So, the adjacency matrix is

(b)

• Only v₁ has a loop that connects it to itself

• v₁ shares 2 edges with v₃ and none with either of v₂ or v₄

• v₂ shares 1 edge with v₃ and none with either v₁ or v₄

• v₃ shares 2 edges with v₁, 1 with v₂, and 2 with v₄

• v₄ shares 2 edges with v₃ and none with either v₁ or v₂

The adjacency matrix is then

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

g

step-by-step explanation: