Explain in detail please! Let Prove by induction that Thinking of each entry of

Question

Explain in detail please!

Let Prove by induction that Thinking of each entry of Mn as a function of t, show that d(Mn)/dt = nMn-1 for n 1 where M degree = I denotes the 2 times 2 identity matrix. Use the result above to show that d(eM)/dt = eM. Give an explicit formula for eM so that the entries of eM are specific functions of the variable t. Consider the oriented graph, labeled 1 and 2.there is an oriented edge from 1 to 2 and each vertex has seven oriented loops from itself, to itself. Draw a diagram of graph G. How many oriented paths of length 137 are there in G starting at vertex 1 and ending at vertex 2?

Explanation / Answer

(A)

(a)

Result is true for n=1

Suppose it is correct for n.

M^(n+1) = M*M^n =

t 1 t^n nt^(n-1)

0 t 0 t^n

=

t^(n+1) nt^n+t^n

0 t*t^n

=

t^(n+1) (n+1)t^n

0 t^(n+1)

So this is correct

(b)

d(M^n) =

nt^(n-1) n(n-1)t^(n-2)

0 nt^(n-1)

= n *

t^(n-1) (n-1)t^(n-2)

0 t^(n-1)

= n*M^(n-1)

(c)

e^M = sum (n>=0)(M^n / n!)

d(e^M) = sum(n>=1) nM^(n-1)/n! = sum(n>=1) M^(n-1)/(n-1)!

d(e^M) = sum(n>=0) M^n/n! (by changing index)

d(e^M) = e^M

(d) since d(e^M)=e^M we need :

e^M =

e^t e^t

0 e^t

B) I let you draw G.

Here we need to compute A^137

Notice A = M(t=7)

So :

A^137 =

7^137 137*7^136

0 7^137

So there is 137*7^136 paths of length 137 from 1->2

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