I need help with discrete math/logic/matrices problems
if anyone is able to help me it is much appreciated.

topics include system of equations, matrices, summations

when you say discrete, i'm thinking more like proofs. but doesnt sound like it. post your problems.

That is part of it, or was for last section

A = [-1 0 1] B = [1 1 -2] C = [ 2 3 ]
[2 1 0 ] [-2 0 1] [ -1 0 ]
[-1 -1 1] [1 1 3 ] [ -1 1 ]

1a)Find AC + BC (distributive law for matrices)

1b)2A - 3A

2)simplify
2 3
Σ Σ(i+3j)
i=0 j=1


solve the systems of equations
x1 + x2= 1
-x1 + x2 + x3 = 2
-1x2 + x3= -1

This post was edited by mcfighter on Jul 7 2013 08:41pm

1a) AC + BC = (A+B)C. so find A + B then multiply by C
1b) 2A - 3A = -1A. so multiply A by -1
3) substitution, elimination, or i assume teacher wants you to do gaussian elimination. set up Ax = b and row-reduce

Thanks, I really appreciate the help, but I was sorta looking more or less for just the answers.

4)
Use basic matrix laws only (don’t solve/prove by examples using 2x2 matrices etc.) to expand (A + B )2.
Explain all steps.
Hint: Use the distributive laws.


5)
Let P(n) be the statement that
1⋅1! + 2⋅2! + 3⋅3! + . . . + n⋅n!= (n + 1)! - 1 is true for all positive integers n ≥ 1.The symbol, !, is read factorial.

(a) Show that P(1) is true, completing the basis step of the proof.
(b) What is the inductive step?
(c) Prove the inductive step.

5) is a generic proof by induction:

Show the statement is true for n=1
Then assume the statement is true for all n in the Natural numbers
Then show P(n+1) is true. **Hint** plug in (n+1) everywhere you see an n and simplify the expression
This will prove it is valid for every positive integer

Thank you for the input, is it possible for you to post the answer?
btw i am willing to pay for this
dont really have time to do this

5)
Let P(n) be the statement that
1⋅1! + 2⋅2! + 3⋅3! + . . . + n⋅n!= (n + 1)! - 1 is true for all positive integers n ≥ 1.The symbol, !, is read factorial.

(a) Show that P(1) is true, completing the basis step of the proof.
(b) What is the inductive step?
(c) Prove the inductive step.

as 'mcfighter' says, it is pretty straightforward

P(1): 1.1! = (1+1)! - 1 which is trivial with 2!=2
next: assume P(n) is true
now prove that under that assumption P(n+1) is true

P(n+1): 1.1!+....+n.n!+(n+1).(n+1)! = (n+2)! - 1 now solve the right part (n+2)! - 1 = (n+2).(n+1)! - 1 = n.(n+1)! + 2.(n+1)! - 1 = n.(n+1)! + 1.(n+1)! + 1.(n+1)! - 1 = (n+1).(n+1)! + (n+1)! - 1

putting it together again 1.1! + ... + n.n! + (n+1).(n+1)! = (n+1).(n+1)! + (n+1)! - 1

now you can subtract (n+1).(n+1)! from both sides and you have the formula for P(n)

qed

thanks, i got the first one..
will pay 70 fg to whoever can answer these questions within 4 hours


(a) Define the function f: R → R by f(x) = 3^3 + 5.
Briefly explain why f is a 1-1 (one-to-one) function. Use numbers. No proof necessary, just an explanation in some detail
(b) Is the function g: R → Z defined by g(n) = [n/2]a one to one function? (Be careful, n/2 means the ceiling function.) Explain.

Briefly explain what f-1 means in general and then find f^-1for the function f in part a.

let A = [1 3]
[1 4]. use the formula for finding the inverse of a 2 x 2 matrix to find A ^-1. Show work. You can check your solution by simply multiplying
AA-1 to see if you obtain I.

you sure that f(x) = 27 + 5?

or should it be f(x) = X^3 + 5?