MATH 203 MIDTERM I

Part 1

Please work out each of the given problems without the use of a calculator.  Credit will be based on the steps that you show towards the final answer.  Show your work. 

Problem 1

Let 

1. Find A^-1
   
    

Solution

We augment the matrix with the identity matrix and put it into row reduced echelon form.

So that the inverse matrix is

2. Find x if

   

    

Solution

The solution to Ax = b is x = A^-1b.  So we multiply

Problem 2

Let

1. Find A² and A³.  

   Solution

   We have

              

2. Make a conjecture about A^k.  

   Solution

   

3. Use induction to prove your conjecture from part B.  

   Solution

   For the case k  =  1, this is trivial

                    

   Assume the statement is true for k.  Then

   

   We need to show that conjecture is true for k + 1.  We have

   

   so by mathematical induction, the theorem is true.

Problem 3 

Let 

Find a 2x1 matrix v such that Av = 2v.

Solution

We write

or

this gives us the two equations

            v₁ + v₂  =  2v₁

            -2v₁ + 4v₂ =2v₂

or

            -v₁ + v₂  =  0

            -2v₁ + 2v₂  =  0

Notice the second equation is a multiple of the first.  We can pick

            v₁  =  1      v₂  =  1

or

               

Part 2

Please work out each of the given problems.  Credit will be based on the steps that you show towards the final answer.  Show your work. 

Problem 4 

Answer the following true or false and explain your reasoning.

A.     If A and B are n x n matrices and AB  =  0, then either A = 0 or B = 0.

Solution

False, for example, let

B.     If A is a nonsingular matrix with

and

        is a solution, then

        is not a solution.

Solution

This is true by the theorem on nonsingular equivalences (TFAE).  Since A is nonsingular, Ax = b has a unique solution, hence there cannot be two distinct solutions.

Problem 5 

A word, v, was sent through a transmission via the matrix

and the code word received was

Determine if the signal is in error and if it is, correct the message and find the word.  If it is not, find the word.

Solution

We find the complete set of words and code words first:

Next, notice that the received message does not match any of the code words, hence the message is in error.  To fix the error, we look at the code word that is closest to the received message.  The number of bits that the received message differs from the code words is 4, 5, 1, and 2.  Hence the original word is (1,0).

Problem 6 

In the city of Digraphville, there are four food-processing plants:  the apple plant, the beet plant, the carrot plant, and the dairy plant.  There are one-way roads from the apple plant to the beet plant and to the dairy plant.  There is also a one-way road from the beet plant to the carrot plant.  There are two-way roads from the apple plant to the carrot plant, from the beet plant to the dairy plant and from the carrot plant to the dairy plant.

A.     Sketch the digraph for this situation  

Solution

B.     Write down the adjacency matrix  

Solution

We have

C.     Use the adjacency matrix to determine how many ways are there to drive from the apple plant to the dairy plant using no more than four roads counted with multiplicity.  

Solution  

The number of ways to drive from the apple plant to the dairy plant using no more than four roads counted with multiplicity is given by 

        [A + A² + A³ + A⁴]₁₄ 

We have

Hence there are 16 ways of getting from the apple plant to the dairy plant using no more than four roads.

Problem 7 

Prove that if A, B, and C are n x n matrices, then

A(B + C)  =  AB + AC

Solution

Proof

We have 

        [A(B + C)]_ij  =  S(a_ik(B + C)_kj)  =  S(a_ik(b_kj + c_kj) )

        =  S(a_ik(b_kj + c_kj) )  =  S(a_ikb_kj + a_ikc_kj)  =  Sa_ikb_kj + Sa_ikc_kj 

        =  (AB)_ij + (AC)_ij 

Problem 8 

Prove that if v and w are solutions to the matrix equation Ax = b and if r + s = 0, then

rv + sw is a solution to the homogeneous equation Ax = 0.  

Solution

  Proof

Since v and w are solutions to the matrix equation equation.  We have

        Av  =  Aw  =  b

We have

        A(rv + sw)  =  A(rv) + A(sw)

        =  rAv + sAw  =  rb + sb  =  (r + s)b  =  (0)b  =  0

Extra Credit:  Write down one thing that your instructor can do to make the class better and one thing that you want to remain the same in the class.

(Any constructive remark will be worth full credit.)