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

       

  1. 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 Ak.  

    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 

Show that if AT = A-1, then |det(A)| = 1.

Solution

From the definition of the inverse, we have:

    AA-1  =  I

Now take the determinate of both sides:

    det(AA-1)  =  det(I)

The determinant of the product is the product of determinant, so the left hand side is

    det(A)det(A-1)

    =  det(A)det(AT)

The determinant of AT is the determinant of A, so

    =  (det(A))2

The right hand side is just 1, since the determinant of the identity matrix is 1, so the equation becomes

       (det(A))2  = 1

so

    det(A) = 1 or det(A) = -1

 

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(aik(B + C)kj)  =  S(aik(bkj + ckj) )

        =  S(aik(bkj + ckj) )  =  S(aikbkj + aikckj)  =  Saikbkj + Saikckj 

        =  (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.)