Note: IML notation is used for matrices in all cases.

1. Find the determinant of the matrix A in each case:

   a) Let A={2 4, 3 1}. 
   b) Let A={1 3, 6 4}.
   c) Let A={3 1 6, 7 4 5, 2 -7 1}.
   d) Let A={3 2 1, 2 1 -3, 4 0 1}.

   Solution:

   a) |A| = 2(1) - 4(3) = -10
   b) |A| = 1(4) - 3(6) = -14
   c) |A| = 3*|4 5, -7 1|(-1)^2 + 1*|7 5, 2 1|(-1)^3 
                    + 6*|7 4, 2 -7|(-1)^4
          = 3(39) - 1(-3) + 6(-57)
          = -222
   d) |A| = 3*|1 -3, 0 1|(-1)^2 + 2*|2 -3, 4 1|(-1)^3
                    + 1*|2 1, 4 0|(-1)^4
          = 3(1) - 2(14) + 1(-4)
          = -29

2. Compute the length of the following vectors:

   a) Let A={2, 4, 3}. 
   b) Let A={1, 3, 2}.
   c) Let A={-2, 1, -1, 2}.
   d) Let A={1, 0, 0, -4}.

   Solution (to 4dp):

   a) L(A)=sqrt(2^2+4^2+3^2)=sqrt(29)=5.3852
   b) L(A)=sqrt(1^2+3^2+2^2)=sqrt(14)=3.7417
   c) L(A)=sqrt((-2)^2+1^2+(-1)^2+2^2)=sqrt(10)=3.1623
   d) L(A)=sqrt(1^2+0^2+0^2+(-4)^2)=sqrt(17)=4.1231

3. Determine if the following vectors are perpendicular. Justify your
   answer.

   a) A={1, 3, 2} and B={-2, 1, -1}
   b) A={1, 2, 0, 1} and B={0, 0, 5, 0}

   Solution:

   a) Since A(T)*B = 1(-2)+3(1)+2(-1)=-1 then A and B are not perpendicular. 
   b) Since A(T)*B = 1(0)+2(0)+0(5)+1(0)=0 then A and B are perpendicular. 

4. Derive the characteristic equation of the following matrices.

   a) Let A={1 2, 4 3}
   b) Let A={13 -4 2, -4 13 -2, 2 -2 10}

   Solution:

   a) |A - lambda(I)| = |(1-lambda) 2, 4 (3-lambda)|
                      = (1-lambda)(3-lambda)-8 
                      = lambda^2 -4lambda - 5 = 0
   b) |A - lambda(I)| = |(13-lambda) -4 2, -4 (13-lambda) -2, 2 -2 (10-lambda)|
                      = (13-lambda)*|(13-lambda) -2, -2 (10-lambda)|(-1)^2 +
                        (-4)*|-4 -2, 2 (10-lambda)|(-1)^3 +
                        (2)*|-4 (13-lambda), 2 -2|(-1)^4
                      = (13-lambda)*((13-lambda)(10-lambda)-4) -
                        (-4)*(-4(10-lambda)+4) +
                        (2)*((-4)(-2) - 2(13-lambda))
                      = -lambda^3 + 36lambda^2 - 405lambda + 1458 = 0

5. Find the eigenvalues in each case.

   a) Let A={1 0, -2 4}
   b) Let A={1 -2, -2 0}
   c) Let A={-1 4, 0 5}
 
   Solution:

   a) |A - lambda(I)| = |(1-lambda) 0, -2 (4-lambda)|
                      = (1-lambda)(4-lambda) = 0
      Hence lambda(1)=1 and lambda(2)=4
   b) |A - lambda(I)| = |(1-lambda) -2, -2 (0-lambda)|
                      = (1-lambda)(0-lambda)-4 
                      = lambda^2 - lambda - 4 = 0
      Hence lambda(1)=(1+sqrt(17))/2 and lambda(2)=(1-sqrt(17))/2
   c) |A - lambda(I)| = |(-1-lambda) 4, 0 (5-lambda)|
                      = (-1-lambda)(5-lambda) = 0
      Hence lambda(1)=-1 and lambda(2)=5
     
   Reminder: The quadratic formula says that x=(-b+/-sqrt(b^2-4ac))/2a