Expert: Sherry Wallin - 12/11/2010

Question
Hi Sherry, last time I asked you a question you requested
that I submit my work and answers for each problem, so I
have and I hope that you can correct any of my mistakes by
showing me the steps in each problem and answers, thanks
Sherry.

1) Find the point of intersection by any method: x+y=3 and
2x-y=9.

I graphed the 2 equations and came up with and intersection
of (4, -1)

2) How many times are y=x-1 and y=x^2-x-1 going to
intersect.

I graphed the 2 equations and say that a line and a parabola
where going to have 2 points of intersection.

3) Find (x,y,z)
x+y+z=5
y-z=3
z=1

Here is my work: y-1=3
z=1
y=3+1
y=4
x+4+1=5
x+5-5
x=0

4) Are these two lines parallel? y=3x=4 and y=6/2x-4
My answer is no, because they are the exact same line

5) Multiply [1  2  0]*[1]
                     [2]
                     [0]

My answer is 1  2  0
            2  4  0
            0  0  0

6)Find x given [12  2] =[2+x  2]
              [-1  9]  [-1   9]

2+x=12
x=12-2
x=10

7) Solve   [-2  2]      [2  5]
        2*[-1  9] + 3* [1  0]

I got [-4   4]   [6  15]   [2  19]
     [-2  18] + [3   0] = [1  18]

8) Solve [-2  2]   [2  -5]   [-4  7]
        [-1  1] - [1   1] = [-2  0]

9) Find the inverse of [2  1]
                      [7  4]

I got [4  -1]
     [-7  2]

10) Find the determinant of [2   2]
                           [7  -1]

I got A= 2.00  2.00
        7.00 -1.00

det(A)= 2.00  2.00
       7.00 -1.00 = -16


Answer
Page~

For #1: Graphing to find a point of intersection is the last thing you want to do. It only works if the answer is an integer. Try using substitution or the addition method to solve the 2 equation system.

eqn 1:  x+y=3
eqn 2: 2x-y=9
multiply eqn 1 by -2 and add to eqn 2:
    -2x -2y = -6
     2x -y  = 9
    ------------
       -3y  = 3
divide by -3
y = -1 and now take this value of y and plug it into either of eqn 1 or eqn 2: I will use eqn 1:  x +(-1) = 3 => x = 4, so your point of intersection is (4,-1) but this only worked the way you did it because -1 and 4 are integers. This would not work if say your coordinates were 1/2, and 2/3.

For #2:  A line and a parabola can intersect at most 2 times but not necessarily two times. Again you need to set them equal to each other and find where they intersect. (You can do this with these two equations because they are both in terms of y):
x-1 = x^2-x-1
-x+1      -x+1
0 =  x^2-2x => x(x-2) = 0 => x = 0 and x-2 = 0 => x = 2
Now plug these values of x into either one of the original equations and find y: Using x - 1: 0-1 = -1 => (0,-1) and 2-1 = 1 => (2,1) as the two points of intersection. I can show you one system with a line and a parabola that does not have 2 points of intersection, in fact, I can find a line and a parabola that do not intersect at all. y = x^2 + 1 will not intersect even once with y = -2 and y = x^2 -1 and x = -1 only intersects at (-1,0).

#3:  looks good
#4: your answer looks good, both lines have the same slope but by the definition of parallel they cannot intersect and these (of course) intersect everywhere
#5: looks good
#6: looks good
#7: Solve [-2  2]  [2  5] I can't tell what operation is going on here
       2*[-1  9] + 3* [1  0] but your answer for this one looks ok
#8: Solve [-2  2]   [2  -5]   [-4  7] again, what is the operation?
       [-1  1] - [1   1] = [-2  0] this one looks fine
#9: This is good, however, you didn't tell me how you got your answer.
   There are a couple of ways to do this and the easiest is to find  the determinant and multiply the original 2x2 matrix by 1/det and in the original matrix interchange the elements 1,1 and 2,2 (these are the elements on the diagonal) and change the signs on the elements on the off diagonal. The other way is to write the original matrix and put the identity to the right of it and solve that matrix and when you are done the inverse will be on the right and the identity on the left.
#10: Your answer is right but why all the zeros? Just multiply the diagonal elements and subtract of the product of the elements on the off diagonal. This only works on 2x2 matrices as well.

Math Prof