Expert: Scott A Wilson - 10/26/2008

Question
Can you please check these for me? I am more confident of my answers this time.

1) Write the augmented matrix for the system of linear equations.

I think its A

3x - 2y + 5z = 31
x + 3y - 3z = -12
-2x - 5y + 3z = 11


a)    [5, -6, 0... 0, 1, 7... 7, 9, 3...10, -12, 0]

b)  [2, 3, 1... 1, -5, -2... 2, -1, -3... 2, 4 -6]

c)  [2, 5, 0.. 11, -2, 1... 8, 9, -3... 0, 1, 6]

d)  [3, 1, 2... -2, 3, -5.. 5, -3, 3... 31, -12, 11]



2) Write the augmented matrix for the system of linear equations.

I think its C

x - 2y + 3z = 9
y + 3z = 5
z = 2

a)    [4, 6, 11... 3, -3, 0... -1, 3, 0... 10, -1, -4]

b)    [3, 4, 5... -2, 0, 1... 6, -9, 11... 0, 1, -2]

c)    [2, 5, 0... 11, -2, 1... 6, -9, 11... 0, -3, -6]

d)    [1, 0, 0... -2, 1, 0... 3, 3, 1.. 9, 5, 2]


3) Write the system of linear equations represented by the following matrix. Use x, y, and z as variables:
  
I think it is C

[7, 0, 2.. 0, 1, 7.. 4, -5, 0.. -13, 11, 6]

a) 7x + 4z = -13
   y - 5z = 11
   2x + 7y = 6

b) 4x - 13y = 0
   x - 5y = 11z
   2x + 7y + 6z = 9

c) 7x + y + 4z = -13
   x + y -5 = 11
   2x + 7y + z = 6

d) 7x + 4 =  -13
   x + 1 = -5z
   2x - 7 = 6y


4) Solve the matrix

I think its B

[1, 0, 0... 2, 1, 0... 1, 0, 1... 0, -2, 3]

a) (2, 1, 1)
b) (0, 0 1)
c) (1, -2, 3)
d) 1, 2, 3)

5) Solve the matrix

I think its A

[1, 0, 0... 1, 1, 0.. 0, 3/2, 1... 3, -2, 0]

a) (5, -2, 0)
b) (1, 3, -2)
c)  (-3,  -5, 1)
d) (1, -2, 1)


6) Perform row operation and write the new matrix

I think its D

[1, 3, 2... -3, 1, -2... 2, -1, 1... 0, 7, 3] -3R^1 + R^2

a) [1, 2, 0... -3, -2, 10... 2, 1, -7... 0, 3, 7]

b) [1, 0, 2.. -3, 10, -2... 2, -7, 1... 0, 7, 3]

c)  [1, -9, 2.. -3, -6, -2... 2, 1, 1... 0, 11, 3]

d)  [2, 10, 1.. -3, -7, 0... 2, 1, 0... 0, -5, 0]


7) Solve the system of equations using matrices.

I think its C

3y - z = -1
x + 5y -z = -4
-3x + 6y + 2z = 11

a) (1, 0, 3)
b) (-2, 4, 3)
c) (0, 0, 4)
d) (-3, 0, 1)


8) Solve the system of equations using matrices.

I think its D

x - 3z = -2
2x + 2y + z = 4
3x + y - 2z = 5

a) (3, 6, 1)
b) (4, -3, 2)
c) 1, -2, -4)
d) (2, -5, 6)


9) Evaluate the determinant in the matrix |4, 5...8, 6|

I think its A

a) 11
b) 23
c) -16
d) -22


10) Evaluate the determinant in the matrix |7, -2... -3, 2|

I think its C

a) 20
b) -11
c) 42
d) -22


11) Using the Cramer's rule, solve the following system:

I think its B

x - 2y = 5
5x - y = -2

a) (3, -5)
b) (0, 3)
c) (-1, -3)
d) (-2, -2)


12) Using the Cramer's rule, solve the following system:

I think its A

3x + 2y = 2
2x + 2y = 3

a) {-1, 5/2}
b) (0, -3
c) {-7/4, -2}
d) (0, 0)


13) Using the Cramer's rule, solve the following system:

I think its B

2x - 9y = 5
3x - 3y = 11

a) (3, 1)
b) {-1/4, -2}
c) {4, 1/3}
d) (-2, 1)


Answer
It looks like you’re guessing at the answers, for almost all of them are wrong.
If you knew what they were talking about, it would be that hard.   

If the matrix is
1   2   3
4   5   6
7   8   9
and the solution is
7
28
46,
augmented just means the combination of the matrix and the solution, so the augmented matrix here is
1   2   3   7
4   5   6   28
7   8   9   46.

By the way, the equations would be x+2y+3z=7, 4x+5y+6z=28 and 7x+8y+0z=46.

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Note that the way they write equations in matrix format is by listing the coefficient of the variables.
If the equations were

x+2y=7
3y+4z=10
5x+6z=21

The augmented matrix would be

1   2   0   7
0   3   4   10
5   0   6   21.

Note that the 0’s were added because there is no variable.  In the first equation, there is no z, so the third element in row 1 has a 0.  In equation 2, there is no x, so the first element in row 2 is a 0.  In equation 3 there is no y.  That puts a zero in the … you get it, right?

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The determinant of a matrix is like this (on a 2x2 matrix):
If the matrix is

6   2
3   7,
the determinant is 6*7 – 3*2 = 42 – 6 = 36.

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Cramer’s rule says to transform the matrix into an upper diagonal and do back substitution.  

Example
The equations are
2x+3y+z=13
4x+5y+7z=29
6x+4y+4z=32

The corresponding matrix is
2   3   1   13
4   5   7   29
6   5   4   32

(When they write out matrices in a single line, the do each of the columns.  
This matrix would be 2 4 6 … 3 5 5… 1 7 4 … 13 29 32)

The first step is to get a 1 at the top and 0’s below in the first column.
Row 1: divide by 2
Row 2: subtract 2*(row 1)
Row 3: subtract 3*(row 1)

1   1.5   0.5   6.5
0   –1   5   3
0   –4   1   –7

Cramer’s rule then says multiply row 2 by –1 and subtract 4*(row 2) from row 3.  The result is
1   1.5   0.5   6.5
0   1   -5   -3
0   0   -19   –19.

Divide row 3 by –19, giving
1   1.5   0.5   6.5
0   1   -5   -3
0   0   1   1.

Now that below the diagonal there are 0’s, do back substitution.
From row 3, z=1.
Row 2 says that y-5z=-3, so y=2.
Row 1 says that x+1.5y+0.5z=6.5,  so x = 3.

When the solution is found, put the answer back in to make sure the problem is write.
If we take (x, y, z) as (3, 2, 1),  and remember the original matrix,
1) 2x+3y+z=13
3) 4x+5y+7z=29
4) 6x+5y+4z=32,
we get (for 1) 2*3 + 3*2 + 1*1 = 6+6+1 = 13 (check),
(for 2) 4*3 + 5*2 + 7*1 = 12 + 10 + 7 = 29 (check), and
(for 3) 6*3 + 5*2 + 4 = 32.

Review these and try your problems again.