Math 1113 Practice for Test 4
x1
 2 x2
 3x3
 6
1. Given the system of equations 5 x1
 7 x2
 2 x3
 0
3x1
(a) Write down the coefficient matrix.
 4 x2

 8
 1
 5

 3
x3
3
7 2 
4
1
2
(b) Write down the augmented matrix.
 1
 5

 3
2
6
0 
8
3
7 2
4
1
(c) Using row operations put the augmented matrix into row echelon form.
R2  R2  5R1




R3
R3  R3  3R1
6
0 17
13 30 
0 2 8 10
 R3  (2)
1
2
3
 1 2 3 6
 0 17 13 30 


 0 1 4 5
R3  R2
 1 2 3 6
 0 1 4 5


 0 17 13 30 
R3  R3  17 R1




R3
6
0
1
4
5
0
0 55 55
 R3  (55)
1
2
3
 1 2 3 6
 0 1 4 5


 0 0 1 1
(d) Use back substitution to solve the system.
x3  1
x2  4 x3  5
x2  4(1)  5
x2  1
x1  2 x2  3 x3  6
x1  2(1)  3(1)  6
x1  1
Answer: (1, 1, 1)
2. Solve the system that has reduced row echelon form using variables x, y and z.
 1 0 4 2
 0 1 1 1


 0 0 0 0 
za
y  z  1
y  a  1
y  1  a
x  4z  2
x  4a  2
x  2  4a
Answer: (2  4a,  1  a, a)
3. Solve the system that has reduced row echelon form
 1 3 0 0 
0 0 1 0 


 0 0 0 1
No solution. The system is inconsistent when there is a leading one in the last column
2
4. Given A    and B  [5 7] compute AB and BA
3
10 14 
AB  
BA=[31]

15 21
5. A pie maker produces three types of pie in two facilities In the matrix A the number of
pies of type i baked at facility j is represented by ai j .
100 400 
A   220 120 
 324 312 
The incomes per pie are represented by the matrix B =[ $3.00 $2.50 $4.25]
Compute BA and interpret the result.
100 400
BA= =[ 3.00 2.50 4.25] 220 120   [ 2227 2826]


324 312 
The total income at facility 1 is $2,227 and the total income at facility 2 is $2826
 1 4 1 1 0 0
3
1 1 0 1 0 
6. (a) Find the inverse of 
.
 1 2 0 0 0 1
R2  R2  3R1
R3  R3  R1
4
1 1 0 0
1
0 11 2 3 1 0 


0 6 1 1 0 1
R2  R2  2 R3
1 1 0
0
1 4
0
1 0 1 1 2 

 0 6 1 1 0
1
R3  R3  6 R2
1 0
0
1 4 1
0 1 0 1 1 2 


0 0 1 7 6 11
R3  ( 1) R3
0
1 4 1 1 0
 0 1 0 1
1 2 

 0 0 1 7 6 11
R1  R1  R3
6 11
 1 4 0 6
 0 1 0 1
1 2 

 0 0 1 7 6
11
R1  R1  4 R2
 1 0 0 2 2 3
 0 1 0 1 1 2 


 0 0 1 7 6 11
 2 2 3
1
A   1 1 2
 7 6 11
1
(b) Solve AX  B , where B  1 , using your answer to part (a)
1
 2 2 3 1  3
X  A B   1 1 2 1   2
 7 6 11 1 12 
1
2 0 0 1
7. Evaluate
1 3 0
0
4 0 3
0
using a cofactor expansion.
0 7 0 11
Expand along the third column
2 0 1
 (1)
3 3
31 3
0
0 7
11
2 0 1
3 1 3
0
0 7 11
Now expand along the first row (or whatever)
3 0
1 0
1 3
 3[2
0
 ( 1)
]
7 11
0 11
0 7
 3[2(33)  7]
 177
8. Use Cramer's rule to find y in the following system:
x  2y  z  1
x 
y  z  0
3x  4 y  z  0
1 1
1
1 0 1
y

3 0
1 2
1 1
1
3 1
4


 2
1 (1)(1)(1)  (2)(1)(3)  (1)(1)(4)  (1)(1)(3)  (1)(1)(4)  (2)(1)(1) 2
1 1 1
3 4
1
9. Find the area of the triangle with vertices (1, 3), (4, 2) and (−1, 5)
x1 y1 1
1
Use area =  x2 y2 1
2
x3 y2 1
1 3 1
4 2 1  (1)(2)(1)  (3)(1)(1)  (1)(4)(5)  (1)(2)(1)  (1)(1)(5)  (3)(4)(1)  4
1 5 1
1
Area = (4)  2
2
10. Use a determinant to find an equation of the line through (−2, 4) and (5, −3).
x y 1
Use x2 y2 1  0
x3 y2 1
x
2
y 1
4 1 0
5 3 1
x
4 1
3 1
y
2 1
5 1
7 x  7 y  14  0
This simplifies to x  y  2  0

2
4
5 3
0
11. A message is encoded using the matrix in Question 6. The coded message is
50 73 28 16 40 12 62 64 29. Decode the message
A B C D E F G H I
1
2
3
4
5
6
7
8
J
K
M
N
O
P
Q
R
S
T
U
V
W
X
Y
Z
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
2 3 
 2

[50 73 28]  1 1 2 
 7 6 11
2 3 
2 3 
 2
 2



[16 40 12]  1 1 2  [62 64 29]  1 1 2 
 7 6 11
 7 6 11
[23 5 12] [12 0 4] [15 14 5]
WELL DONE
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