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Quiz 6 Solutions for Math 121 - November 16, 2006, Quizzes of Algebra

The answers and hints for quiz 6 of math 121, which was held on november 16, 2006. It includes the solutions for finding the product of two matrices, solving a system of linear equations using matrix inverses, and evaluating a sum.

Typology: Quizzes

Pre 2010

Uploaded on 08/16/2009

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Math 121, Quiz 6, 16 November 2006
Name: Hints and Answers
1. Let A=21 4
3 1 0 ,B=1 3
21. Find the following, if possible, if an operation
is not possible, state why it is not possible.
(a) AB (b) BA
Answer. (a) Not possible because the number of columns of Ais 3 which is not equal to the
number of rows of Bwhich is 2.
(b) BA exists and is found as follows
BA =1 3
21 21 4
3 1 0
=(1)(2) + (3)(3) (1)(1) + (3)(1) (1)(4) + (3)(0)
(2)(2) + (1)(3) (2)(1) + (1)(1) (2)(4) + (1)(0) =7 2 4
73 8
2. Consider the following system of equations:
x+y+ 2z=2
2x+ 3y+ 3z= 3
3x+ 3y+ 7z=1
.
Solve this system using the fact that the inverse of
1 1 2
2 3 3
3 3 7
is
12 13
5 1 1
3 0 1
.
Answer. Multiply the inverse of Aby the matrix of constants as follows:
x
y
z
=
12 13
5 1 1
3 0 1
2
3
1
=
24
12
5
Thus the solution is x=24, y = 12, z = 5.
3. Evaluate the sum
5
X
k=3
2k1.
Answer.
5
X
k=3
2k1 = 2(3) 1 + 2(4) 1 + 2(5) 1 = 5 + 7 + 9 = 21.

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Math 121, Quiz 6, 16 November 2006

Name: Hints and Answers

  1. Let A =

[ 2 − 1 4

]

, B =

[ 1

]

. Find the following, if possible, if an operation is not possible, state why it is not possible. (a) AB (b) BA

Answer. (a) Not possible because the number of columns of A is 3 which is not equal to the number of rows of B which is 2. (b) BA exists and is found as follows

BA =

[ 1

] [ 2 − 1 4

]

[ (1)(2) + (3)(−3) (1)(−1) + (3)(1) (1)(4) + (3)(0)

]

[ − 7 2 4

]

  1. Consider the following system of equations:

x + y + 2 z = − 2 2 x + 3 y + 3 z = 3 3 x + 3 y + 7 z = − 1

Solve this system using the fact that the inverse of

 (^) is

Answer. Multiply the inverse of A by the matrix of constants as follows:  

x y z

Thus the solution is x = − 24 , y = 12, z = 5.

  1. Evaluate the sum

∑^5

k=

2 k − 1.

Answer.

∑^5

k=

2 k − 1 = 2(3) − 1 + 2(4) − 1 + 2(5) − 1 = 5 + 7 + 9 = 21.