Math 121, Quiz 6, 16 November 2006
Name: Hints and Answers
1. Let A=2−1 4
−3 1 0 ,B=1 3
2−1. 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 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
2−1 2−1 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
7−3 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 −1−3
−5 1 1
−3 0 1
.
Answer. Multiply the inverse of Aby the matrix of constants as follows:
x
y
z
=
12 −1−3
−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
2k−1.
Answer.
5
X
k=3
2k−1 = 2(3) −1 + 2(4) −1 + 2(5) −1 = 5 + 7 + 9 = 21.