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Mathematics, 21.10.2019 23:30 ykluhredd

Let a = 3
2 5
b = βˆ’2 7
3 βˆ’4
c = βˆ’5 3
2 βˆ’1
= 0
0 0
and = 0
0 1
evaluate the following.
a. a + b
b. b + a
c. a + (b + c)
d. (a + b) + c
e. a +
f. a +
g. a β‹…
h. β‹… a
i. β‹… a
j. a β‹… b
k. b β‹… a
l. a β‹… c
m. c β‹… a
n. a β‹… b + a β‹… c
o. a β‹… (b + c)
p. a β‹… b β‹… c
q. c β‹… b β‹… a
r. a β‹… c β‹… b
s. det(a)
t. det(b)
u. det(c)
v.
w.
x. det(a β‹… b β‹… c)
y. det(c β‹… b β‹… a)

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Answers: 1

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You know the right answer?
Let a = 3
2 5
b = βˆ’2 7
3 βˆ’4
c = βˆ’5 3
2 βˆ’1
= 0
0 0
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