2x + 3y = 13 4x - y = -2
2 answers:
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The imaginary unit belongs to the set of complex numbers, denoted by
. These numbers take the form
, where
are any real numbers.
The set of real numbers, , is a subset of
, where each number in
can be obtained by taking
and letting
be any real number.
But any number in with non-zero imaginary part is not a real number. This includes
.
- "is it possible that i can use an imaginary number for a real number"
I'm not sure what you mean by this part of your question. It is possible to represent any real number as a complex number, but not a purely imaginary one. All real numbers are complex, but not all complex numbers are real. For example, 2 is real and complex because .
There are some operations that you can carry out on purely imaginary numbers to get a purely real number. A famous example is raising to the
-th power. Since
, we have
A) 4a - 3b + 2c
4(2, -1, 5) - 3(4, 3 , -2) + 2(5, 4, 0) = (8, -4, 20) - (12, 9, - 6) + (10, 8, 0) =
= (8 - 12 + 10 , -4 - 9 + 8 , 20 + 6 + 0) = (6, - 5, 26)
Answer: (6, - 5, 26)
b) magnitude of vector b
c) vector of length 7 parallel to vector c
=> m(5,4,0) = (5m,4m,0)
=>
=> m = 7 / √41 ≈ 1.093
=> 1.093 (5, 4, 0) = (5.465 , 4.372, 0)
Answer: (5.465 , 4.372 , 0)
4(2, -1, 5) - 3(4, 3 , -2) + 2(5, 4, 0) = (8, -4, 20) - (12, 9, - 6) + (10, 8, 0) =
= (8 - 12 + 10 , -4 - 9 + 8 , 20 + 6 + 0) = (6, - 5, 26)
Answer: (6, - 5, 26)
b) magnitude of vector b
c) vector of length 7 parallel to vector c
=> m(5,4,0) = (5m,4m,0)
=>
=> m = 7 / √41 ≈ 1.093
=> 1.093 (5, 4, 0) = (5.465 , 4.372, 0)
Answer: (5.465 , 4.372 , 0)