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3 years ago

2i^4−5i^3+3i^2+√−81 rewritten in a+bi form?

Mathematics

1 answer:

Margarita [4]3 years ago

5 0

Answer:

-1 + 14i.

Step-by-step explanation:

i^2 = -1.

i^4 = i^2 * i^2

= -1 * -1 = 1.

i^3 = i^2 * i = -i.

√−81 = i√81 = 9i.

2i^4−5i^3+3i^2+√−81

= 2 - (5 * -i) - 3 + 9i

= -1 + 5i + 9i

= -1 + 14i.

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4 years ago

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3 years ago

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xeze [42]

Answer:

The correct answer is A..

Step-by-step explanation:

From the Invertible Matrix Theorem (IMT) we have a set of equivalent conditions to determine if a square matrix is invertible or not. In particular, it says that a square matrix of dimension tex]n\times n[/tex] is invertible if and only if, its columns span the vector space tex]R^n[/tex].

In the particular case of this exercise we have a matrix of dimension tex]5\times 5[/tex]. So, by the Invertible Matrix Theorem its columns must span the vector space tex]R^5[/tex]. Now, according to the statement of the exercise this condition does not hold. Hence, the given matrix cannot be invertible.

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3 years ago