Solutions of x^5-32i=0 in rectangular form.

1 answer:

6 0

$\bf i^5\implies i^4i\implies 1i\implies i
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32=2^5\\\\
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x^5-32i=0\implies x^5=32i\implies x=\sqrt[5]{32i}\implies x=\sqrt[5]{2^5i^5}
\\\\\\
x=\sqrt[5]{(2i)^5}\implies x=2i\implies x=0+2i\implies (0~,~2)$

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disa [49]

The correct choice is C .

5 0

3 years ago

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(1 point) A, P and D are n×n matrices. Check the true statements below: A. A is diagonalizable if and only if A has n eigenvalue

yarga [219]

Answer:

Following are the answers to this questions:

A) False.

B) False.

C) True.

D) True.

Step-by-step explanation:

In option A,

Let A is $3 \times 3$ diagonal matrices that is,

$\left[\begin{array}{ccc}0&1&0\\0&0&6\\0&0&0\end{array}\right]$

here $A^3$ = 0, In this A is the nilpotent matrix and A has three values that are equal to 0 but it is not a diagonalizable matrix.

In option B,

If $A =\left[\begin{array}{ccc}0&1\\0&0\end{array}\right]$ in given value A elements are 0 and 1 as a distinct eigen value, that's why A is diagonalizable but |A| =0 its means A is not invertible, that's why it is false.

In option C,

It is given that $A= PDP^{-1}$ where A~ D, which is A is similar to diagonal matrices, that's why it is true.

In option D,

It is given that A has eigen vector, which leases from $R^n$ . A has eigen vector and A is $n \times n$ matrix, the number of A eigen values = given vector of A, that's why it is correct.