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4 years ago

Dada a Matriz A= [1 -2 1 ][-3 1 4] [0 6 8] , escreva uma matriz B diferente de A, tal que detA =detB.

Mathematics

1 answer:

Mariulka [41]4 years ago

3 0

Use o fato de que o determinante de qualquer matriz quadrada é o mesmo da sua transposta.

$\mathbf A=\begin{bmatrix}1&-2&1\\-3&1&4\\0&6&8\end{bmatrix}$

$\mathbf B=\mathbf A^\top=\begin{bmatrix}1&-3&0\\-2&1&6\\1&4&8\end{bmatrix}$

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prohojiy [21]

Answer:

answer is 11.

Step-by-step explanation:

the median is always the middle

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3 years ago

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In an arithmetic sequence, the nth term an is given by the formula an=a1+(n−1)d, where a1 is the first term and d is the commo

Dmitry_Shevchenko [17]

Answer:

$a_{10} = \frac{10}{65536}$

Step-by-step explanation:

The first step to solving this problem is verifying if this sequence is an arithmetic sequence or a geometric sequence.

This sequence is arithmetic if:

$a_{3} - a_{2} = a_{2} - a_{1}$

We have that:

$a_{3} = 40, a_{2} = 10, a_{3} = \frac{5}{2}$

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$\frac{5}{2} - 10 = 10 - 40$

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This sequence is geometric if:

$\frac{a_{3}}{a_{2}} = \frac{a_{2}}{a_{1}}$

$\frac{\frac{5}[2}}{10} = \frac{10}{40}$

$\frac{5}{20} = \frac{1}{4}$

$\frac{1}{4} = \frac{1}{4}$

This is a geometric sequence, in which:

The first term is 40, so $a_{1} = 40$

The common ratio is $\frac{1}{4}$ , so $r = \frac{1}{4}$ .

We have that:

$a_{n} = a_{1}*r^{n-1}$

The 10th term is $a_{10}$ . So:

$a_{10} = a_{1}*r^{9}$

$a_{10} = 40*(\frac{1}{4})^{9}$

$a_{10} = \frac{40}{262144}$

Simplifying by 4, we have:

$a_{10} = \frac{10}{65536}$

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The 4th selection is appropriat:
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Answer:

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or you can do:

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