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2 years ago

8) 4.8.12.5.10. -17.

Mathematics

1 answer:

ASHA 777 [7]2 years ago

8 0

Answer:

1) 1 3 7 13 21 31 43 ( +2, +4, +6, +8, +10 и т.д.)

2) 255 127 63 31 15 7 3 (число -1, а затем полученное число :2)

3)92 46 44 22 20 10 8 4 ( :2, -2, :2, -2, :2, -2 и т.д.)

4) 1 2 4 7 28 33 198 (+1, *2, +3, *4, +5, *6, +7, *8 и т.д)

5)3 7 6 7 9 7 12 7 15 7 (первое, третье, пятое и т.д. числа +3; между ними 7)

6) 6 8 11 13 16 18 21 23 (+2, +3, +2, +3 и т.д.)

7)1 4 8 13 19 26 34 43 (+3, +4, +5, +6, +7, +8 и т.д.)

8)128 64 32 16 8 4 2 (:2, :2, :2, :2 и т.д.)

9)3 4 8 17 33 58 94 (+1, +4, +9, +16, +25, +36 и т.д. (т.е. то, что прибавляется-увеличивается на +3, +5, +7, +9, +11, и т.д.))

10)47 39 32 26 21 17 14 12 (-8, -7, -6, -5, -4, -3 и т.д.)

11)256 37 64 31 16 25 4 19 (первое, третье, пятое и т.д. числа :4; а второе, четвертое, шестое и т.д. числа -6)

12)5 9 13 17 21 25 29 (+4, +4, +4, +4, +4 и т.д.)

13)1 4 3 6 5 8 7 10 (+3, -1, +3, -1, +3, -1 и т.д.)

14)2 4 6 12 14 28 30 60 (*2, +2, *2, +2, *2, +2 и т.д.)

15)11 12 10 13 9 14 8 15 (+1, -2, +3, -4, +5, -6, +7, -8 и т.д.)

Step-by-step explanation:

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The question is incomplete. The complete question is as follows:

Solve for X. Assume X is a 2x2 matrix and I denotes the 2x2 identity matrix. Do not use decimal numbers in your answer. If there are fractions, leave them unevaluated.

$\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ · X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ =I.

First, we have to identify the matrix I. As it was said, the matrix is the identiy matrix, which means

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

So, $\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ · X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

Isolating the X, we have

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ - $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

Resolving:

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}2-1&8-0\\-6-0&-9-1\end{array}\right]$

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Now, we have a problem similar to A.X=B. To solve it and because we don't divide matrices, we do X=A⁻¹·B. In this case,

X= $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ ⁻¹· $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Now, a matrix with index -1 is called Inverse Matrix and is calculated as: A . A⁻¹ = I.

So,

$\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ · $\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]$ = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

9a - 3b = 1

7a - 6b = 0

9c - 3d = 0

7c - 6d = 1

Resolving these equations, we have a= $\frac{2}{11}$ ; b= $\frac{7}{33}$ ; c= $\frac{-1}{11}$ and d= $\frac{-3}{11}$ . Substituting:

X= $\left[\begin{array}{ccc}\frac{2}{11} &\frac{-1}{11} \\\frac{7}{33}&\frac{-3}{11} \end{array}\right]$ · $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$