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saul85 [17]
2 years ago
12

Matrix multiplication was used to encode a message using the given encoding matrix: [i 47 A=1 |-1 -3] The original message was c

onverted to row matrices of size: 1x 2 and each was multiplied by A. sp 0 A 1 B 2 C 3 D 4 E F G H I J K L M N O P Q R 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 14 | 15 | 16 | 17 | 18 S T U V W X Y Z 19 20 21 22 23 24 25 26 Coded message: 11 52 -8 -9 -13 -39 5 20 12 56 5 20 -2 7 9 41 25 100 Find A and use it to decode the message

Mathematics
1 answer:
nadya68 [22]2 years ago
3 0

Answer:

A^{-1}=\left[\begin{array}{cc}-3&-4\\1&1\end{array}\right] message SHOW_ME_THE_MONEY_  

Step-by-step explanation:

The matrix

A=\left[\begin{array}{cc}1&4\\-1&-3\end{array}\right]\rightarrow |A|=(1 \times -3)-(-1\times 4)=1\\\rightarrow A^{-1}=\left[\begin{array}{cc}-3&-4\\1&1\end{array}\right] \\

We can check that in fact A*A^⁻1=I_2 the identity matrix of size 2 x 2.

Now the message was divided in 1 x 2 matrices, then we have that the sequence given is the result of multiplying m by A, so to get m again we multiply now by A^⁻1. and we get the next table

Encoded message    Decoded message    message in letters by association

11 52        19 8    S H

-8 -9         15 23  O W

-13 -39        0 13   _ M

5 20         5 0   E _

12 56        20 8    T H

5 20         5 0    E _

-2 7           13 15  M O

9 41         14 5   N E

25 100       25 0    Y _

Then the message decoded is SHOW_ME_THE_MONEY_                                

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Step-by-step explanation:

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Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7a
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Answer:

  • a) 3/5·((-2)^n + 4·3^n)
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Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form a[n]=r^n, then it will satisfy ...

  r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}

Rearranging and dividing by r^{n-2}, we get the quadratic ...

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The quadratic formula tells us values of r that satisfy this are ...

  r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

  a[n]=pr_1^n+qr_2^n

We can find p and q by solving the initial condition equations:

\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]

These have the solution ...

p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}

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Using these formulas on the first recurrence relation, we get ...

a)

c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n

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The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

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For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

  a[n]=(p+qn)r^n

The initial condition equations are now ...

\left[\begin{array}{cc}1&0\\r&r\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]

and the solutions for p and q are ...

p=a[0]\\\\q=\dfrac{a[1]-a[0]r}{r}

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Using these formulas on problem (d), we get ...

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c_1=2,\ c_2=-1,\ a[0]=4,\ a[1]=1\\\\r=\dfrac{2+\sqrt{2^2+4(-1)}}{2}=1\\\\p=4,\ q=\dfrac{1-4(1)}{1}=-3\\\\a[n]=4-3n

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And for problem (f), we get ...

f)

c_1=-6,\ c_2=-9,\ a[0]=3,\ a[1]=-3\\\\r=\dfrac{-6+\sqrt{6^2+4(-9)}}{2}=-3\\\\p=3,\ q=\dfrac{-3-3(-3)}{-3}=-2\\\\a[n]=(3-2n)(-3)^n

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<em>Comment on problem g</em>

Yes, the bases of the exponential terms are conjugate irrational numbers. When the terms are evaluated, they do resolve to rational numbers.

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Answer:

I think the other angles are 40 degrees excpet the "x" which is 100 also

Step-by-step explanation:

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