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pychu [463]
3 years ago
10

Wyatt is going to a carnival that has games and rides. Each game costs $1.25 and each ride costs $2.75. Wyatt spent $20.25 altog

ether at the carnival and the number of rides he went on is twice the number of games he played. Determine the number of games Wyatt played and the number of rides Wyatt went on.
Mathematics
1 answer:
Strike441 [17]3 years ago
5 0

Answer:

Wyatt played 3 games and went on 6 Rides

Step-by-step explanation:

Let games be represented as x

Let ride be represented as y

Each game cost $1.25

Each ride cost $2.75

Total money spent by Wyatt= $20.25

x1.25 + y2.75 = 20.25 ..... Equation 1

the number of rides he went on is twice the number of games he played

Y= 2x ... Equation 2

Substituting the value of y into equation 1

x1.25 + y2.75 = 20.25

x1.25 + 2(x)2.75 = 20.25

x1.25 + x5.5 = 20.25

x6.75= 20.25

x= 20.25/6.75

X= 3

Y= 2x

Y= 2(3)

Y= 6

Wyatt played 3 games and went on 6 Rides

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It looks like given matrices are supposed to be

\begin{array}{ccccccc}\begin{bmatrix}3&2\\2&3\end{bmatrix} & & \begin{bmatrix}1&-1\\2&-1\end{bmatrix} & & \begin{bmatrix}1&2&3\\0&2&3\\0&0&3\end{bmatrix} & & \begin{bmatrix}3&1&1\\1&3&1\\1&1&3\end{bmatrix}\end{array}

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• det(A) = determinant of A = product of eigenvalues

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and

λ₁ + λ₂ = 6   ⇒   λ₁ λ₂ = λ₁ (6 - λ₁) = 5

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⇒   λ₁ = 5 or λ₁ = 1

To find the corresponding eigenvectors, we solve for the vector v in Av = λv, or equivalently (A - λI) v = 0.

• For λ = 1, we have

\begin{bmatrix}3-1&2\\2&3-1\end{bmatrix}v = \begin{bmatrix}2&2\\2&2\end{bmatrix}v = 0

With v = (v₁, v₂)ᵀ, this equation tells us that

2 v₁ + 2 v₂ = 0

so that if we choose v₁ = -1, then v₂ = 1. So Av = v for the eigenvector v = (-1, 1)ᵀ.

• For λ = 5, we would end up with

\begin{bmatrix}-2&2\\2&-2\end{bmatrix}v = 0

and this tells us

-2 v₁ + 2 v₂ = 0

and it follows that v = (1, 1)ᵀ.

Then the decomposition of A into PDP⁻¹ is obtained with

P = \begin{bmatrix}-1 & 1 \\ 1 & 1\end{bmatrix}

D = \begin{bmatrix}1 & 0 \\ 0 & 5\end{bmatrix}

where the n-th column of P is the eigenvector associated with the eigenvalue in the n-th row/column of D.

(b) Consult part (a) for specific details. You would find that the eigenvalues are i and -i, as in i = √(-1). The corresponding eigenvectors are (1 + i, 2)ᵀ and (1 - i, 2)ᵀ, so that A = PDP⁻¹ if

P = \begin{bmatrix}1+i & 1-i\\2&2\end{bmatrix}

D = \begin{bmatrix}i&0\\0&i\end{bmatrix}

(c) For a 3×3 matrix, I'm not aware of any shortcuts like above, so we proceed as usual:

\det(A-\lambda I) = \det\begin{bmatrix}1-\lambda & 2 & 3 \\ 0 & 2-\lambda & 3 \\ 0 & 0 & 3-\lambda\end{bmatrix} = 0

Since A - λI is upper-triangular, the determinant is exactly the product the entries on the diagonal:

det(A - λI) = (1 - λ) (2 - λ) (3 - λ) = 0

and it follows that the eigenvalues are λ₁ = 1, λ₂ = 2, and λ₃ = 3. Now solve for v = (v₁, v₂, v₃)ᵀ such that (A - λI) v = 0.

• For λ = 1,

\begin{bmatrix}0&2&3\\0&1&3\\0&0&2\end{bmatrix}v = 0

tells us we can freely choose v₁ = 1, while the other components must be v₂ = v₃ = 0. Then v = (1, 0, 0)ᵀ.

• For λ = 2,

\begin{bmatrix}-1&2&3\\0&0&3\\0&0&1\end{bmatrix}v = 0

tells us we need to fix v₃ = 0. Then -v₁ + 2 v₂ = 0, so we can choose, say, v₂ = 1 and v₁ = 2. Then v = (2, 1, 0)ᵀ.

• For λ = 3,

\begin{bmatrix}-2&2&3\\0&-1&3\\0&0&0\end{bmatrix}v = 0

tells us if we choose v₃ = 1, then it follows that v₂ = 3 and v₁ = 9/2. To make things neater, let's scale these components by a factor of 2, so that v = (9, 6, 2)ᵀ.

Then we have A = PDP⁻¹ for

P = \begin{bmatrix}1&2&9\\0&1&6\\0&0&2\end{bmatrix}

D = \begin{bmatrix}1&0&0\\0&2&0\\0&0&3\end{bmatrix}

(d) Consult part (c) for all the details. Or, we can observe that λ₁ = 2 is an eigenvalue, since subtracting 2I from A gives a matrix of only 1s and det(A - 2I) = 0. Then using the eigen-facts,

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and we find λ₂ = 2 and λ₃ = 5.

I'll omit the details for finding the eigenvector associated with λ = 5; I ended up with v = (1, 1, 1)ᵀ.

• For λ = 2,

\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}v = 0

tells us that if we fix v₃ = 0, then v₁ + v₂ = 0, so that we can pick v₁ = 1 and v₂ = -1. So v = (1, -1, 0)ᵀ.

• For the repeated eigenvalue λ = 2, we find the generalized eigenvector such that (A - 2I)² v = 0.

\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}^2 v = \begin{bmatrix}3&3&3\\3&3&3\\3&3&3\end{bmatrix}v = 0

This time we fix v₂ = 0, so that 3 v₁ + 3 v₃ = 0, and we can pick v₁ = 1 and v₃ = -1. So v = (1, 0, -1)ᵀ.

Then A = PDP⁻¹ if

P = \begin{bmatrix}1 & 1 & 1 \\ 1 & -1 & 0 \\ 1 & 0 & -1\end{bmatrix}

D = \begin{bmatrix}5&0&0\\0&2&0\\0&2&2\end{bmatrix}

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