Math Problem Statement

A is a 2 βœ• 2 matrix with eigenvectors v1 =

1

βˆ’1 and v2 =

1

1 corresponding to eigenvalues πœ†1 = 1 2 and πœ†2 = 2, respectively, and x =

5

βˆ’1 . Find A9x.

Solution

To solve for A9xA^9x, we can leverage the fact that AA is a diagonalizable matrix. Given the eigenvectors v1v_1 and v2v_2, and their corresponding eigenvalues Ξ»1\lambda_1 and Ξ»2\lambda_2, the matrix AA can be represented as:

A=PDPβˆ’1A = PDP^{-1}

where PP is the matrix of eigenvectors and DD is the diagonal matrix of eigenvalues.

Given:

  • Eigenvectors: v1=(1βˆ’1)v_1 = \begin{pmatrix} 1 \\ -1 \end{pmatrix} and v2=(11)v_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}
  • Eigenvalues: Ξ»1=12\lambda_1 = \frac{1}{2} and Ξ»2=2\lambda_2 = 2
  • Vector x=(5βˆ’1)x = \begin{pmatrix} 5 \\ -1 \end{pmatrix}
  1. Construct the matrix PP and DD:

P=(11βˆ’11),D=(12002)P = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}, \quad D = \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & 2 \end{pmatrix}

  1. Find the inverse of PP:

Pβˆ’1=12(1βˆ’111)P^{-1} = \frac{1}{2} \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}

  1. Compute A9A^9:

Since A=PDPβˆ’1A = PDP^{-1}, then A9=PD9Pβˆ’1A^9 = PD^9P^{-1}.

D9=((12)90029)=(151200512)D^9 = \begin{pmatrix} \left(\frac{1}{2}\right)^9 & 0 \\ 0 & 2^9 \end{pmatrix} = \begin{pmatrix} \frac{1}{512} & 0 \\ 0 & 512 \end{pmatrix}

A9=P(151200512)Pβˆ’1A^9 = P \begin{pmatrix} \frac{1}{512} & 0 \\ 0 & 512 \end{pmatrix} P^{-1}

  1. Calculate A9xA^9x:

A9x=PD9Pβˆ’1xA^9x = PD^9P^{-1}x

First, find Pβˆ’1xP^{-1}x:

Pβˆ’1x=12(1βˆ’111)(5βˆ’1)=12(5βˆ’(βˆ’1)5+(βˆ’1))=12(64)=(32)P^{-1}x = \frac{1}{2} \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 5 \\ -1 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 5 - (-1) \\ 5 + (-1) \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 6 \\ 4 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}

Then, multiply by D9D^9:

D9(32)=(151200512)(32)=(35121024)D^9 \begin{pmatrix} 3 \\ 2 \end{pmatrix} = \begin{pmatrix} \frac{1}{512} & 0 \\ 0 & 512 \end{pmatrix} \begin{pmatrix} 3 \\ 2 \end{pmatrix} = \begin{pmatrix} \frac{3}{512} \\ 1024 \end{pmatrix}

Finally, multiply by PP:

A9x=(11βˆ’11)(35121024)=(3512+1024βˆ’3512+1024)=(1024.0058593751023.994140625)A^9x = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} \frac{3}{512} \\ 1024 \end{pmatrix} = \begin{pmatrix} \frac{3}{512} + 1024 \\ -\frac{3}{512} + 1024 \end{pmatrix} = \begin{pmatrix} 1024.005859375 \\ 1023.994140625 \end{pmatrix}

So,

A9xβ‰ˆ(1024.011023.99)A^9x \approx \begin{pmatrix} 1024.01 \\ 1023.99 \end{pmatrix}

Would you like more details on any part of this process?

Related Questions:

  1. How do you diagonalize a matrix?
  2. What are the properties of eigenvalues and eigenvectors in matrix multiplication?
  3. How do you compute matrix powers using eigenvalues and eigenvectors?
  4. What are some practical applications of diagonalization in engineering?
  5. How can one find the inverse of a 2x2 matrix quickly?

Tip:

When dealing with matrix exponentiation, diagonalization is a powerful tool that simplifies the computation, especially for high powers.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Matrix Diagonalization
Eigenvalues and Eigenvectors
Matrix Exponentiation

Formulas

A = PDP^{-1}
A^9 = PD^9P^{-1}
D^9 = diag(Ξ»1^9, Ξ»2^9)

Theorems

Diagonalization Theorem

Suitable Grade Level

College Level