Math Problem Statement

When A =  5 3 −4 −2  find matrices D and P in a diagonalization of A given that λ1 > λ2.

  1. D =  2 0 0 1  , P =  −1 4 1 −3 
  2. D =  2 0 0 1  , P =  1 −3 −1 4 
  3. D =  −1 0 0 −2  , P =  1 −3 −1 4 
  4. D =  −1 0 0 −2  , P =  −1 3 −1 4 
  5. D =  −1 0 0 −2  , P =  −1 4 1 −3 
  6. D =  2 0 0 1  , P =  −1 3 −1 4 

Solution

To determine the correct matrices DD and PP in the diagonalization of matrix AA, we need to identify:

  1. The eigenvalues of AA, which will make up the entries in DD.
  2. The corresponding eigenvectors of AA, which will make up the columns of PP.

Given: A=(5342)A = \begin{pmatrix} 5 & 3 \\ -4 & -2 \end{pmatrix}

Step 1: Finding Eigenvalues

The eigenvalues λ\lambda are solutions to the characteristic polynomial: det(AλI)=0\det(A - \lambda I) = 0 Calculate det(AλI)\det(A - \lambda I): AλI=(5λ342λ)A - \lambda I = \begin{pmatrix} 5 - \lambda & 3 \\ -4 & -2 - \lambda \end{pmatrix} The determinant is: (5λ)(2λ)(3)(4)=λ23λ2(5 - \lambda)(-2 - \lambda) - (3)(-4) = \lambda^2 - 3\lambda - 2 Solving gives eigenvalues

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Matrix Diagonalization
Eigenvalues
Eigenvectors

Formulas

Characteristic polynomial: det(A - λI) = 0
Determinant of 2x2 matrix: det([[a, b], [c, d]]) = a*d - b*c

Theorems

Diagonalization theorem
Eigenvalue and Eigenvector theorems

Suitable Grade Level

Grades 11-12, College Level