Math Problem Statement
Find an invertible matrix P and a diagonal matrix D such that P^−1AP=D. A = 4x4 matrix where the first row was 1, 4, 16, 0. The second row is 0, −1, 0, 0. The third row is 0, 0, −1, 0, The fourth row is 0, 6, 18, 2. What is the matrix of P and D
Solution
To find an invertible matrix and a diagonal matrix such that , we need to determine the eigenvalues and eigenvectors of the given matrix . The matrix is:
1 & 4 & 16 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 6 & 18 & 2 \end{bmatrix}$$ ### Step 1: Find the Eigenvalues of A The eigenvalues of a matrix are found by solving the characteristic equation: $$\text{det}(A - \lambda I) = 0$$ where $$I$$ is the identity matrix of the same dimension as $$A$$, and $$\lambda$$ are the eigenvalues. Subtracting $$\lambda$$ from the diagonal entries of $$A$$, we have: $$A - \lambda I = \begin{bmatrix} 1-\lambda & 4 & 16 & 0 \\ 0 & -1-\lambda & 0 & 0 \\ 0 & 0 & -1-\lambda & 0 \\ 0 & 6 & 18 & 2-\lambda \end{bmatrix}$$ The determinant of $$A - \lambda I$$ can be calculated directly due to its upper block triangular nature: - The determinant is simply the product of its diagonal components: $$(1-\lambda)((-1-\lambda)(-1-\lambda))(2-\lambda)$$ This implies the eigenvalues are the solutions to: $$(1-\lambda)(-1-\lambda)^2(2-\lambda) = 0$$ Which gives: - $$\lambda = 1$$ - $$\lambda = -1$$ (with multiplicity 2) - $$\lambda = 2$$ ### Step 2: Find the Eigenvectors of A #### For $$\lambda = 1$$: Solve $$(A - I)x = 0$$: $$\begin{bmatrix} 0 & 4 & 16 & 0 \\ 0 & -2 & 0 & 0 \\ 0 & 0 & -2 & 0 \\ 0 & 6 & 18 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$ A possible solution is $$x_1 = 1, x_2 = 0, x_3 = 0, x_4 = 0$$. #### For $$\lambda = -1$$: Solve $$(A + I)x = 0$$: $$\begin{bmatrix} 2 & 4 & 16 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 6 & 18 & 3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$ Solutions include: - $$x_2 = -3, x_3 = 1, x_4 = 0$$ - $$x_2 = 0, x_3 = 0, x_4 = 1$$ #### For $$\lambda = 2$$: Solve $$(A - 2I)x = 0$$: $$\begin{bmatrix} -1 & 4 & 16 & 0 \\ 0 & -3 & 0 & 0 \\ 0 & 0 & -3 & 0 \\ 0 & 6 & 18 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$ A possible solution is $$x_1 = 0, x_2 = 1, x_3 = 0, x_4 = 0$$. ### Eigenvectors Matrix $$P$$ Arrange $$P$$ using found eigenvectors as columns: $$P = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -3 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$ ### Diagonal Matrix $$D$$ Arrange eigenvalues on the diagonal of $$D$$: $$D = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 2 \end{bmatrix}$$ To ensure correctness, check if $$P^{-1}AP = D$$ by computing these matrices using standard operations, and verify if these conditions hold true.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Matrix Diagonalization
Eigenvalues and Eigenvectors
Formulas
Characteristic Equation: det(A - λI) = 0
Matrix Diagonalization: P^−1AP = D
Theorems
Eigenvalue-Eigenvector Theorem
Diagonalizability of Matrices
Suitable Grade Level
Undergraduate Mathematics