Q19E Question: For \(A = \left( {\beg... [FREE SOLUTION]

91影视

Linear Algebra and its Applications

David C. Lay, Steven R. Lay and Judi J. McDonald

$Math Studyset 91影视 Explanations$ Math

5th Edition 路 483 pages

ISBN: 978-03219822384

Chapter 5: Q19E (page 267)

Question: For $A = \left( {\begin{array}{*{20}{c}}1&2&3\\1&2&3\\1&2&3\end{array}} \right)$, find the eigenvalues, with no calculation. Justify your answer.

Short Answer

Expert verified

Eigenvalues of the given matrix is 0, as the determinant of the matrix is 0, so it is non-invertible, which imply $A{\bf{x}} = 0$ has a nontrivial solution.

Step by step solution

Definition

Eigenvalue: Let $\lambda $ is a scaler, $A$ is an $n \times n$ matrix and ${\bf{x}}$ is an eigenvector corresponding to $\lambda $, $\lambda $ is said to an eigenvalue of the matrix $A$ if there exists a nontrivial solution ${\bf{x}}$ of $A{\bf{x}} = \lambda {\bf{x}}$.

Find Eigenvalues

The given matrix is $A = \left( {\begin{array}{*{20}{c}}1&2&3\\1&2&3\\1&2&3\end{array}} \right)$.

As per the definition of eigenvalues, if $A{\bf{x}} = \lambda {\bf{x}}$ has a nontrivial solution, then $\lambda $ is the eigenvalue, and the matrix $A$ is non-invertible if and only if $A{\bf{x}} = 0$ has nontrivial solution and 0 is the eigenvalue.

Here, the matrix is invertible if its determinant is not 0. So, check, the matrix is invertible or not by checking its determinant.

The determinant of the determinant is 0, as all the rows of the matrix are identical. Which imply the matrix is non-invertible. Hence, the eigenvalue will be 0.