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Q3E

Expert-verifiedFound in: Page 267

Book edition
5th

Author(s)
David C. Lay, Steven R. Lay and Judi J. McDonald

Pages
483 pages

ISBN
978-03219822384

**Question: Is \(\left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right)\) an eigenvalue of \(\left( {\begin{array}{*{20}{c}}{ - 3}&1\\{ - 3}&8\end{array}} \right)\)? If so, find the eigenvalue.**

No, \(\left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right)\) is not the eigenvector of the given matrix \(\left( {\begin{array}{*{20}{c}}{ - 3}&1\\{ - 3}&8\end{array}} \right)\).

If there exists a non-zero vector \({\bf{x}}\) which satisfies \(A{\bf{x}} = \lambda {\bf{x}}\) for some scaler \(\lambda \), then \({\bf{x}}\) be the eigenvector of an \(n \times n\) matrix \(A\).

Denote the given matrix by \(A\) and the given vector by \({\bf{x}}\).

\(A = \left( {\begin{array}{*{20}{c}}{ - 3}&1\\{ - 3}&8\end{array}} \right)\), \({\bf{x}} = \left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right)\)

According to the definition of eigenvalue, \({\bf{x}} = \left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right)\) is the eigenvector of the matrix \(A\), if \(A{\bf{x}} = \lambda {\bf{x}}\).

Find the product of \(A\), and \({\bf{x}}\).

\(\begin{array}{c}A{\bf{x}} = \left( {\begin{array}{*{20}{c}}{ - 3}&1\\{ - 3}&8\end{array}} \right)\left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1\\{29}\end{array}} \right)\end{array}\)

It can be observed that \(\left( {\begin{array}{*{20}{c}}1\\{29}\end{array}} \right) \ne \lambda \left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right)\)

So, \(\left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right)\) is not the eigenvector of the given matrix \(\left( {\begin{array}{*{20}{c}}{ - 3}&1\\{ - 3}&8\end{array}} \right)\).

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