4.2.1: Exercises 4.2

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In Exercises \(\PageIndex{1}\) – \(\PageIndex{6}\), a matrix \(A\) is given. For each,

Find the eigenvalues of \(A\), and for each eigenvalue, find an eigenvector.
Do the same for \(A^{T}\).
Do the same for \(A^{−1}\)
Find \(\text{tr}(A)\).
Find \(\text{det}(A)\). Use Theorem 4.2.1 to verify your results.

Exercise \(\PageIndex{1}\)

\(\left[\begin{array}{cc}{0}&{4}\\{-1}&{5}\end{array}\right]\)

Answer

\(\lambda_{1}=1\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{4}\\{1}\end{array}\right];\)
\(\lambda_{2}=4\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{1}\\{1}\end{array}\right]\)
\(\lambda_{1}=1\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{-1}\\{1}\end{array}\right];\)
\(\lambda_{2}=4\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{-1}\\{4}\end{array}\right]\)
\(\lambda_{1}=1/4\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{1}\\{1}\end{array}\right];\)
\(\lambda_{2}=1\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{4}\\{1}\end{array}\right]\)
\(5\)
\(4\)

Exercise \(\PageIndex{2}\)

\(\left[\begin{array}{cc}{-2}&{-14}\\{-1}&{3}\end{array}\right]\)

Answer

\(\lambda_{1}=-4\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{7}\\{1}\end{array}\right];\)
\(\lambda_{2}=5\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{-2}\\{1}\end{array}\right]\)
\(\lambda_{1}=-4\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{1}\\{2}\end{array}\right];\)
\(\lambda_{2}=5\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{-1}\\{7}\end{array}\right]\)
\(\lambda_{1}=-1/4\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{7}\\{1}\end{array}\right];\)
\(\lambda_{2}=1/5\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{-2}\\{1}\end{array}\right]\)
\(1\)
\(-20\)

Exercise \(\PageIndex{3}\)

\(\left[\begin{array}{cc}{5}&{30}\\{-1}&{-6}\end{array}\right]\)

Answer

\(\lambda_{1}=-1\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{-5}\\{1}\end{array}\right];\)
\(\lambda_{2}=0\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{-6}\\{1}\end{array}\right]\)
\(\lambda_{1}=-1\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{1}\\{6}\end{array}\right];\)
\(\lambda_{2}=0\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{1}\\{5}\end{array}\right]\)
\(A\) is not invertible.
\(-1\)
\(0\)

Exercise \(\PageIndex{4}\)

\(\left[\begin{array}{cc}{-4}&{72}\\{-1}&{13}\end{array}\right]\)

Answer

\(\lambda_{1}=4\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{9}\\{1}\end{array}\right];\)
\(\lambda_{2}=5\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{8}\\{1}\end{array}\right]\)
\(\lambda_{1}=4\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{-1}\\{8}\end{array}\right];\)
\(\lambda_{2}=5\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{-1}\\{9}\end{array}\right]\)
\(\lambda_{1}=1/4\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{9}\\{1}\end{array}\right];\)
\(\lambda_{2}=1/5\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{8}\\{1}\end{array}\right]\)
\(9\)
\(20\)

Exercise \(\PageIndex{5}\)

\(\left[\begin{array}{ccc}{5}&{-9}&{0}\\{1}&{-5}&{0}\\{2}&{4}&{3}\end{array}\right]\)

Answer

\(\lambda_{1}=-4\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{-7}\\{-7}\\{6}\end{array}\right];\)
\(\lambda_{2}=3\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{0}\\{0}\\{1}\end{array}\right]\)
\(\lambda_{3}=4\text{ with }\vec{x_{3}}=\left[\begin{array}{c}{9}\\{1}\\{22}\end{array}\right]\)
\(\lambda_{1}=-4\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{-1}\\{9}\\{0}\end{array}\right];\)
\(\lambda_{2}=3\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{-20}\\{26}\\{7}\end{array}\right]\)
\(\lambda_{3}=4\text{ with }\vec{x_{3}}=\left[\begin{array}{c}{-1}\\{1}\\{0}\end{array}\right]\)
\(\lambda_{1}=-1/4\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{-7}\\{-7}\\{6}\end{array}\right];\)
\(\lambda_{2}=1/3\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{0}\\{0}\\{1}\end{array}\right]\)
\(\lambda_{3}=1/4\text{ with }\vec{x_{3}}=\left[\begin{array}{c}{9}\\{1}\\{22}\end{array}\right]\)
\(3\)
\(-48\)

Exercise \(\PageIndex{6}\)

\(\left[\begin{array}{ccc}{0}&{25}&{0}\\{1}&{0}&{0}\\{1}&{1}&{-3}\end{array}\right]\)

Answer

\(\lambda_{1}=-5\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{-5}\\{1}\\{2}\end{array}\right];\)
\(\lambda_{2}=-3\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{0}\\{0}\\{1}\end{array}\right]\)
\(\lambda_{3}=5\text{ with }\vec{x_{3}}=\left[\begin{array}{c}{20}\\{4}\\{3}\end{array}\right]\)
\(\lambda_{1}=-5\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{-1}\\{5}\\{0}\end{array}\right];\)
\(\lambda_{2}=-3\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{1}\\{-11}\\{8}\end{array}\right]\)
\(\lambda_{3}=5\text{ with }\vec{x_{3}}=\left[\begin{array}{c}{1}\\{5}\\{0}\end{array}\right]\)
\(\lambda_{1}=-1/5\text{ with }\vec{x_{1}}=\left[\begin{array}{c}{-5}\\{1}\\{2}\end{array}\right];\)
\(\lambda_{2}=-1/3\text{ with }\vec{x_{2}}=\left[\begin{array}{c}{0}\\{0}\\{1}\end{array}\right]\)
\(\lambda_{3}=1/5\text{ with }\vec{x_{3}}=\left[\begin{array}{c}{20}\\{4}\\{3}\end{array}\right]\)
\(-3\)
\(75\)