Question

In Exercises 13-20, find an invertible matrix \(P\) and a matrix \(C\) of the form \(\left( {\begin{aligned}{}a&{}&{ - b}\\b&{}&a\end{aligned}} \right)\) such that the given matrix has the form \(A = PC{P^{ - 1}}\). For Exercises 13-16, use information from Exercises 1-4.

15.\(\left( {\begin{aligned}{}1&{}&5\\{ - 2}&{}&3\end{aligned}} \right)\)

Short answer

The invertible matrix \(P\) and matrix \(C\) are \(P = \left( {\begin{aligned}{}1&{}&3\\2&{}&0\end{aligned}} \right),C = \left( {\begin{aligned}{}2&{}&{ - 3}\\3&{}&2\end{aligned}} \right)\).

Step-by-step solution

Finding the matrix \(P\) and the matrix \(C\)  such that  \(A = PC{P^{ - 1}}\)

Let \(A\) be a \(2 \times 2\) with eigenvalues of the form \(a \pm bi\) , and let \(v\) be the eigenvector corresponding to the eigenvalue \(a - bi\), then the matrix \(P\) will be \(P = \left( {\begin{aligned}{}{{\mathop{\rm Re}\nolimits} v}&{{\mathop{\rm Im}\nolimits} v}\end{aligned}} \right)\).

The matrix \(C\) can be obtained by \(C = {P^{ - 1}}AP\).

Find the Invertible matrix

From Exercise 3, the eigenvalues of\(A\)are\(\lambda = 2 \pm 3i\), and \(v = \left( {\begin{aligned}{}{1 + 3i}\\2\end{aligned}} \right)\) be the eigenvector correspond to the eigenvalue \(\lambda = 2 - 3i\).

By theorem 9,

\(\begin{aligned}{}P &= \left( {\begin{aligned}{}{{\mathop{\rm Re}\nolimits} v}&{}&{{\mathop{\rm Im}\nolimits} v}\end{aligned}} \right)\\P &= \left( {\begin{aligned}{}1&{}&3\\2&{}&0\end{aligned}} \right)\end{aligned}\)

Substitute the value of\({P^{ - 1}}\), \(P\) and \(A\) into \(C = {P^{ - 1}}AP\) to obtain matrix \(C\).

\(\begin{aligned}{}C &= {P^{ - 1}}AP\\C &= \frac{1}{6}\left( {\begin{aligned}{}0&{}&{ - 3}\\{ - 2}&{}&1\end{aligned}} \right)\left( {\begin{aligned}{}1&{}&5\\{ - 2}&{}&3\end{aligned}} \right)\left( {\begin{aligned}{}1&{}&3\\2&{}&0\end{aligned}} \right)\\C &= \left( {\begin{aligned}{}2&{}&{ - 3}\\3&{}&2\end{aligned}} \right)\end{aligned}\)

Hence \(P = \left( {\begin{aligned}{}1&{}&3\\2&{}&0\end{aligned}} \right)\) and \(C = \left( {\begin{aligned}{}2&{}&{ - 3}\\3&{}&2\end{aligned}} \right)\).