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.

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

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 \(C = {P^{ - 1}}AP\).

Given that \(A = \left( {\begin{aligned}{}1&{}&{ - 2}\\1&{}&3\end{aligned}} \right)\) .

Find the eigenvalues by using \(\det \left( {A - \lambda I} \right) = 0\).

\(\begin{aligned}{}\det \left( {A - \lambda I} \right) &= 0\\\left( {1 - \lambda } \right)\left( {3 - \lambda } \right) + 2 &= 0\\3 - \lambda - 3\lambda + {\lambda ^2} + 5 &= 0\\{\lambda ^2} - 4\lambda + 5 &= 0\\\lambda &= \frac{{4 \pm \sqrt {16 - 20} }}{2}\\\lambda &= 2 \pm i\end{aligned}\)

Thus, the eigenvalues are \(\lambda = 2 \pm i\).

For the eigenvalue \(\lambda = 2 - i\)we have,

\(\begin{aligned}{}\left( {A - \lambda I} \right)v &= 0\\\left( {\left( {\begin{aligned}{}1&{}&{ - 2}\\1&{}&3\end{aligned}} \right) - \left( {2 - i} \right)\left( {\begin{aligned}{}1&{}&0\\0&{}&1\end{aligned}} \right)} \right)\left( {\begin{aligned}{}x\\y\end{aligned}} \right) &= 0\\\left( {\begin{aligned}{}{ - 1 + i}&{}&{ - 2}\\1&{}&{1 + i}\end{aligned}} \right)\left( {\begin{aligned}{}x\\y\end{aligned}} \right) &= 0\end{aligned}\)

Now considering the augmented matrix we get,

\(\begin{aligned}{}\left( {\begin{aligned}{}{ - 1 + i}&{}&{ - 2}&{}&|&{}&0\\1&{}&{1 + i}&{}&|&{}&0\end{aligned}} \right){\rm{ }}{R_2} \to \left( { - 1 + i} \right){R_2}\\ \sim \left( {\begin{aligned}{}{ - 1 + i}&{}&{ - 2}&{}&|&{}&0\\{ - 1 + i}&{}&{ - 2}&{}&|&{}&0\end{aligned}} \right){\rm{ }}{R_2} \to {R_2} - {R_1}\\ \sim \left( {\begin{aligned}{}{ - 1 + i}&{}&{ - 2}&{}&|&{}&0\\0&{}&0&{}&|&{}&0\end{aligned}} \right)\end{aligned}\)

Hence, we get\(y\)is a free variable.

Let\(y = 1\)then,

\(\begin{aligned}{}\left( { - 1 + i} \right)x - 2 &= 0\\x &= \frac{2}{{ - 1 + i}}\\x &= \frac{{2\left( {1 + i} \right)}}{{ - 2}}\\x &= - 1 - i\end{aligned}\)

Hence \(v = \left( {\begin{aligned}{}x\\y\end{aligned}} \right) \Rightarrow v = \left( {\begin{aligned}{}{ - 1 - i}\\1\end{aligned}} \right)\).

Now \(P = \left( {\begin{aligned}{}{{\mathop{\rm Re}\nolimits} v}&{}&{{\mathop{\rm Im}\nolimits} v}\end{aligned}} \right) \Rightarrow P = \left( {\begin{aligned}{}{ - 1}&{}&{ - 1}\\1&{}&0\end{aligned}} \right)\).

Write the matrix\({P^{ - 1}}\).

\({P^{ - 1}} = \frac{1}{2}\left( {\begin{aligned}{}0&{}&1\\1&{}&{ - 1}\end{aligned}} \right)\)

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

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

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