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.

20. \(\left( {\begin{aligned}{}{ - 1.64}&{}&{ - 2.4}\\{1.92}&{}&{2.2}\end{aligned}} \right)\)

Short answer

The invertible matrix and matrix are \(C = \left( {\begin{aligned}{}{.28}&{}&{ - .96}\\{.96}&{}&{.28}\end{aligned}} \right),\;{\rm{and}}\,P = \left( {\begin{aligned}{}{ - 2}&{}&{ - 1}\\2&{}&0\end{aligned}} \right)\)

Step-by-step solution

Finding the matrix \(P\) and the matrix  \(C\)

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

Given that \(A = \left( {\begin{aligned}{}{ - 1.64}&{}&{ - 2.4}\\{1.92}&{}&{2.2}\end{aligned}} \right)\).

The characteristic polynomial is \({\lambda ^2} - .56\lambda + 1\). Hence the eigenvalues of \(A\) are \(\lambda = - .28 \pm .96i\).

To find an eigenvector corresponding to \( - .28 - .96i\) we compute,

\(A - \left( { - .28 - .96i} \right)I = \left( {\begin{aligned}{}{1.92 + .96i}&{}&{ - 2.4}\\{1.92}&{}&{ - 1.92 + .96i}\end{aligned}} \right)\)

The equation \(\left( {A - \left( { - .28 - .96i} \right)I} \right)x = 0\) has a system of equations,

\(\begin{aligned}{}\left( {1.92 + .96i} \right){x_1} - 2.4{x_2} &= 0\\1.92{x_1} + ( - 1.92 + .96i){x_2} &= 0\end{aligned}\)

Hence \({x_1} = \frac{{ - 2 - i}}{2}{x_2}\) , \({x_2}\)is a free variable.

Find the matrix further

Thus, the eigen vector corresponds to the eigenvalue\( - .28 - .96i\).

\(v = \left( {\begin{aligned}{}{ - 2 - i}\\2\end{aligned}} \right)\)

By Theorem 9,

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

And

\(\begin{aligned}{}C &= {P^{ - 1}}AP\\ &= \frac{1}{2}\left( {\begin{aligned}{}0&{}&1\\{ - 2}&{}&{ - 2}\end{aligned}} \right)\left( {\begin{aligned}{}{ - 1.64}&{}&{ - 2.4}\\{1.92}&{}&{2.2}\end{aligned}} \right)\left( {\begin{aligned}{}{ - 2}&{}&{ - 1}\\2&{}&0\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}{.28}&{}&{ - .96}\\{.96}&{}&{.28}\end{aligned}} \right)\end{aligned}\)

Thus, the invertible matrix and matrix are \(C = \left( {\begin{aligned}{}{.28}&{}&{ - .96}\\{.96}&{}&{.28}\end{aligned}} \right),\;{\rm{and}}\;P = \left( {\begin{aligned}{}{ - 2}&{}&{ - 1}\\2&{}&0\end{aligned}} \right)\).