Exam-Style Problem

9231 P13 - Nov 2012 - Q12 - 11 marks

6521

Answer only one of the following two alternatives.

EITHER
The vector \(\mathbf{e}\) is an eigenvector of each of the \(n \times n\) matrices \(\mathbf{A}\) and \(\mathbf{B}\), with corresponding eigenvalues \(\lambda\) and \(\mu\) respectively. Prove that \(\mathbf{e}\) is an eigenvector of the matrix \(\mathbf{A B}\) with eigenvalue \(\lambda \mu\).

It is given that the matrix \(\mathbf{A}\), where
\(\mathbf{A}=\left(\begin{array}{rrr} 3 & 2 & 2 \\ -2 & -2 & -2 \\ 1 & 2 & 2 \end{array}\right),\)
has eigenvectors \(\left(\begin{array}{r}0 \\ 1 \\ -1\end{array}\right)\) and \(\left(\begin{array}{r}1 \\ 0 \\ -1\end{array}\right)\). Find the corresponding eigenvalues.

Given that 2 is also an eigenvalue of \(\mathbf{A}\), find a corresponding eigenvector.

The matrix \(\mathbf{B}\), where
\(\mathbf{B}=\left(\begin{array}{rrr} -1 & 2 & 2 \\ 2 & 2 & 2 \\ -3 & -6 & -6 \end{array}\right),\)
has the same eigenvectors as \(\mathbf{A}\). Given that \(\mathbf{A B}=\mathbf{C}\), find a non-singular matrix \(\mathbf{P}\) and a diagonal matrix \(\mathbf{D}\) such that
\(\mathbf{P}^{-1} \mathbf{C}^{2} \mathbf{P}=\mathbf{D} .\)

OR
Obtain the general solution of the differential equation
\(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+6 \frac{\mathrm{~d} x}{\mathrm{~d} t}+13 x=75 \cos 2 t\)

Given that \(x=5\) and \(\frac{\mathrm{d} x}{\mathrm{~d} t}=0\) when \(t=0\), find \(x\) in terms of \(t\).

Show that, for large positive values of \(t\) and for any initial conditions,
\(x \approx 5 \cos (2 t-\phi),\)
where the constant \(\phi\) is such that \(\tan \phi=\frac{4}{3}\).

Solution

Answer: \(\mathbf e\) is an eigenvector of \(\mathbf{AB}\) with eigenvalue \(\lambda\mu\).

For \(\mathbf A\), the eigenvalues corresponding to \(\begin{pmatrix}0\\1\\-1\end{pmatrix}\) and \(\begin{pmatrix}1\\0\\-1\end{pmatrix}\) are \(0\) and \(1\).

A corresponding eigenvector for eigenvalue \(2\) is \(\begin{pmatrix}2\\-1\\0\end{pmatrix}\).

One suitable choice is

\(\mathbf P=\begin{pmatrix}0&1&2\\1&0&-1\\-1&-1&0\end{pmatrix},\qquad \mathbf D=\begin{pmatrix}0&0&0\\0&9&0\\0&0&16\end{pmatrix},\)

so that \(\mathbf P^{-1}\mathbf C^2\mathbf P=\mathbf D\).

If \(\mathbf A\mathbf e=\lambda\mathbf e\) and \(\mathbf B\mathbf e=\mu\mathbf e\), then

\(\mathbf{AB}\mathbf e=\mathbf A(\mathbf B\mathbf e)=\mathbf A(\mu\mathbf e)=\mu\mathbf A\mathbf e=\mu\lambda\mathbf e=\lambda\mu\mathbf e.\)

Therefore \(\mathbf e\) is an eigenvector of \(\mathbf{AB}\) with eigenvalue \(\lambda\mu\).

For

\(\mathbf A=\begin{pmatrix}3&2&2\\-2&-2&-2\\1&2&2\end{pmatrix},\)

using \(\mathbf v_1=\begin{pmatrix}0\\1\\-1\end{pmatrix}\),

\(\mathbf A\mathbf v_1=\begin{pmatrix}0\\0\\0\end{pmatrix}=0\mathbf v_1,\)

so the corresponding eigenvalue is \(0\).

Using \(\mathbf v_2=\begin{pmatrix}1\\0\\-1\end{pmatrix}\),

\(\mathbf A\mathbf v_2=\begin{pmatrix}1\\0\\-1\end{pmatrix}=1\mathbf v_2,\)

so the corresponding eigenvalue is \(1\).

For eigenvalue \(2\), solve

\((\mathbf A-2\mathbf I)\mathbf v=\mathbf0.\)

One suitable eigenvector is

\(\mathbf v_3=\begin{pmatrix}2\\-1\\0\end{pmatrix},\)

since

\(\mathbf A\mathbf v_3=\begin{pmatrix}4\\-2\\0\end{pmatrix}=2\begin{pmatrix}2\\-1\\0\end{pmatrix}.\)

Now \(\mathbf B\) has the same eigenvectors. Direct multiplication gives the corresponding \(\mathbf B\)-eigenvalues

\(0,\quad -3,\quad -2.\)

Hence \(\mathbf C=\mathbf{AB}\) has corresponding eigenvalues

\(0\cdot0=0,\qquad 1\cdot(-3)=-3,\qquad 2\cdot(-2)=-4.\)

Therefore \(\mathbf C^2\) has eigenvalues

\(0,\quad 9,\quad 16.\)

Taking the common eigenvectors as columns gives

\(\mathbf P=\begin{pmatrix}0&1&2\\1&0&-1\\-1&-1&0\end{pmatrix}.\)

Thus one suitable diagonal matrix is

\(\mathbf D=\begin{pmatrix}0&0&0\\0&9&0\\0&0&16\end{pmatrix},\)

and

\(\mathbf P^{-1}\mathbf C^2\mathbf P=\mathbf D.\)