Exam-Style Problem

9231 P11 - Jun 2014 - Q9 - 10 marks

6275

The matrix \(\mathbf{M}\), where
\(\mathbf{M}=\left(\begin{array}{rrr} -2 & 2 & 2 \\ 2 & 1 & 2 \\ -3 & -6 & -7 \end{array}\right),\)
has an eigenvector \(\left(\begin{array}{r}0 \\ 1 \\ -1\end{array}\right)\). Find the corresponding eigenvalue.

It is given that if the eigenvalues of a general \(3 \times 3\) matrix \(\mathbf{A}\), where
\(\mathbf{A}=\left(\begin{array}{lll} a & b & c \\ d & e & f \\ g & h & i \end{array}\right),\)
are \(\lambda_{1}, \lambda_{2}\) and \(\lambda_{3}\) then
\(\lambda_{1}+\lambda_{2}+\lambda_{3}=a+e+i\)
and
the determinant of \(\mathbf{A}\) has the value \(\lambda_{1} \lambda_{2} \lambda_{3}\).
Use these results to find the other two eigenvalues of the matrix \(\mathbf{M}\), and find corresponding eigenvectors.

Solution

Answer: The corresponding eigenvalue is \(-1\).

The other eigenvalues are \(-3\) and \(-4\). Corresponding eigenvectors may be taken as \(\begin{pmatrix}2\\-1\\0\end{pmatrix}\) for \(-3\), and \(\begin{pmatrix}1\\0\\-1\end{pmatrix}\) for \(-4\).

Let \(\mathbf{v}=\begin{pmatrix}0\\1\\-1\end{pmatrix}\). Then

\(\mathbf{M}\mathbf{v}=\begin{pmatrix}-2 & 2 & 2\\2 & 1 & 2\\-3 & -6 & -7\end{pmatrix}\begin{pmatrix}0\\1\\-1\end{pmatrix}=\begin{pmatrix}0\\-1\\1\end{pmatrix}= -1\begin{pmatrix}0\\1\\-1\end{pmatrix}.\)

So the corresponding eigenvalue is \(-1\).

Now let the other two eigenvalues be \(\lambda_2\) and \(\lambda_3\). The trace of \(\mathbf{M}\) is

\(-2+1-7=-8.\)

Hence

\(\lambda_1+\lambda_2+\lambda_3=-8.\)

Since \(\lambda_1=-1\),

\(\lambda_2+\lambda_3=-7.\)

Also the determinant is the product of the eigenvalues. Compute

\(\det\mathbf{M}=\begin{vmatrix}-2 & 2 & 2\\2 & 1 & 2\\-3 & -6 & -7\end{vmatrix}.\)

Expanding along the first row,

\(\det\mathbf{M}=-2\begin{vmatrix}1 & 2\\-6 & -7\end{vmatrix}-2\begin{vmatrix}2 & 2\\-3 & -7\end{vmatrix}+2\begin{vmatrix}2 & 1\\-3 & -6\end{vmatrix}.\)

\(\det\mathbf{M}=-2( -7+12 )-2( -14+6 )+2( -12+3 )=-2(5)-2(-8)+2(-9)=-10+16-18=-12.\)

Therefore

\(\lambda_1\lambda_2\lambda_3=-12.\)

With \(\lambda_1=-1\), this gives

\(\lambda_2\lambda_3=12.\)

So \(\lambda_2\) and \(\lambda_3\) satisfy

\(t^2-(\lambda_2+\lambda_3)t+\lambda_2\lambda_3=0\),

that is

\(t^2+7t+12=0.\)

Hence

\((t+3)(t+4)=0,\)

so the other eigenvalues are \(-3\) and \(-4\).

For \(\lambda=-3\), solve \((\mathbf{M}+3\mathbf{I})\mathbf{x}=\mathbf{0}\):

\(\mathbf{M}+3\mathbf{I}=\begin{pmatrix}1 & 2 & 2\\2 & 4 & 2\\-3 & -6 & -4\end{pmatrix}.\)

From the first equation, \(x+2y+2z=0\). Taking \(z=0\) and \(x=2\) gives \(y=-1\), so an eigenvector is

\(\begin{pmatrix}2\\-1\\0\end{pmatrix}.\)

For \(\lambda=-4\), solve \((\mathbf{M}+4\mathbf{I})\mathbf{x}=\mathbf{0}\):

\(\mathbf{M}+4\mathbf{I}=\begin{pmatrix}2 & 2 & 2\\2 & 5 & 2\\-3 & -6 & -3\end{pmatrix}.\)

Taking \(y=0\) and \(x=1\) gives \(z=-1\), so an eigenvector is

\(\begin{pmatrix}1\\0\\-1\end{pmatrix}.\)