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{\bf Question}

Prove that $$\left| \begin{array}{ccc} \cos \theta & 1 & 0 \\ 1 &
2\cos \theta & 1 \\ 0 & 1& 2\cos \theta \end{array} \right| = \cos
3\theta.$$  Evaluate $$\left| \begin{array}{ccc} 2\cos \theta & 1
& 0 \\ 1 & 2\cos \theta & 1 \\ 0 & 1& 2\cos \theta \end{array}
\right|.$$

For which values of $\theta$ are the corresponding 3x3 matrices
singular.  Write down the inverse when they exist.

\vspace{.25in}

{\bf Answer}

\begin{eqnarray*} A & = & \left| \begin{array}{ccc} \cos \theta & 1
& 0 \\ 1 & 2\cos \theta & 1 \\ 0 & 1& 2\cos \theta \end{array}
\right| \\ & = & \cos \theta (4 \cos^2 \theta - 1) - 2\cos \theta
\\ & = & 4cos^3\theta - 3\cos \theta \\ & = & \cos 3\theta
\end{eqnarray*}

\begin{eqnarray*} B & = & \left| \begin{array}{ccc} \cos \theta & 1 & 0 \\ 1 &
2\cos \theta & 1 \\ 0 & 1& 2\cos \theta \end{array} \right|  \\ &
= & 2\cos \theta(4\cos^2 \theta - 1) -\cos \theta \\ & = & 2\cos
\theta(4\cos^2 - 2) \\ & = & 4\cos \theta \cos 2 \theta
\\ & = & \frac{\sin 4 \theta}{\sin \theta} \hspace{.2in} (\sin
\theta \not=0) \end{eqnarray*}


The matrix of $A$ is singular if $\cos 3\theta = 0$ so $\theta =
(2n+1)\frac{\pi}{6}$

The matrix $B$ is singular if $\cos \theta = 0$ or $\cos 2 \theta
= 0$  So $\theta = (2n+1)\frac{\pi}{2}$ and $\theta =
(2n+1)\frac{\pi}{4}$.



${}$

The inverse of $A$ is $$\frac{1}{\cos 3\theta} \left(
\begin{array}{ccc} 4 \cos^2 \theta -1 & -2 \cos \theta & 1 \\
-2\cos \theta & 2 \cos^2 \theta & -\cos \theta \\ 1 & -\cos \theta
& 2 \cos^2 \theta - 1 \end{array} \right)$$

The inverse of $B$ is $$\frac{1}{4 \cos \theta \cos 2\theta}
\left( \begin{array}{ccc} 4\cos^2 \theta - 1 & -2\cos \theta & 1
\\ -2\cos \theta & 4 \cos^2 \theta & -2 \cos \theta \\ 1 & -2 \cos
\theta & 4\cos^2\theta -1 \end{array} \right)$$

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