\documentclass{beamer}
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\usepackage{beamerthemeshadow,graphics,amsmath,amscd,mathtools}
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\begin{document}
\author{M.\ Nielsen}
\title{Lineær Algebra, 2010}

\frame[shrink]{
\frametitle{Cramer's regel}
\begin{block}{Løsning til $A\vx=\vb$ for $A$ invertibel}
For en invertibel $n\times n$-matrix $A$ og $\vb\in\bR^n$, er den entydige løsning $\vu=A^{-1}\vb$ til $A\vx=\vb$ givet ved
\vskip -.5cm$$\vu=(u_1,u_2,\ldots,u_n)^T,\qquad u_j=\frac{\det(M_j)}{\det(A)},\quad j=1,2,\ldots,n,$$
\vskip -.1cm hvor $M_j$ er den matrix der fremkommer ved at udskift søjle $j$ i $A$ med $\vb$.
\end{block}
\begin{exampleblock}{Eksempel}
Løs
$$\begin{bmatrix}
  1 & 1 \\
  1 & 3 \\
  \end{bmatrix}\begin{bmatrix}
  x_1 \\
  x_2 \\
  \end{bmatrix}
=\begin{bmatrix}
  1 \\
  -1 \\
  \end{bmatrix}$$
Vi har,
\vskip -.5cm
$$x_1=\frac{\det 
\begin{bmatrix}
  1 & 1 \\
  -1 & 3 \\
  \end{bmatrix}}
  {\det \begin{bmatrix}
  1 & 1 \\
  1 & 3 \\
  \end{bmatrix}}=
  \frac{4}{2}=2,\quad
  x_2=\frac{\det 
\begin{bmatrix}
  1 & 1 \\
  1 & -1 \\
  \end{bmatrix}}
  {\det \begin{bmatrix}
  1 & 1 \\
  1 & 3 \\
  \end{bmatrix}}=
  \frac{-2}{2}=-1.
  $$
\end{exampleblock}
}

\end{document}