\documentclass{beamer}
\mode<presentation>{
\usetheme{boadilla}
%% \setbeamercovered{transparent}
\setbeamercovered{invisible}
\usefonttheme{professionalfonts} %%% usual maths fonts
}
\beamertemplatenavigationsymbolsempty %% navigation bars
%%%%%%%%%%%%%%%%%%%%%%%%%%%
\setbeamertemplate{theorem}[ams style]
\setbeamertemplate{theorems}[numbered]
\newenvironment{example*}
{\addtocounter{theorem}{-1}\example}
{\endexample}
%%%%%%%%%%%%%%%%%%%%%%%%%%%
\newcommand{\qiq}{\quad \implies \quad}
\newcommand{\qaq}{\quad \text{and} \quad}
%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%% for vectors
\renewcommand{\vec}[1]{\mathbf{#1}}
\newcommand{\veca}{\vec{a}}
\newcommand{\vecb}{\vec{b}}
\newcommand{\vecc}{\vec{c}}
\newcommand{\veci}{\vec{i}}
\newcommand{\vecj}{\vec{j}}
\newcommand{\veck}{\vec{k}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% content of the slides %%
%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
%%%%%%%%% Example 2
\begin{example}
Find the eigenvalues and corresponding eigenvectors for
$
\vec{B}
=
\begin{pmatrix}
3 & 2 & 4 \\
1 & 2 & 0 \\
1 & -2 & 1
\end{pmatrix}
$.
\end{example}
% \begin{block}{Solution:}
% \pause
\[
\begin{aligned}
\det(\vec{B}-\lambda\vec{I}) &= 0 \\
\begin{vmatrix}
3-\lambda & 2 & 4 \\
1 & 2-\lambda & 0 \\
1 & -2 & 1-\lambda
\end{vmatrix}
&= 0 \\
(3-\lambda)
\begin{vmatrix}
2-\lambda & - \\
-2 & 1-\lambda
\end{vmatrix}
-2
\begin{vmatrix}
1 & 0 \\
1 & 1-\lambda
\end{vmatrix}
+4
\begin{vmatrix}
1 & 2-\lambda \\
1 & -2
\end{vmatrix}
&= 0 \\
(3-\lambda)(2-\lambda)(1-\lambda)-2(1-\lambda)+4(-4+\lambda)
&= 0 \\
(\lambda+1)(\lambda-3)(\lambda-4) &= 0 \\
\lambda &= -1,3,4
\end{aligned}
\]
\framebreak
\[
\vec{B}x = \lambda x
\qiq
\begin{pmatrix}
3 & 2 & 4 \\
1 & 2 & 0 \\
1 & -2 & 1
\end{pmatrix}
\begin{pmatrix}
x \\ y \\ z
\end{pmatrix}
=
\lambda
\begin{pmatrix}
x \\ y \\ z
\end{pmatrix}
\qiq
\begin{array}{rcl}
3x+2y+4z &=& \lambda x \\
x+2y &=& \lambda y \\
x-2y+z &=& \lambda z
\end{array}
\]
When $\lambda=-1$: $x=-3y$ and $z=\frac{5}{2}y$.
An eigenvector is
$
\begin{pmatrix}
-6 \\
2 \\
5
\end{pmatrix}
$. \\
When $\lambda=3$: $x=y$ and $y=-2z$.
An eigenvector is
$
\begin{pmatrix}
-2 \\ -2 \\ 1
\end{pmatrix}
$. \\
When $\lambda=4$: $x=2y$ and $z=0$.
An eigenvector is
$
\begin{pmatrix}
2 \\ 1 \\ 0
\end{pmatrix}
$.
% \end{block}
\end{frame}
\end{document}
解决方案从开始line 47，我希望从中进一步打破它line 80。但是\框架中断似乎不起作用。
做到这一点的最好方法是什么？
答案1
Beamer 块通常不可破坏。但是您可以使用tcolorbox内部主题，用 tcolorboxes 替换普通的 Beamer 块。然后您可以像这样使它们可破坏：
\documentclass{beamer}
\mode<presentation>{
\usetheme{boadilla}
%% \setbeamercovered{transparent}
\setbeamercovered{invisible}
\usefonttheme{professionalfonts} %%% usual maths fonts
}
\beamertemplatenavigationsymbolsempty %% navigation bars
%%%%%%%%%%%%%%%%%%%%%%%%%%%
\setbeamertemplate{theorem}[ams style]
\setbeamertemplate{theorems}[numbered]
\newenvironment{example*}
{\addtocounter{theorem}{-1}\example}
{\endexample}
%%%%%%%%%%%%%%%%%%%%%%%%%%%
\newcommand{\qiq}{\quad \implies \quad}
\newcommand{\qaq}{\quad \text{and} \quad}
%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%% for vectors
\renewcommand{\vec}[1]{\mathbf{#1}}
\newcommand{\veca}{\vec{a}}
\newcommand{\vecb}{\vec{b}}
\newcommand{\vecc}{\vec{c}}
\newcommand{\veci}{\vec{i}}
\newcommand{\vecj}{\vec{j}}
\newcommand{\veck}{\vec{k}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% content of the slides %%
%%%%%%%%%%%%%%%%%%%%%%%%%%%
\usepackage[most]{tcolorbox}
\useinnertheme{tcolorbox}
\tcbset{breakable}
\begin{document}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[allowframebreaks]
%%%%%%%%% Example 2
\begin{example}
Find the eigenvalues and corresponding eigenvectors for
$
\vec{B}
=
\begin{pmatrix}
3 & 2 & 4 \\
1 & 2 & 0 \\
1 & -2 & 1
\end{pmatrix}
$.
\end{example}
\begin{block}{Solution:}
% \pause
\[
\begin{aligned}
\det(\vec{B}-\lambda\vec{I}) &= 0 \\
\begin{vmatrix}
3-\lambda & 2 & 4 \\
1 & 2-\lambda & 0 \\
1 & -2 & 1-\lambda
\end{vmatrix}
&= 0 \\
(3-\lambda)
\begin{vmatrix}
2-\lambda & - \\
-2 & 1-\lambda
\end{vmatrix}
-2
\begin{vmatrix}
1 & 0 \\
1 & 1-\lambda
\end{vmatrix}
+4
\begin{vmatrix}
1 & 2-\lambda \\
1 & -2
\end{vmatrix}
&= 0 \\
(3-\lambda)(2-\lambda)(1-\lambda)-2(1-\lambda)+4(-4+\lambda)
&= 0 \\
(\lambda+1)(\lambda-3)(\lambda-4) &= 0 \\
\lambda &= -1,3,4
\end{aligned}
\]
\framebreak
\[
\vec{B}x = \lambda x
\qiq
\begin{pmatrix}
3 & 2 & 4 \\
1 & 2 & 0 \\
1 & -2 & 1
\end{pmatrix}
\begin{pmatrix}
x \\ y \\ z
\end{pmatrix}
=
\lambda
\begin{pmatrix}
x \\ y \\ z
\end{pmatrix}
\qiq
\begin{array}{rcl}
3x+2y+4z &=& \lambda x \\
x+2y &=& \lambda y \\
x-2y+z &=& \lambda z
\end{array}
\]
When $\lambda=-1$: $x=-3y$ and $z=\frac{5}{2}y$.
An eigenvector is
$
\begin{pmatrix}
-6 \\
2 \\
5
\end{pmatrix}
$. \\
When $\lambda=3$: $x=y$ and $y=-2z$.
An eigenvector is
$
\begin{pmatrix}
-2 \\ -2 \\ 1
\end{pmatrix}
$. \\
When $\lambda=4$: $x=2y$ and $z=0$.
An eigenvector is
$
\begin{pmatrix}
2 \\ 1 \\ 0
\end{pmatrix}
$.
\end{block}
\end{frame}
\end{document}
