diff --git a/mb/Uebung1.pdf b/mb/Uebung1.pdf
index 7f33645..99a6421 100644
--- a/mb/Uebung1.pdf
+++ b/mb/Uebung1.pdf
Binary files differ
diff --git a/mb/Uebung1.tex b/mb/Uebung1.tex
index 04960ec..8526c57 100644
--- a/mb/Uebung1.tex
+++ b/mb/Uebung1.tex
@@ -80,40 +80,35 @@
     	\textit{Jan-Peter Hohloch}\\ \textit{Maximus Mutschler}
     }{SS 15}{4}
     \vspace{1cm}
-    
+
     \Aufgabe{}{7}
 	    \begin{enumerate}[(a)]
-		    \item $ \mathbf{R_z(\alpha)}= % %\begin{pmatrix}
-		   % &&&\\
-		   % &&&\\
-		   % &&&\\
-		   % &&&\\
-		   % \end{pmatrix} 
-		   \begin{pmatrix}
-		   cos(\alpha)&-sin(\alpha)&0&0\\
-		   sin(\alpha)&cos(\alpha)&0&0\\
-		   0&0&1&0\\
-		   0&0&0&1
-		   \end{pmatrix}\\
-		   \\
-		   \mathbf{Trans}=
-		   \begin{pmatrix}
-		   1&0&0&t_x\\
-		   0&1&0&t_y\\
-		   0&0&1&t_z\\
-		   0&0&0&1
-		   \end{pmatrix}$
+		    \item $ \mathbf{R_z(\alpha)}=
+    		   \begin{pmatrix}
+    		   \cos(\alpha)&-\sin(\alpha)&0&0\\
+    		   \sin(\alpha)&\cos(\alpha)&0&0\\
+    		   0&0&1&0\\
+    		   0&0&0&1
+    		   \end{pmatrix}\\
+    		   \\
+    		   \mathbf{Trans}=
+    		   \begin{pmatrix}
+    		   1&0&0&t_x\\
+    		   0&1&0&t_y\\
+    		   0&0&1&t_z\\
+    		   0&0&0&1
+    		   \end{pmatrix}$
 		   \item $\mathbf{T}_b= \mathbf{Trans}\cdot\mathbf{R_z(\alpha)}=
 		   		   \begin{pmatrix}
-		   		   cos(\alpha)&-sin(\alpha)&0&t_x\\
-		   		   sin(\alpha)&cos(\alpha)&0&t_y\\
+		   		   \cos(\alpha)&-\sin(\alpha)&0&t_x\\
+		   		   \sin(\alpha)&\cos(\alpha)&0&t_y\\
 		   		   0&0&1&t_z\\
 		   		   0&0&0&1
 		   		   \end{pmatrix}\\$
 		   \item  $\mathbf{T}_c= \mathbf{R_z(\alpha)\cdot\mathbf{Trans}}=
 		   			   \begin{pmatrix}
-		   			   cos(\alpha)&-sin(\alpha)&0&0\\
-		   			   sin(\alpha)&cos(\alpha)&0&0\\
+		   			   \cos(\alpha)&-\sin(\alpha)&0&0\\
+		   			   \sin(\alpha)&\cos(\alpha)&0&0\\
 		   			   0&0&1&0\\
 		   			   0&0&0&1
 		   			   \end{pmatrix}\cdot
@@ -124,46 +119,53 @@
 		   			    0&0&0&1
 		   			    \end{pmatrix} =\\
 		   			    \begin{pmatrix}
-		   			    cos(\alpha)&-sin(\alpha)&0&xcos(\alpha)-ysin(\alpha)\\
-		   			    sin(\alpha)&cos(\alpha)&0&ycos(\alpha)+xsin(\alpha)\\
-		   			    0&0&1&z\\
+		   			    \cos(\alpha)&-\sin(\alpha)&0&t_x\cos(\alpha)-t_y\sin(\alpha)\\
+		   			    \sin(\alpha)&\cos(\alpha)&0&t_y\cos(\alpha)+t_x\sin(\alpha)\\
+		   			    0&0&1&t_z\\
 		   			    0&0&0&1
 		   			    \end{pmatrix} $
-		 \item $\mathbf{^BT_A} =\mathbf{Trans}(2,-1,1)\cdot\mathbf{R_x}(\frac{\pi}{2})\cdot\mathbf{R_z}(\frac{\pi}{2})\\=
-		 	    \begin{pmatrix}
-		 	    1&0&0&2\\
-		 	    0&1&0&-1\\
-		 	    0&0&1&1\\
-		 	    0&0&0&1
-		 	    \end{pmatrix} \cdot
-		 	    \begin{pmatrix}
-		 	     1&0&0&0\\
-		 	     0&0&-1&0\\
-		 	     0&1&0&0\\
-		 	     0&0&0&1\\
-		 	    \end{pmatrix}  \cdot
-		 	     \begin{pmatrix}
-		 	     0&-1&0&0\\
-		 	     1&0&0&0\\
-		 	     0&0&1&0\\
-		 	     0&0&0&1\\
-		 	     \end{pmatrix}\\=
-		 	      \begin{pmatrix}
-		 	      0&-1&0&2\\
-		 	      0&0&-1&-1\\
-		 	      1&0&0&1\\
-		 	      0&0&0&1\\
-		 	      \end{pmatrix}\\$
-      \item $ \mathbf{^AT_B} = 
+		 \item \begin{align*}
+            \mathbf{^BT_A} &=\mathbf{Trans}(2,-1,1)\cdot\mathbf{R_x}(90^\circ)\cdot\mathbf{R_y}(-90^\circ)\\
+            &=\mathbf{Trans}(2,-1,1)\cdot \begin{pmatrix}
+                1&0&0&0\\
+                0&0&-1&0\\
+                0&1&0&0\\
+                0&0&0&1
+            \end{pmatrix}\cdot \begin{pmatrix}
+                0&0&-1&0\\
+                0&1&0&0\\
+                1&0&0&0\\
+                0&0&0&1
+            \end{pmatrix}\\
+            &=\begin{pmatrix}
+                1&0&0&2\\
+                0&1&0&-1\\
+                0&0&1&1\\
+                0&0&0&1
+            \end{pmatrix}\cdot \begin{pmatrix}
+                0&0&-1&0\\
+                -1&0&0&0\\
+                0&1&0&0\\
+                0&0&0&1
+            \end{pmatrix}\\
+            &=\begin{pmatrix}
+                0&-1&0&-1\\
+                0&0&1&-1\\
+                -1&0&0&2\\
+                0&0&0&1
+            \end{pmatrix}
+             \end{align*}
+
+      \item $ \mathbf{^AT_B} =
       \begin{pmatrix}
       0&0&1&-1\\
       -1&0&0&2\\
       0&-1&0&-1\\
       0&0&0&1\\
       \end{pmatrix}\\$
-		 
+
 	    \end{enumerate}
-	    	
+
     \Aufgabe{}{3}
      \begin{enumerate}[(a)]
      \item{Grid Based Representation}\\
@@ -175,27 +177,27 @@
      -  limited number of samples
      \item {Parameter Based Representation}\\
      +p(x) directly determinable for each x \\- possible low accuracy due to approximation
-     
+
      \end{enumerate}
     \Aufgabe{}{5}
     \begin{enumerate}[(a)]
-   
+
     \item $ \mu_S =\begin{pmatrix}
     \frac{7}{3}\\
     \frac{7}{3}\\
     \frac{17}{6}\\
      \end{pmatrix}\\$
 
-\item $\Sigma_S=    
+\item $\Sigma_S=
     \begin{pmatrix}
     	\frac{8}{3}&\frac{16}{15}&\frac{22}{15}\\
     	\frac{16}{15}&\frac{34}{15}&\frac{7}{15}\\
     	\frac{22}{15}&\frac{7}{15}&\frac{89}{30}\\
     \end{pmatrix}\\$
- 
+
   \item $ N(\mu,\Sigma) = \frac{1}{(2\pi)^(\frac{\pi}{2})|\Sigma|^(\frac{1}{2})}e^{(-\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu) )}\\
-	|\Sigma_S|= \frac{264}{25}\\
-	\Sigma^{-1} = 
+	|\Sigma_S|= \frac{264}{25}\\%TODO: wrong exponents?
+	\Sigma^{-1} =
 	 \begin{pmatrix}
 	 \frac{61}{99}&\frac{-31}{132}&\frac{-53}{198}\\
 	 \frac{-31}{132}&\frac{6}{11}&\frac{1}{33}\\
@@ -213,26 +215,43 @@
 2\\3
 \end{pmatrix}\\
 \Sigma_y=\begin{pmatrix}
-0,5&1\\2&0.5
+1&0.5\\0.5&2
 \end{pmatrix}\\$
 \item $\mu_y=\begin{pmatrix}
 4\\-2
 \end{pmatrix}\\
 \Sigma_y=\begin{pmatrix}
--2&4\\8&-2
+4&-2\\-2&8
 \end{pmatrix}\\$
 \item$ \mu_y=\begin{pmatrix}
 \frac{2}{\Delta t}\\\frac{1}{\Delta t}
 \end{pmatrix}\\
 \Sigma_y=\begin{pmatrix}
-\frac{1}{2\Delta t^2}&\frac{1}{\Delta t^2}\\
-\frac{2}{\Delta t^2}&\frac{1}{2\Delta t^2}
+\frac{1}{(\Delta t)^2}&\frac{1}{2(\Delta t)^2}\\
+\frac{1}{2(\Delta t)^2}&\frac{2}{(\Delta t)^2}
 \end{pmatrix}\\$
-\item $\mu_y=?\\
-\Sigma_y=?$
+\item $\mu_y=2\mu_x+3\mu_z=\begin{pmatrix}
+    4\\2
+\end{pmatrix}+3\mu_z\\
+\Sigma_y=\begin{pmatrix}
+    2&0\\0&2
+\end{pmatrix}\begin{pmatrix}
+    1&0.5\\0.5&2
+\end{pmatrix}\begin{pmatrix}
+    2&0\\0&2
+\end{pmatrix}+\begin{pmatrix}
+    2&0\\0&2
+\end{pmatrix}\Sigma_z \begin{pmatrix}
+    2&0\\0&2
+\end{pmatrix}=\begin{pmatrix}
+    4&2\\2&8
+\end{pmatrix}+\begin{pmatrix}
+    2&0\\0&2
+\end{pmatrix}\Sigma_z \begin{pmatrix}
+    2&0\\0&2
+\end{pmatrix}$
 
 \end{enumerate}
-    %TODO Aufgaben bearbeiten
 
 \end{document}