=for timestamp
Mi Mär 22 18:08:41 CET 2006

=head2 65. Hausaufgabe

=head3 Geometrie-Buch Seite 40, Aufgabe 7

Berechne und vereinfache.

=over

=item a)

M<\begin{vmatrix}1&1&0\\1&1+a&0\\1&1&1+b\end{vmatrix} = \left(1 + b\right)
\begin{vmatrix}1&1\\1&1+a\end{vmatrix} = \left(1 + b\right) \left(1 + a -
1\right) = a + ab;>

=item b)

M<\begin{vmatrix}1&a&-b\\-a&1&c\\b&-c&1\end{vmatrix} =
\begin{vmatrix}1&c\\-c&1\end{vmatrix} + a
\begin{vmatrix}a&-b\\-c&1\end{vmatrix} + b
\begin{vmatrix}a&-b\\1&c\end{vmatrix} = \\
1 + c^2 + a^2 - abc + abc + b^2 = 1 + a^2 + b^2 + c^2;>

=item c)

M<\begin{vmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{vmatrix} =
\begin{vmatrix}b&c\\b^2&c^2\end{vmatrix} -
\begin{vmatrix}a&c\\a^2&c^2\end{vmatrix} +
\begin{vmatrix}a&b\\a^2&b^2\end{vmatrix} = \\
bc^2 - b^2c - ac^2 + a^2c + ab^2 - a^2b = 
a^2 \left(c - b\right) + b^2 \left(a - c\right) + c^2 \left(b - a\right);>

=item d)

M<\begin{vmatrix}a&b&a+b\\b&a+b&a\\a+b&a&b\end{vmatrix} = a
\begin{vmatrix}a+b&a\\a&b\end{vmatrix} - b
\begin{vmatrix}b&a+b\\a&b\end{vmatrix} + \left(a + b\right)
\begin{vmatrix}b&a+b\\a+b&a\end{vmatrix} = a \left(ab + b^2 - a^2\right) - b
\left(b^2 - a^2 - ab\right) + \left(a + b\right)\left(ab - a^2 - 2ab -
b^2\right) = -2a^3 - 2b^3;>

=item e)

M<\begin{vmatrix}\sin\alpha&\cos\alpha\tan\beta&\cos\alpha\\-\cos\alpha&\sin\alpha\tan\beta&\sin\alpha\\0&-1&\tan\beta\end{vmatrix}
= \begin{vmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{vmatrix} +
\tan\beta
\begin{vmatrix}\sin\alpha&\cos\alpha\tan\beta\\-\cos\alpha&\sin\alpha\tan\beta\end{vmatrix}
= \sin^2 \alpha + \cos^2 \alpha + \tan^2 \beta \left(\sin^2 \alpha + \cos^2
\alpha\right) = 1 + \tan^2 \beta;>

=back

=head3 Geometrie-Buch Seite 40, Aufgabe 9

Löse die Gleichungssysteme mit der Cramer-Regel.

=over

=item a)

M<\begin{array}{rcrcrcl}
{} 2 x_1 &+&   x_2 &+& 5 x_3 &=& 1; \\
{} 2 x_1 &+& 4 x_2 &+&   x_3 &=& 1; \\
{}   x_1 &+&   x_2 &+& 2 x_3 &=& 1;
\end{array}>

M<D = \begin{vmatrix}2&1&5\\2&4&1\\1&1&2\end{vmatrix} =
\begin{vmatrix}1&5\\4&1\end{vmatrix} - \begin{vmatrix}2&5\\2&1\end{vmatrix} + 2
\begin{vmatrix}2&1\\2&4\end{vmatrix} = -19 + 8 + 12 = 1 \neq 0;>

M<x_1 = \frac{D_1}{D} =
\frac{\begin{vmatrix}1&1&5\\1&4&1\\1&1&2\end{vmatrix}}{D} =
\begin{vmatrix}4&1\\1&2\end{vmatrix} - \begin{vmatrix}1&5\\1&2\end{vmatrix} +
\begin{vmatrix}1&5\\4&1\end{vmatrix} = 7 + 3 - 19 = -9;>

M<x_2 = \frac{D_2}{D} =
\frac{\begin{vmatrix}2&1&5\\2&1&1\\1&1&2\end{vmatrix}}{D} =
-\begin{vmatrix}2&1\\1&2\end{vmatrix} + \begin{vmatrix}2&5\\1&2\end{vmatrix} -
\begin{vmatrix}2&5\\2&1\end{vmatrix} = -3 - 1 + 8 = 4;>

M<x_3 = \frac{D_3}{D} =
\frac{\begin{vmatrix}2&1&1\\2&4&1\\1&1&1\end{vmatrix}}{D} =
\begin{vmatrix}2&4\\1&1\end{vmatrix} - \begin{vmatrix}2&1\\1&1\end{vmatrix} +
\begin{vmatrix}2&1\\2&4\end{vmatrix} = -2 - 1 + 6 = 3;>

M<(x_1, x_2, x_3) = (-9, 4, 3);>

=item b)

M<\begin{array}{rcrcrcl}
{} 3 x_1 &+& 5 x_2 &+& 3 x_3 &=&  1; \\
{} 2 x_1 &+&  -x_2 &+&  -x_3 &=& -2; \\
{}   x_1 &+& 3 x_2 &+& 2 x_3 &=& -1;
\end{array}>

M<D = \begin{vmatrix}1&5&3\\2&-1&-1\\1&3&2\end{vmatrix} =
-2 \begin{vmatrix}5&3\\3&2\end{vmatrix} - \begin{vmatrix}3&3\\1&2\end{vmatrix} + 2
\begin{vmatrix}3&5\\1&3\end{vmatrix} = -2 -3 + 4 = -1 \neq 0;>

M<x_1 = \frac{D_1}{D} =
\frac{\begin{vmatrix}1&5&3\\-2&-1&-1\\-1&3&2\end{vmatrix}}{D} = -\left(
2 \begin{vmatrix}5&3\\3&2\end{vmatrix} - \begin{vmatrix}1&3\\-1&2\end{vmatrix} +
\begin{vmatrix}1&5\\-1&3\end{vmatrix}\right) = -\left(2 - 5 + 8\right) = -5;>

M<x_2 = \frac{D_2}{D} =
\frac{\begin{vmatrix}3&1&3\\2&-1&-1\\1&-1&2\end{vmatrix}}{D} = -\left(
\begin{vmatrix}1&3\\-2&-1\end{vmatrix} + \begin{vmatrix}3&3\\2&-1\end{vmatrix} +
2 \begin{vmatrix}3&1\\2&-2\end{vmatrix}\right) = -\left(5 - 9 - 16\right) =
20;>

M<x_3 = \frac{D_3}{D} =
\frac{\begin{vmatrix}3&5&1\\2&-1&-2\\1&3&-1\end{vmatrix}}{D} = -\left(
\begin{vmatrix}2&-1\\1&3\end{vmatrix} + 2 \begin{vmatrix}3&5\\1&3\end{vmatrix} -
\begin{vmatrix}3&5\\2&-1\end{vmatrix}\right) = -\left(7 + 8 + 13\right) = -28;>

M<(x_1, x_2, x_3) = (-5, 20, -28);>

=back