oblicz wyznacznik
: 05 lut 2011, 13:39
Umie ktos to rozwiazac?
-1 1 0 2
0 2 -1 1
-1 0 1 -1
2 1 1 0
-1 1 0 2
0 2 -1 1
-1 0 1 -1
2 1 1 0
\(\left|
\begin{array}{llll}
-1 & 1 & 0 & 2 \\
0 & 2 & -1 & 1 \\
-1 & 0 & -1 & 1 \\
2 & 1 & 1 & 0
\end{array}
\right| ^{K_{4}=K_{4}+2K_{1}}=\left|
\begin{array}{llll}
-1 & 1 & 0 & 0 \\
0 & 2 & -1 & 1 \\
-1 & 0 & -1 & -1 \\
2 & 1 & 1 & 4
\end{array}
\right| ^{K_{2}=K_{2}+K_{1}}=\\ \left|
\begin{array}{llll}
-1 & 0 & 0 & 0 \\
0 & 2 & -1 & 1 \\
-1 & -1 & -1 & -1 \\
2 & 3 & 1 & 4
\end{array}
\right| ^{W_{2}=W_{2}+W_{3}}=\left|
\begin{array}{llll}
-1 & 0 & 0 & 0 \\
-1 & 1 & -2 & 0 \\
-1 & -1 & -1 & -1 \\
2 & 3 & 1 & 4
\end{array}
\right| ^{K_{3}=K_{3}+2K_{2}}=\\=\left|
\begin{array}{llll}
-1 & 0 & 0 & 0 \\
-1 & 1 & 0 & 0 \\
-1 & -1 & -3 & -1 \\
2 & 3 & 7 & 4
\end{array}
\right| ^{K_{4}=K_{4}-\frac{1}{3}K_{3}}=\left|
\begin{array}{llll}
-1 & 0 & 0 & 0 \\
-1 & 1 & 0 & 0 \\
-1 & -1 & -3 & 0 \\
2 & 3 & 7 & \frac{5}{3}
\end{array}
\right| =(-1) \cdot 1 \cdot (-3) \cdot \frac{5}{3}=5\)