macierze

[wiola89]

Korzystając z twierdzenia Cramera rozwiązać następujące układy równań liniowych:

2x -y = 1
x + 3y = 18


x +2y + 3z = 6
2x + y + z = 4
3x + y - 4z = o

[agulka1987]

wzór Cramera \(\displaystyle{ x _{i} = \frac{detA _{i} }{detA}}\)

\(\displaystyle{ det A= ft[\begin{array}{cc}2$ -1\\1$ 3\end{array} \right] = 7}\)

\(\displaystyle{ detA_{1} ft[\begin{array}{cc}1$ -1\\18$ 3\end{array}\right] =21}\)

\(\displaystyle{ x_{1}= \frac{detA _{1} }{detA} = \frac{21}{7}=3}\)

\(\displaystyle{ detA_{2} ft[\begin{array}{cc}2$ 1\\1$ 18\end{array}\right] =35}\)

\(\displaystyle{ x_{2}= \frac{detA _{2} }{detA} = \frac{35}{7}=5}\)

czyli x= 3, y= 5





\(\displaystyle{ det A= ft[\begin{array}{ccc}1$ 2 $3\\2$ 1 $1\\3$ 1 $-4\end{array} \right] = 14}\)

\(\displaystyle{ det A _{1} = ft[\begin{array}{ccc}6$ 4 $0\\2$ 1 $1\\3$ 1 $-4\end{array} \right] = 14}\)

\(\displaystyle{ det A _{2} = ft[\begin{array}{ccc}1$ 2 $3\\6$ 4 $0\\3$ 1 $-4\end{array} \right] = 14}\)

\(\displaystyle{ det A _{3} = ft[\begin{array}{ccc}1$ 2 $3\\2$ 1 $1\\6$ 4 $0\end{array} \right] = 14}\)

\(\displaystyle{ x=y=z = 1}\)