zad 3
rozwiazac rownanie macierzowe
\(\displaystyle{ \left[\begin{array}{cc}-1&2\\-3&4\end{array}\right]X= ft[\begin{array}{cc}4&1\\2&3\end{array}\right]+2\left[\begin{array}{cc}-1&1\\0&-2\end{array}\right]}\)
zad 4
rozwiazac uklad rownan Cramera korzystajac ze wzorow Cramera lub stosujac macierz odwrotna
\(\displaystyle{ \left\{\begin{array}{l} \quad x_1-x_2+4x_3=-4\\-2x_1+x_2-x_3=0\\3x_1+2x_2+x_3=4 \end{array}}\)
zad 5
rozwiazac uklad rownan liniowych
\(\displaystyle{ \left\{\begin{array}{l} x_1-2x_2+2x_3-x_4=3\\5x_1+2x_2+3x_3+x_4=1\\-2x_1+3x_2+2x_3+x_4=-1\end{array}}\)
rownanie macierzowe i dwa uklady rownan
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rownanie macierzowe i dwa uklady rownan
zad.1
\(\displaystyle{ \left[\begin{array}{cc}-1&2\\-3&4\end{array}\right]X= ft[\begin{array}{cc}4&1\\2&3\end{array}\right]+2\left[\begin{array}{cc}-1&1\\0&-2\end{array}\right]}\)
\(\displaystyle{ \left[\begin{array}{cc}-1&2\\-3&4\end{array}\right]X= ft[\begin{array}{cc}4&1\\2&3\end{array}\right]+\left[\begin{array}{cc}-2&2\\0&-4\end{array}\right]}\)
\(\displaystyle{ \left[\begin{array}{cc}-1&2\\-3&4\end{array}\right]X= ft[\begin{array}{cc}2&3\\2&-4\end{array}\right]}\)
\(\displaystyle{ X=\left[\begin{array}{cc}-1&2\\-3&4\end{array}\right]^{-1}\cdot ft[\begin{array}{cc}2&3\\2&-4\end{array}\right]}\)
\(\displaystyle{ X=\left[\begin{array}{cc}2&-1\\\frac{3}{2}&-\frac{1}{2}\end{array}\right]\cdot ft[\begin{array}{cc}2&3\\2&-4\end{array}\right]}\)
\(\displaystyle{ X=\left[\begin{array}{cc}2&10\\2&\frac{13}{2}\end{array}\right]}\)
\(\displaystyle{ \left[\begin{array}{cc}-1&2\\-3&4\end{array}\right]X= ft[\begin{array}{cc}4&1\\2&3\end{array}\right]+2\left[\begin{array}{cc}-1&1\\0&-2\end{array}\right]}\)
\(\displaystyle{ \left[\begin{array}{cc}-1&2\\-3&4\end{array}\right]X= ft[\begin{array}{cc}4&1\\2&3\end{array}\right]+\left[\begin{array}{cc}-2&2\\0&-4\end{array}\right]}\)
\(\displaystyle{ \left[\begin{array}{cc}-1&2\\-3&4\end{array}\right]X= ft[\begin{array}{cc}2&3\\2&-4\end{array}\right]}\)
\(\displaystyle{ X=\left[\begin{array}{cc}-1&2\\-3&4\end{array}\right]^{-1}\cdot ft[\begin{array}{cc}2&3\\2&-4\end{array}\right]}\)
\(\displaystyle{ X=\left[\begin{array}{cc}2&-1\\\frac{3}{2}&-\frac{1}{2}\end{array}\right]\cdot ft[\begin{array}{cc}2&3\\2&-4\end{array}\right]}\)
\(\displaystyle{ X=\left[\begin{array}{cc}2&10\\2&\frac{13}{2}\end{array}\right]}\)