Řešení - Základní operace s maticemi

$$\in v\left(\left[\begin{array}{cc}2& 4\\ 5& -4\end{array}\right]\right)$$

MatrixCoreOperationsStep2TransitionInvPlan

$$\in v\left(\left[\begin{array}{cc}2& 4\\ 5& -4\end{array}\right]\right)$$

MatrixCoreOperationsStep2TransitionInvCheck

$$\in v\left(\left[\begin{array}{cc}2& 4\\ 5& -4\end{array}\right]\right)$$

R1 <-> R2

$$\left[\begin{array}{cccc}5& -4& 0& 1\\ 2& 4& 1& 0\end{array}\right]$$

R1 <- 1/5R1

$$\left[\begin{array}{cccc}1& -0.8& 0& 0.2\\ 2& 4& 1& 0\end{array}\right]$$

R2 <- R2 - 2R1

$$\left[\begin{array}{cccc}1& -0.8& 0& 0.2\\ 0& 5.6& 1& -0.4\end{array}\right]$$

R2 <- 5/28R2

$$\left[\begin{array}{cccc}1& -0.8& 0& 0.2\\ 0& 1& 0.178571& -0.071429\end{array}\right]$$

R1 <- R1 + 4/5R2

$$\left[\begin{array}{cccc}1& 0& 0.142857& 0.142857\\ 0& 1& 0.178571& -0.071429\end{array}\right]$$

MatrixCoreOperationsStep2TextUnitInv

$$\left[\begin{array}{cccc}0& 142857& 0& 142857\\ 0& 178571& -0& 071429\end{array}\right]$$

MatrixCoreOperationsStep3TransitionConclusionInv

$$\left[\begin{array}{cccc}0& 142857& 0& 142857\\ 0& 178571& -0& 071429\end{array}\right]$$

MatrixCoreOperationsStep3TransitionStudyTipInv

$$\left[\begin{array}{cccc}0& 142857& 0& 142857\\ 0& 178571& -0& 071429\end{array}\right]$$

Zobrazte konečný maticový nebo skalární výsledek v kanonickém tvaru pro stabilní směrování a kontrolu.