Řešení - Základní operace s maticemi

$$\in v\left(\left[\begin{array}{cc}2& 4\\ 4& 1\end{array}\right]\right)$$

MatrixCoreOperationsStep2TransitionInvPlan

$$\in v\left(\left[\begin{array}{cc}2& 4\\ 4& 1\end{array}\right]\right)$$

MatrixCoreOperationsStep2TransitionInvCheck

$$\in v\left(\left[\begin{array}{cc}2& 4\\ 4& 1\end{array}\right]\right)$$

R1 <-> R2

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

R1 <- 1/4R1

$$\left[\begin{array}{cccc}1& 0.25& 0& 0.25\\ 2& 4& 1& 0\end{array}\right]$$

R2 <- R2 - 2R1

$$\left[\begin{array}{cccc}1& 0.25& 0& 0.25\\ 0& 3.5& 1& -0.5\end{array}\right]$$

R2 <- 2/7R2

$$\left[\begin{array}{cccc}1& 0.25& 0& 0.25\\ 0& 1& 0.285714& -0.142857\end{array}\right]$$

R1 <- R1 - 1/4R2

$$\left[\begin{array}{cccc}1& 0& -0.071429& 0.285714\\ 0& 1& 0.285714& -0.142857\end{array}\right]$$

MatrixCoreOperationsStep2TextUnitInv

$$\left[\begin{array}{cccc}-0& 071429& 0& 285714\\ 0& 285714& -0& 142857\end{array}\right]$$

MatrixCoreOperationsStep3TransitionConclusionInv

$$\left[\begin{array}{cccc}-0& 071429& 0& 285714\\ 0& 285714& -0& 142857\end{array}\right]$$

MatrixCoreOperationsStep3TransitionStudyTipInv

$$\left[\begin{array}{cccc}-0& 071429& 0& 285714\\ 0& 285714& -0& 142857\end{array}\right]$$

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