Řešení - Základní operace s maticemi

$$\in v\left(\left[\begin{array}{cc}5& -1\\ 4& 2\end{array}\right]\right)$$

MatrixCoreOperationsStep2TransitionInvPlan

$$\in v\left(\left[\begin{array}{cc}5& -1\\ 4& 2\end{array}\right]\right)$$

MatrixCoreOperationsStep2TransitionInvCheck

$$\in v\left(\left[\begin{array}{cc}5& -1\\ 4& 2\end{array}\right]\right)$$

R1 <- 1/5R1

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

R2 <- R2 - 4R1

$$\left[\begin{array}{cccccc}1& -0& 2& 0& 2& 0\\ 0& 2& 8& -0& 8& 1\end{array}\right]$$

R2 <- 5/14R2

$$\left[\begin{array}{cccc}1& -0.2& 0.2& 0\\ 0& 1& -0.285714& 0.357143\end{array}\right]$$

R1 <- R1 + 1/5R2

$$\left[\begin{array}{cccc}1& 0& 0.142857& 0.071429\\ 0& 1& -0.285714& 0.357143\end{array}\right]$$

MatrixCoreOperationsStep2TextUnitInv

$$\left[\begin{array}{cccc}0& 142857& 0& 071429\\ -0& 285714& 0& 357143\end{array}\right]$$

MatrixCoreOperationsStep3TransitionConclusionInv

$$\left[\begin{array}{cccc}0& 142857& 0& 071429\\ -0& 285714& 0& 357143\end{array}\right]$$

MatrixCoreOperationsStep3TransitionStudyTipInv

$$\left[\begin{array}{cccc}0& 142857& 0& 071429\\ -0& 285714& 0& 357143\end{array}\right]$$

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