Řešení - Základní operace s maticemi

$$\in v\left(\left[\begin{array}{cc}-1& 2\\ 2& 3\end{array}\right]\right)$$

MatrixCoreOperationsStep2TransitionInvPlan

$$\in v\left(\left[\begin{array}{cc}-1& 2\\ 2& 3\end{array}\right]\right)$$

MatrixCoreOperationsStep2TransitionInvCheck

$$\in v\left(\left[\begin{array}{cc}-1& 2\\ 2& 3\end{array}\right]\right)$$

R1 <-> R2

$$\left[\begin{array}{cccc}2& 3& 0& 1\\ -1& 2& 1& 0\end{array}\right]$$

R1 <- 1/2R1

$$\left[\begin{array}{cccc}1& 1.5& 0& 0.5\\ -1& 2& 1& 0\end{array}\right]$$

R2 <- R2 + R1

$$\left[\begin{array}{cccc}1& 1.5& 0& 0.5\\ 0& 3.5& 1& 0.5\end{array}\right]$$

R2 <- 2/7R2

$$\left[\begin{array}{cccc}1& 1.5& 0& 0.5\\ 0& 1& 0.285714& 0.142857\end{array}\right]$$

R1 <- R1 - 3/2R2

$$\left[\begin{array}{cccc}1& 0& -0.428571& 0.285714\\ 0& 1& 0.285714& 0.142857\end{array}\right]$$

MatrixCoreOperationsStep2TextUnitInv

$$\left[\begin{array}{cccc}-0& 428571& 0& 285714\\ 0& 285714& 0& 142857\end{array}\right]$$

MatrixCoreOperationsStep3TransitionConclusionInv

$$\left[\begin{array}{cccc}-0& 428571& 0& 285714\\ 0& 285714& 0& 142857\end{array}\right]$$

MatrixCoreOperationsStep3TransitionStudyTipInv

$$\left[\begin{array}{cccc}-0& 428571& 0& 285714\\ 0& 285714& 0& 142857\end{array}\right]$$

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