Řešení - Základní operace s maticemi

$$\in v\left(\left[\begin{array}{cc}-3& 5\\ 4& -2\end{array}\right]\right)$$

MatrixCoreOperationsStep2TransitionInvPlan

$$\in v\left(\left[\begin{array}{cc}-3& 5\\ 4& -2\end{array}\right]\right)$$

MatrixCoreOperationsStep2TransitionInvCheck

$$\in v\left(\left[\begin{array}{cc}-3& 5\\ 4& -2\end{array}\right]\right)$$

R1 <-> R2

$$\left[\begin{array}{cccc}4& -2& 0& 1\\ -3& 5& 1& 0\end{array}\right]$$

R1 <- 1/4R1

$$\left[\begin{array}{cccc}1& -0.5& 0& 0.25\\ -3& 5& 1& 0\end{array}\right]$$

R2 <- R2 + 3R1

$$\left[\begin{array}{cccc}1& -0.5& 0& 0.25\\ 0& 3.5& 1& 0.75\end{array}\right]$$

R2 <- 2/7R2

$$\left[\begin{array}{cccc}1& -0.5& 0& 0.25\\ 0& 1& 0.285714& 0.214286\end{array}\right]$$

R1 <- R1 + 1/2R2

$$\left[\begin{array}{cccc}1& 0& 0.142857& 0.357143\\ 0& 1& 0.285714& 0.214286\end{array}\right]$$

MatrixCoreOperationsStep2TextUnitInv

$$\left[\begin{array}{cccc}0& 142857& 0& 357143\\ 0& 285714& 0& 214286\end{array}\right]$$

MatrixCoreOperationsStep3TransitionConclusionInv

$$\left[\begin{array}{cccc}0& 142857& 0& 357143\\ 0& 285714& 0& 214286\end{array}\right]$$

MatrixCoreOperationsStep3TransitionStudyTipInv

$$\left[\begin{array}{cccc}0& 142857& 0& 357143\\ 0& 285714& 0& 214286\end{array}\right]$$

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