Řešení - Základní operace s maticemi

$$\in v\left(\left[\begin{array}{cc}-2& 5\\ 3& 1\end{array}\right]\right)$$

MatrixCoreOperationsStep2TransitionInvPlan

$$\in v\left(\left[\begin{array}{cc}-2& 5\\ 3& 1\end{array}\right]\right)$$

MatrixCoreOperationsStep2TransitionInvCheck

$$\in v\left(\left[\begin{array}{cc}-2& 5\\ 3& 1\end{array}\right]\right)$$

R1 <-> R2

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

R1 <- 1/3R1

$$\left[\begin{array}{cccc}1& 0.333333& 0& 0.333333\\ -2& 5& 1& 0\end{array}\right]$$

R2 <- R2 + 2R1

$$\left[\begin{array}{cccc}1& 0.333333& 0& 0.333333\\ 0& 5.666667& 1& 0.666667\end{array}\right]$$

R2 <- 3/17R2

$$\left[\begin{array}{cccc}1& 0.333333& 0& 0.333333\\ 0& 1& 0.176471& 0.117647\end{array}\right]$$

R1 <- R1 - 1/3R2

$$\left[\begin{array}{cccc}1& 0& -0.058824& 0.294118\\ 0& 1& 0.176471& 0.117647\end{array}\right]$$

MatrixCoreOperationsStep2TextUnitInv

$$\left[\begin{array}{cccc}-0& 058824& 0& 294118\\ 0& 176471& 0& 117647\end{array}\right]$$

MatrixCoreOperationsStep3TransitionConclusionInv

$$\left[\begin{array}{cccc}-0& 058824& 0& 294118\\ 0& 176471& 0& 117647\end{array}\right]$$

MatrixCoreOperationsStep3TransitionStudyTipInv

$$\left[\begin{array}{cccc}-0& 058824& 0& 294118\\ 0& 176471& 0& 117647\end{array}\right]$$

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