Giải pháp - Cac phep toan cot loi cua ma tran

$$\in v\left(\left[\begin{array}{cc}5& 1\\ 4& 0\end{array}\right]\right)$$

Ap dung phep bien doi hang hoac phep toan ma tran de tao ket qua duoc yeu cau.

$$\in v\left(\left[\begin{array}{cc}5& 1\\ 4& 0\end{array}\right]\right)$$

Ap dung phep bien doi hang hoac phep toan ma tran de tao ket qua duoc yeu cau.

$$\in v\left(\left[\begin{array}{cc}5& 1\\ 4& 0\end{array}\right]\right)$$

R1 <- 1/5R1

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

R2 <- R2 - 4R1

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

R2 <- -5/4R2

$$\left[\begin{array}{cccc}1& 0.2& 0.2& 0\\ 0& 1& 1& -1.25\end{array}\right]$$

R1 <- R1 - 1/5R2

$$\left[\begin{array}{cccc}1& 0& 0& 0.25\\ 0& 1& 1& -1.25\end{array}\right]$$

Ap dung phep bien doi hang hoac phep toan ma tran de tao ket qua duoc yeu cau.

$$\left[\begin{array}{ccc}0& 0& 25\\ 1& -1& 25\end{array}\right]$$

Trinh bay ket qua ma tran hoac vo huong cuoi cung o dang chuan.

$$\left[\begin{array}{ccc}0& 0& 25\\ 1& -1& 25\end{array}\right]$$

Trinh bay ket qua ma tran hoac vo huong cuoi cung o dang chuan.

$$\left[\begin{array}{ccc}0& 0& 25\\ 1& -1& 25\end{array}\right]$$

Trinh bay ket qua ma tran hoac vo huong cuoi cung o dang chuan.