Question

inverse the matrix(2 0 1,3 2 -5,1 -1 0) by gauss jordan method

Answer 1

Inverse the matrix $\begin{pmatrix} 2 & 0 & 1 \\ 3 & 2 & -5 \\ 1 & -1 & 0 \end{pmatrix}$ by Gauss Jordan method.

Solution:

$$\begin{pmatrix} 2 & 0 & 1 & 1 & 0 & 0 \\ 3 & 2 & -5 & 0 & 1 & 0 \\ 1 & -1 & 0 & 0 & 0 & 1 \end{pmatrix}$$

Exchange row 1 and row 3.

$$\begin{pmatrix} 1 & -1 & 0 & 0 & 0 & 1 \\ 3 & 2 & -5 & 0 & 1 & 0 \\ 2 & 0 & 1 & 1 & 0 & 0 \end{pmatrix} \sim \begin{pmatrix} 1 & -1 & 0 & 0 & 0 & 1 \\ 0 & 5 & -5 & 0 & 1 & -3 \\ 0 & 2 & 1 & 1 & 0 & -2 \end{pmatrix} \sim \begin{pmatrix} 1 & -1 & 0 & 0 & 0 & 1 \\ 0 & 1 & -1 & 0 & 1 & -3 \\ 0 & 2 & 1 & 1 & 0 & -2 \end{pmatrix} \sim$$ $$\begin{pmatrix} 1 & -1 & 0 & 0 & 0 & 1 \\ 0 & 1 & -1 & 0 & 5 & -3 \\ 0 & 0 & 3 & 1 & -2 & -4 \\ 5 & & & & 5 & -4 \end{pmatrix} \sim \begin{pmatrix} 1 & -1 & 0 & 0 & 0 & 1 \\ 0 & 1 & -1 & 0 & 1 & -3 \\ 0 & 0 & 1 & 1 & 5 & -5 \\ 1 & -2 & 4 & 1 & -4 \\ 3 & -2 & 5 & -4 \\ 15 & & & & 15 \end{pmatrix}$$ $$\begin{pmatrix} 1 & -1 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 3 & 15 & -13 \\ 0 & 0 & 1 & 1 & -2 & -4 \\ 3 & & & 15 & -4 \\ 15 & & & & 15 \end{pmatrix} \sim \begin{pmatrix} 1 & 0 & 0 & 1 \\ 1 & 1 & 15 & -13 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 3 & -2 & 4 & 15 \end{pmatrix}$$

So inverse matrix is $\begin{pmatrix} \frac{1}{3} & \frac{1}{15} & \frac{2}{15} \\ \frac{1}{3} & \frac{1}{15} & -\frac{13}{15} \\ \frac{1}{3} & -\frac{2}{15} & -\frac{4}{15} \end{pmatrix}$

Answer: $\begin{pmatrix} \frac{1}{3} & \frac{1}{15} & \frac{2}{15} \\ \frac{1}{3} & \frac{1}{15} & -\frac{13}{15} \\ \frac{1}{3} & -\frac{2}{15} & -\frac{4}{15} \end{pmatrix}.$