Answer to Question #169347 in Linear Algebra for Mohammad Hossain

Question

Find the row echelon form of [

1 2 − 2 − 2 −3

1 3 − 2 0 −4

3 8 − 7 − 2 −11

2 1 − 9 − 10 − 3

]. 

Answer 1

Solution:

Given: $\begin{bmatrix}1&2&-2&-2&-3\\ 1&3&-2&0&-4\\ 3&8&-7&-2&-11\\ 2&1&-9&-10&-3\end{bmatrix}$

$R_1\:\leftrightarrow \:R_3$

$=\begin{bmatrix}3&8&-7&-2&-11\\ 1&3&-2&0&-4\\ 1&2&-2&-2&-3\\ 2&1&-9&-10&-3\end{bmatrix}$

$R_2\:\rightarrow \:R_2-\frac{1}{3}\cdot \:R_1$

$=\begin{bmatrix}3&8&-7&-2&-11\\ 0&\frac{1}{3}&\frac{1}{3}&\frac{2}{3}&-\frac{1}{3}\\ 1&2&-2&-2&-3\\ 2&1&-9&-10&-3\end{bmatrix}$

$R_3\:\rightarrow \:R_3-\frac{1}{3}\cdot \:R_1$

$=\begin{bmatrix}3&8&-7&-2&-11\\ 0&\frac{1}{3}&\frac{1}{3}&\frac{2}{3}&-\frac{1}{3}\\ 0&-\frac{2}{3}&\frac{1}{3}&-\frac{4}{3}&\frac{2}{3}\\ 2&1&-9&-10&-3\end{bmatrix}$

$R_4\:\rightarrow \:R_4-\frac{2}{3}\cdot \:R_1$

$=\begin{bmatrix}3&8&-7&-2&-11\\ 0&\frac{1}{3}&\frac{1}{3}&\frac{2}{3}&-\frac{1}{3}\\ 0&-\frac{2}{3}&\frac{1}{3}&-\frac{4}{3}&\frac{2}{3}\\ 0&-\frac{13}{3}&-\frac{13}{3}&-\frac{26}{3}&\frac{13}{3}\end{bmatrix}$

$R_2\:\leftrightarrow \:R_4$

$=\begin{bmatrix}3&8&-7&-2&-11\\ 0&-\frac{13}{3}&-\frac{13}{3}&-\frac{26}{3}&\frac{13}{3}\\ 0&-\frac{2}{3}&\frac{1}{3}&-\frac{4}{3}&\frac{2}{3}\\ 0&\frac{1}{3}&\frac{1}{3}&\frac{2}{3}&-\frac{1}{3}\end{bmatrix}$

$R_3\:\rightarrow \:R_3-\frac{2}{13}\cdot \:R_2 \\ =\begin{bmatrix}3&8&-7&-2&-11\\ 0&-\frac{13}{3}&-\frac{13}{3}&-\frac{26}{3}&\frac{13}{3}\\ 0&0&1&0&0\\ 0&\frac{1}{3}&\frac{1}{3}&\frac{2}{3}&-\frac{1}{3}\end{bmatrix}$

$R_4\:\rightarrow \:R_4+\frac{1}{13}\cdot \:R_2 \\ =\begin{bmatrix}3&8&-7&-2&-11\\ 0&-\frac{13}{3}&-\frac{13}{3}&-\frac{26}{3}&\frac{13}{3}\\ 0&0&1&0&0\\ 0&0&0&0&0\end{bmatrix}$

This is required reduce matrix to row echelon form.