Chapter 7.3, Problem 14E

Solution

To identify: what elementary row operations applied $\left[\begin{array}{cccc}0& -3& 5& 2\\ 1& -2& 3& 0\\ 3& 1& -1& 2\end{array}\right]$ to will yield the given matrix $\left[\begin{array}{cccc}-2& 1& -1& 2\\ 1& -2& 3& 0\\ 3& 1& -1& 2\end{array}\right]$ .

2 times the second row is added to first row.

  $\left(2\right)R_2+R_1=\left[\begin{array}{cccc}0& -3& 5& 2\\ 1& -2& 3& 0\\ 3& 1& -1& 2\end{array}\right]$

Given information:

The given matrix is.

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

Explanation:

Consider the given information.

In order to yield the matrix $\left[\begin{array}{cccc}0& -3& 5& 2\\ 1& -2& 3& 0\\ 3& 1& -1& 2\end{array}\right]$ to $\left[\begin{array}{cccc}-2& 1& -1& 2\\ 1& -2& 3& 0\\ 3& 1& -1& 2\end{array}\right]$ first elementary operation is $R_2\to \left(2\right)R_2$ that is second row is multiplied by 2.

Hence,

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

  $R_2\to \left(2\right)R_2$

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

Next elementary operation applied is $R_1\to R_1+R_2$ i.e. obtained second row in above step the is added to first row:

  $\left[\begin{array}{cccc}0& -3& 5& 2\\ 2\left(1\right)& 2\left(-2\right)& 2\left(3\right)& 2\left(0\right)\\ 3& 1& -1& 2\end{array}\right]\underrightarrow{R_1\to R_1+2R_2}\left[\begin{array}{cccc}2\left(1\right)+\left(-2\right)& 2\left(-2\right)+1& 2\left(3\right)+\left(-1\right)& 2\left(0\right)+2\\ 1& -2& 3& 0\\ 3& 1& -1& 2\end{array}\right]$ which yield $\left[\begin{array}{cccc}0& -3& 5& 2\\ 1& -2& 3& 0\\ 3& 1& -1& 2\end{array}\right]$ .

So, the elementary row operations are $R_2\to \left(2\right)R_2\text{and}{R}_1\to R_1+2R_2$ .