## Elementary Linear Algebra 7th Edition

Published by Cengage Learning

# Chapter 1 - Systems of Linear Equations - 1.2 Gaussian Elimination and Gauss-Jordan Elimination - 1.2 Exercises - Page 22: 10

#### Answer

The first row remained unchanged. The second row is transformed: The first row multiplied by $2$ is added to it. The third row is transformed: The first row multiplied by $5$ is added to it.

#### Work Step by Step

The first row remained unchanged. The second and the third rows were transformed $$\text{Row 1 Original} = \left[\begin{matrix}-1&-2&3&-2\end{matrix}\right]\Longrightarrow \text{Row 1 Transformed} = \left[\begin{matrix}-1&-2&3&-2\end{matrix}\right];$$ $$\text{Row 2 Original} = \left[\begin{matrix}2&-5&1&-7\end{matrix}\right]\Longrightarrow \text{Row 2 Transformed} = \left[\begin{matrix}0&-9&7&-11\end{matrix}\right].$$ $$\text{Row 3 Original} = \left[\begin{matrix}5&4&-7&6\end{matrix}\right]\Longrightarrow \text{Row 3 Transformed} = \left[\begin{matrix}0&-6&8&-4\end{matrix}\right].$$ To transform a row we can only multiply each element in it by some number and add the element in the same column of another row multiplied by some number to it. This can be achieved here by Step 1: Adding to the second row the first one multiplied by $2$. Indeed, we will have checking this for each element: $$2+2\times(-1) = 0\quad -5+2(-2) = -9\quad 1+2\times3 = 7\quad -7+2\times(-2) = -11$$ Step 2: Adding the first row multiplied by $5$ to the third row. Again, checking for each element we have: $$5+5\times(-1) = 0\quad 4+5\times(-2) = -6\quad -7+5\times3 = 8\quad 6+5\times(-2) = -4.$$

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