solve system of equations using row operations on augmented matrix

Hey guys I really need some help on solving 2 algebra problems.

The question states: Solve the system of equations using row operations on the augmented matrix.

Problem 1.

x+y-z=5

x+2y-3z=9

x-y+3z=3

Problem 2.

x-y+z=14

3x+2y+z=19

-2x+y-z=-21

I just want to know the steps on how to solve it one by one and I could solve the problem but as of now I am clueless on how to solve it. If you guys could help me out I would really appreciate it. Thank you in advance.

Re: solve system of equations using row operations on augmented matrix

[1 1 -1 : 5]

[1 2 -3 : 9]

[1 -1 3 : 3]

The left side of the : is the coefficients of x,y,z in our system and the numbers on the right side are our values. We need to be this is row echloen form.

First replace row 2 by (-row 2 + row 1). Then

[1 1 -1 : 5]

[0 -1 2 : -4]

[1 -1 3 : 3]

Now, replace row 3 by (-row 3 + row 1). Then

[1 1 -1 : 5]

[0 -1 2 : -4]

[0 2 -4 : -2]

Now multiply row 2 by -1. Then

[1 1 -1 : 5]

[0 1 -2 : 4]

[0 2 -4 : -2]

Now, replace row 3 by (-1/2 row 3 + row 2). Then

[1 1 -1 : 5]

[0 1 -2 : 4]

[0 0 0 : 5]

This third row implies that 0x + 0y + 0z = 5 i.e. 0 = 5 which is impossible. No solution!

Re: solve system of equations using row operations on augmented matrix

Hello, irv1234!

Here's the second one.

Quote:

$\displaystyle \begin{array}{ccc}x-y+z&=&14 \\3x+2y+z&=&19 \\ \text{-}2x+y-z&=&\text{-}21 \end{array}$

$\displaystyle \text{We have: }\:\left|\begin{array}{ccc|c}1&\text{-}1&1&14 \\ 3&2&1&19 \\ \text{-}2&1&\text{-}1 & \text{-}21 \end{array}\right| $

$\displaystyle \begin{array}{c}\text{Switch} \\ R_2\:\&\,R_3 \\ \end{array}\left|\begin{array}{ccc|c}1&\text{-}1&1&14 \\ \text{-}2&1&\text{-}1 & \text{-}21 \\ 3&2&1&19 \end{array}\right| $

$\displaystyle \begin{array}{c} \\ R_2+2R_1 \\ R_3-3R_1 \end{array}\left|\begin{array}{ccc|c}1&\text{-}1&1&14 \\ 0&\text{-}1&1&7 \\ 0&5&\text{-}2&\text{-}23\end{array}\right|$

$\displaystyle \begin{array}{c}R_1-R_2 \\ \\ R_3+5R_2\end{array}\left|\begin{array}{ccc|c}1&0&0 &7 \\ 0&\text{-}1&1&7 \\ 0&0&3&12\end{array}\right|$

. . . $\displaystyle \begin{array}{c} \\ \text{-}1R_2 \\ \frac{1}{3}R_3 \end{array} \left|\begin{array}{ccc|c}1&0&0&7 \\ 0&1&\text{-}1 & \text{-}7 \\ 0&0&1&4 \end{array}\right|$

$\displaystyle \begin{array}{c} \\ R_2+R_3 \\ \\ \end{array} \left|\begin{array}{ccc|c}1&0&0&7 \\ 0&1&0&\text{-}3 \\ 0&0&1&4 \end{array}\right|$

$\displaystyle \text{Therefore: }\:\begin{Bmatrix}x &=& 7 \\ y &=& \text{-}3 \\ z &=& 4 \end{Bmatrix}$