# How do you solve #x-y+z=3# and #2y-z=1# and #-x+2y= -1# using matrices?

##### 1 Answer

#### Explanation:

Your linear equation can be written as the following matrix (first column representing the

#( (1, -1, 1, |, 3),(0, 2, -1, |, 1),(-1, 2, 0, |, -1) ) " " {: (I),(II),(III) :}#

To eliminate the

# rArr ( (1, -1, 1, |, 3),(0, 2, -1, |, 1),(color(blue)(0), color(blue)(1), color(blue)(1), |, color(blue)(2)) ) " " {: (I),(II),(color(blue)(III)) :}#

To eliminate the

# rArr ( (1, -1, 1, |, 3),(color(red)(0), color(red)(0), color(red)(-3), |,color(red)( -3)),(0, 1, 1, |, 2) ) " " {: (I),(color(red)(II)),(III) :}#

Compute

# rArr ( (1, -1, 1, |, 3),(color(green)(0), color(green)(0), color(green)(1), |, color(green)(1)),(0, 1, 1, |, 2) ) " " {: (I),(color(green)(II)),(III) :}#

Now, use the second row to eliminate the

compute

# rArr ( (1, -1, 1, |, 3),(0, 0, 1, |, 1),(color(orange)(0), color(orange)(1), color(orange)(0), |, color(orange)(1)) ) " " {: (I),(II),(color(orange)(III)) :}#

We are almost done. Now, you need to use the third row to eliminate the

Let's do it step by step: first,

# rArr ( (color(violet)(1), color(violet)(0), color(violet)(1), |, color(violet)(4)),(0, 0, 1, |, 1),(0, 1, 0, |, 1) ) " " {: (color(violet)(I)),(II),(III) :}#

Last step: compute

# rArr ( (color(gray)(1), color(gray)(0), color(gray)(0), |, color(gray)(3)),(0, 0, 1, |, 1),(0, 1, 0, |, 1) ) " " {: (color(gray)(I)),(II),(III) :}#

If you would like, you can also swap

The solution is