# How do you find the inverse of #A=##((3, 1, 0), (1, -1, 2), (1, 1, 1))#?

##### 1 Answer

#### Answer:

#### Explanation:

You can take your matrix

#( (3, 1, 0, |, 1, 0, 0),(1, -1, 2, |, 0, 1, 0),(1, 1, 1, |, 0, 0, 1) )#

Your goal is to transform the matrix at the left side into the identity matrix.

If you perform the same operations on the identity matrix at your right side, you will get the inverse matrix at the right.

#(A | I) ->...-> (I | A^(-1) )#

Let's start.

#( (3, 1, 0, |, 1, 0, 0),(1, -1, 2, |, 0, 1, 0),(1, 1, 1, |, 0, 0, 1) )#

... swap rows

# rArr ( (1, 1, 1, |, 0, 0, 1),(1, -1, 2, |, 0, 1, 0),(3, 1, 0, |, 1, 0, 0) )#

... eliminate the

# rArr ( (1, 1, 1, |, 0, 0, 1),(color(blue)(0), color(blue)(-2), color(blue)(1), |, color(blue)(0), color(blue)(1), color(blue)(-1)),(3, 1, 0, |, 1, 0, 0) )#

... eliminate the

# rArr ( (1, 1, 1, |, 0, 0, 1),(0, -2, 1, |, 0, 1, -1),(color(orange)(0), color(orange)(-2), color(orange)(-3), |, color(orange)(1), color(orange)(0), color(orange)(-3)) )#

... eliminate the

# rArr ( (1, 1, 1, |, 0, 0, 1),(0, -2, 1, |, 0, 1, -1),(color(green)(0), color(green)(0), color(green)(-4), |, color(green)(1), color(green)(-1), color(green)(-2)) )#

... create a

# rArr ( (1, 1, 1, |, 0, 0, 1),(0, -2, 1, |, 0, 1, -1),(color(red)(0), color(red)(0), color(red)(1), |, color(red)(-1/4), color(red)(1/4), color(red)(1/2)) )#

The third row is ready. Now, let's take care of the second row:

First of all, create a

# rArr ( (1, 1, 1, |, 0, 0, 1),(color(gray)(0), color(gray)(1), color(gray)(-1/2), |, color(gray)(0), color(gray)(-1/2), color(gray)(1/2)),(0, 0, 1, |, -1/4, 1/4, 1/2) )#

... eliminate the

# rArr ( (1, 1, 1, |, 0, 0, 1),(color(purple)(0), color(purple)(1), color(purple)(0), |, color(purple)(-1/8), color(purple)(-3/8), color(purple)(3/4)),(0, 0, 1, |, -1/4, 1/4, 1/2) )#

The second row is now also ready. There are two more operations left to take care of the first row:

... eliminate the

# rArr ( (color(brown)(1), color(brown)(0), color(brown)(1), |, color(brown)(1/8), color(brown)(3/8), color(brown)(1/4)),(0, 1, 0, |, -1/8, -3/8, 3/4),(0, 0, 1, |, -1/4, 1/4, 1/2) )#

Last but not least, eliminate the

# rArr ( (color(lightgreen)(1), color(lightgreen)(0), color(lightgreen)(0), |, color(lightgreen)(3/8), color(lightgreen)(1/8), color(lightgreen)(-1/4)),(0, 1, 0, |, -1/8, -3/8, 3/4),(0, 0, 1, |, -1/4, 1/4, 1/2) )#

Thus, the result of the computation is

# ( (1, 0, 0, |, 3/8, 1/8, -1/4),(0, 1, 0, |, -1/8, -3/8, 3/4),(0, 0, 1, |, -1/4, 1/4, 1/2) )# ,

and the inverse matrix is:

#A^(-1) = ( (3/8, 1/8, -1/4),(-1/8, -3/8, 3/4),( -1/4, 1/4, 1/2) )# .

To validate the result, you can compute