Question 190355
<Font size = 8 color = "red"><b>Edwin's solution:</b></font>
<pre><font size = 4 color = "indigo"><b>
{{{A=(matrix(3,3,1,1,1,2,1,1,2,2,3))}}}

We begin by augmenting that matrix on the right with the
identity matrix, and we have

{{{(matrix(3,7,1,1,1,"|",1,0,0,2,1,1,"|",0,1,0,2,2,3,"|",0,0,1 )     )}}}

We want to use row operations and end up with a matrix 
that has the identity of the left looking like this:

{{{(matrix(3,7,1,0,0,"|",a,b,c,0,1,0,"|",d,e,f,0,0,1,"|",g,h,i  )     )}}}

where there are numbers where the letters "a" through "i"
are.  The inverse will then be

{{{A^(-1)=(matrix(3,3,a,b,c,d,e,f,g,h,i))}}}

We start with this:

{{{(matrix(3,7,1,1,1,"|",1,0,0,2,1,1,"|",0,1,0,2,2,3,"|",0,0,1 )     )}}}

We already have a 1 in the upper left corner.  We need to
get a 0 right under it.  So we multiply row 1 {{{(matrix(1,7,1,1,1,"|",1,0,0))}}}
by -2 and get {{{(matrix(1,7,-2,-2,-2,"|",-2,0,0))}}}
 and add it to row 2 which is {{{(matrix(1,7,2,1,1,"|",0,1,0))}}}
 and get {{{(matrix(1,7,0,-1,-1,"|",-2,1,0))}}}
.
Now we replace row 2 by that and we now have:

{{{(matrix(3,7,1,1,1,"|",1,0,0,0,-1,-1,"|",-2,1,0,2,2,3,"|",0,0,1 )     )}}}

We need to
get a 0 where the 2 is in the lower left corner.  
So we multiply row 1 {{{(matrix(1,7,1,1,1,"|",1,0,0))}}}
again by -2 and get {{{(matrix(1,7,-2,-2,-2,"|",-2,0,0))}}}
 and add it to row 3 which is {{{(matrix(1,7,2,2,3,"|",0,0,1))}}}
 and get {{{(matrix(1,7,0,0,1,"|",-2,0,1))}}}
.
Now we replace row 3 by that and we now have:

{{{(matrix(3,7,1,1,1,"|",1,0,0,0,-1,-1,"|",-2,1,0,0,0,1,"|",-2,0,1 )     )}}}

We need to
get a 0 where the second 1 is in the top row.  
So we just add row 2 {{{(matrix(1,7,0,-1,-1,"|",-2,1,0))}}}
to row 1 which is {{{(matrix(1,7,1,1,1,"|",1,0,0))}}}
 and get {{{(matrix(1,7,1,0,0,"|",-1,0,0))}}} 

Now we replace row 1 by that and we now have:

{{{(matrix(3,7,1,0,0,"|",-1,0,0,0,-1,-1,"|",-2,1,0,0,0,1,"|",-2,0,1 )     )}}}

We need to
get a 1 where the -1 is in row column 2, so we just
divide that row through by -1, which means we just
change the signs of all the elements on row 2:

{{{(matrix(3,7,1,0,0,"|",-1,0,0,0,1,1,"|",2,-1,0,0,0,1,"|",-2,0,1 )     )}}}

Finally we need to
get a 0 where 1 is in row 2, column 3.
 
So we multiply row 3 {{{(matrix(1,7,0,0,1,"|",-2,0,1))}}}
again by -1 and get {{{(matrix(1,7,0,0,-1,"|",2,0,-1))}}}
 and add it to row 2 which is {{{(matrix(1,7,0,1,1,"|",2,-1,0))}}}
 and get {{{(matrix(1,7,0,1,0,"|",4,-1,-1))}}}
.
Now we replace row 3 by that and we now have: 

{{{(matrix(3,7,1,0,0,"|",-1,0,0,0,1,0,"|",4,-1,-1,0,0,1,"|",-2,0,1 )     )}}} 

Therefore the inverse of {{{A}}} is the 3x3
matrix on the right of the partition bar:

{{{A^(-1)=(matrix(3,3,-1,1,0,4,-1,-1,-2,0,1))}}}

Edwin</pre>