Edwin's solution:
We begin by augmenting that matrix on the right with the
identity matrix, and we have
We want to use row operations and end up with a matrix
that has the identity of the left looking like this:
where there are numbers where the letters "a" through "i"
are. The inverse will then be
We start with this:
We already have a 1 in the upper left corner. We need to
get a 0 right under it. So we multiply row 1
by -2 and get
and add it to row 2 which is
and get
.
Now we replace row 2 by that and we now have:
We need to
get a 0 where the 2 is in the lower left corner.
So we multiply row 1
again by -2 and get
and add it to row 3 which is
and get
.
Now we replace row 3 by that and we now have:
We need to
get a 0 where the second 1 is in the top row.
So we just add row 2
to row 1 which is
and get
Now we replace row 1 by that and we now have:
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:
Finally we need to
get a 0 where 1 is in row 2, column 3.
So we multiply row 3
again by -1 and get
and add it to row 2 which is
and get
.
Now we replace row 3 by that and we now have:
Therefore the inverse of is the 3x3
matrix on the right of the partition bar:
Edwin
