SOLUTION: can someone please show me how to do this problem: Find the inverse, if it exists, of the given matrix [2 -1 0] [3 -2 0] [-2 3 1] a. [2 -1 0] [3 -2 0] [-5 4 -1]

Augment it with the identity matrix:

 

Now do row operations to get the identity on the left
of the partition and then the inverse will be on the right
of the partition.  We will place all the 0's first,
then we will place the 1's.



To get a 0 where the 3 in the 2nd row is, 
multiply the 1st row by -3
multiply the second row by 2
Add -3 times the 1st row to 2 times the 2nd row
Keep the 1st row as it is.
This work can be done mentally if you will write -3
to the left of the 1st row and 2 to the left of the
2nd row.  Then fill in the next matrix with a new 2nd
row.  The other two rows will remain the same.





To get a 0 where the -2 in the 3rd row is, 
multiply the 1st row by 1 (leave it as it is)
multiply the second row by 1 (leave it as it is)
Add 1 times the 1st row to 1 times the 3rd row
Keep the 1st row as it is.
This work can be done mentally easily if you'll
put 1's left of the 1st and 3rd rows.
Then fill in the next matrix with a new 3rd
row.   The other two rows will remain the same.





To get a 0 where the -1 in the 1st row is, 
multiply the 2nd row by -1
multiply the 1st row by 1 (leave it as it is)
Add -1 times the 2nd row to 1 times the 1st row
Keep the 2nd row as it is.
This work can be done mentally if you will write -1
to the left of the 2nd row and 1 to the left of the
1st row.  Then fill in the next matrix with a new 1st
row. The other two rows will remain the same.





To get a 0 where the 2 in the 3rd row is, 
multiply the 2nd row by 2
multiply the 3rd row by 1 (leave it as it is)
Add 2 times the 2nd row to 1 times the 3rd row
Keep the 2nd row as it is.
This work can be done mentally if you will write 2
to the left of the 2nd row and 1 to the left of the
3rd row.  Then fill in the next matrix with a new 3rd
row. The other two rows will remain the same.





Now we have all the 0's placed. We need to get a 1 where 
the 2 in in the 1st row.  To do that we multiply the 1st
row through by . Or you can think of it as 
dividing the row through by 2:





We need to get a 1 where 
the -1 in in the 2nd row.  To do that we multiply the 2nd
row through by . Or you can think of it as 
changing all the signs in the 2nd row:





The inverse of the original matrix is found right of 
the partition.  So the answer is

, choice d.

Edwin