Expert: Sherry Wallin - 7/30/2009

Question
Hi, this is probably the longest question you have been ever asked..but please bear with me here

Find the inverse of matrix (if there is one) ALGEBRAICALLY. if there is not an inverse, say that the matrix is singular.

[3 1 2]
[3 2 -1]
[1 1  1]

Ok here is my 3 pages of work, i would have scanned it but im out of ink..

i used the Row echelon form..

[3 1 2]  [1 0 0]
[3 2 -1] [0 1 0]
[1 1 1]  [0 0 1]

using r1= -2r3+r1: Obtained 1 in row 1, column 1 :

[1 -1 0]  [1 0 -2]
[3  2 -1] [0 1  0]
[1  1  1] [0 0  1]

using r2= -3r3 + r2, obtained 0 in row 2, column 1 :

[1 -1 0] [1 0 -2]
[0 -1 -4][0 1 -3]
[1  1  1][0 0  1]

using r3 = -r1 + r3 obtained 0 in row 3, column 1 :
  
[1 -1 0] [ 1 0 -2]
[0 -1 -4][ 0 1 -3]
[0  2  1][-1 0  3]

using r1= -r2+r1 i obtained 0 in row 1, column 2:

[1  0  4] [1 -1  1]
[0 -1 -4] [0  1 -3]
[0  2  1] [-1 0  3]

using r2= -r2, i obtained 1 in row 2, column 2:

[1 0 4][1 -1 1]
[0 1 4][0 -1 3]
[0 2 1][-1 0 3]

Using r3= -2r2 + r3 , i obtained 0 in row 3, column 2:

[1 0  4][0 -1  1]
[0 1  4][0 -1  3]
[0 0 -7][-1 2 -3]

using r3= -1/7r3, i obtained 1 in row 3, column 3

[1 0 4][0   -1    1]
[0 1 4][0   -1    3]
[0 0 1][2/7 -2/7 3/7]

using r1= -r2 + r1, i obtained 0 in row 1, column 3:

[1 0 0][0    0   -2]
[0 1 4][0   -1    3]
[0 0 1][2/7 -2/7 3/7]

using r2= -4r3 + r2, i obtained 0 in row 1, column 3:

[1 0 0][0     0   -2]
[0 1 0][-8/7 1/7 9/7]
[0 0 1][2/7 -2/7 3/7]

so i got the inverse from this:

[0    0   -2]
[-8/7 1/7 9/7]
[2/7 -2/7 3/7]

is that right ? please help ! thanks a bunch !!

Answer
Hi, no it isn't right unless there is no solution because you have arrived at 0*x + 0*y = -2 and there are no values of x and y that that will ever happen with. If what you want is for me to find your mistake (or your first mistake then I will do that!) Nonetheless all you have to do is to substitute your 'possible' solution and if it makes all the equations true then it is a solution, otherwise it is not. So let me check what you have:

The first error you made is in step 7:

using r3= -1/7r3, i obtained 1 in row 3, column 3

[1 0 4][0   -1    1]
[0 1 4][0   -1    3]
[0 0 1][2/7 -2/7 3/7]

(-1/7)(-1)=(1/7) not (2/7) as you have in row3 col4

Math Prof