Solving 3x3 Matrices Manually

Date: 01/27/98 at 22:53:04
From: Miranda
Subject: 3X3 Matrices - Solving Manually

I need to figure out how to find the inverse of a 3x3 matrix manually 
(without pushing the inverse button on my calculator). How do I find 
this kind of info? I know how to find the determinant and inverse of a 
2x2 matrix, but the 3x3 has got me stumped.

Date: 01/31/98 at 10:47:23
From: Doctor Luis
Subject: Re: 3X3 Matrices - Solving Manually

First you need to find the determinant of the 3x3 matrix. You know how 
to do that, right? 

If the determinant is zero, you know your matrix is singular (it has 
no inverse). If the determinant is one (unitary matrix), the inverse
of your matrix will be what is called the "adjoint of A" (denoted by 
adj(A)). Note that the adjoint of an NxN matrix is also an N x N 
matrix.

 In general, the inverse of an  N x N matrix A is

                1
    A^(-1) = -------- * adj(A)
              det(A)

 
A little explanation of the formula is required. You are familiar with 
the determinant: you compute it fairly easily (although it is 
computationally intensive when N gets a little large). Say you have 
the following matrix:

           [ 1  0  1 ]
           [         ]
      A =  [ 0  2  1 ]  =  ( a(i,j) )      i,j = 1,2,3
           [         ]
           [ 1  1  1 ]

and you want to evaluate the determinant of A. You choose to expand 
the determinant by minors along the top row

                  [ 2  1 ]            [ 0  1 ]            [ 0  2 ] 
 det A  =  1*det( [      ] ) - 0*det( [      ] ) + 1*det( [      ] )
                  [ 1  1 ]            [ 1  1 ]            [ 1  1 ]

        = 1*(2*1-1*1) - 0*(0*1-1*1) + 1*(0*1-1*2)

        = 1 - 0 - 2

 det(A) = -1

There are two things worth mentioning. First, note that the elements 
of the top row (which we used to evaluate the determinant) are:

   a(1,1) = 1   (first row, first column)
   a(1,2) = 0   (first row, second col)
   a(1,3) = 1   (first row, third col) 

Second, notice the determinants that we multiplied these elements by. 
They are called cofactors of A. They are:

                [ 2  1 ]
  C(1,1) = det( [      ] )   -- cofactor of a(1,1)
                [ 1  1 ]


                [ 0  1 ]
  C(1,2) = det( [      ] )  -- cofactor of a(1,2)
                [ 1  1 ]


                [ 0  2 ]
  C(1,3) = det( [      ] )  -- cofactor of a(1,3)
                [ 1  1 ]

So you can see that there are nine cofactors for our 3x3 matrix, one 
for each element. Essentially, to obtain the cofactor C(i,j) of the 
ith row, and jth column, element a(i,j), you cross out the ith row and 
the jth column of your matrix A, and take the determinant of the 
smaller matrix (minor) that's left.

Notice that these cofactors are useful in evaluating the determinant 
of a matrix, since you can go along any row or column, multiply each 
a(i,j) by C(i,j) (just along the row or column you chose of course), 
and add all those together with the corresponding sign, as determined 
by the following:


  [ +  -  + ]
  [         ]
  [ -  +  - ]   (notice that the sign is also determined by (-1)^(i+j) )
  [         ]
  [ +  -  + ]


Now you know how to evaluate determinants. What about the adjoint 
matrix? Well, the elements of the adjoint matrix (of A) are nothing 
more than the cofactors of A transposed!


       [ a(1,1) a(1,2) a(1,3) ]
       [                      ]
   A = [ a(2,1) a(2,2) a(2,3) ]
       [                      ]
       [ a(3,1) a(3,2) a(3,3) ]

 so that,

          [ C(1,1) C(2,1) C(3,1) ]
          [                      ]
 adj(A) = [ C(1,2) C(2,2) C(3,2) ]
          [                      ]
          [ C(1,3) C(2,3) C(3,3) ]
 

where C(i,j) is the cofactor of a(i,j).

Now, for our original matrix,


           [ 1  0  1 ]
           [         ]
      A =  [ 0  2  1 ] 
           [         ]
           [ 1  1  1 ]


you should verify that,


           [ 1  1 -2 ]
           [         ]
 adj(A) =  [ 1  0 -1 ] 
           [         ]
           [-2 -1  2 ]
 

To get the inverse, you just divide by the determinant!

det(A) = -1, remember? so,


                             1
 inverse of A = A^(-1) = --------- * adj(A)
                          det(A)

    
                       [ 1  1 -2 ]
                   1   [         ]
              =  --- * [ 1  0 -1 ] 
                  -1   [         ]
                       [-2 -1  2 ]


                 [-1 -1  2 ]
                 [         ]
              =  [-1  0  1 ] 
                 [         ]
                 [ 2  1 -2 ]
 

That's essentially it. Of course this holds too for square matrices of 
higher rank (N x N matrices), not just 3x3's.

There are other methods of finding the inverse matrix, like augmenting 
the matrix by the identity matrix and then trying to make the original 
matrix into the identity matrix by applying row and column operations 
to the augmented matrix, and so on.

If you want to read more on the properties of matrices, including the 
proof of the theorem we used here to find the inverse, take a look at 
Dr. Mark V. Sapir's Linear Algebra Webnotes:

 http://www.math.vanderbilt.edu/~msapir/M314HNotes/frame.html   

--Doctor Luis,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: 05/08/00 at 05:42:32
From: Anshuman Mukherjee
Subject: Re: 3X3 Matrices - Solving Manually

Quoting from above, "the elements of the adjoint matrix (of A) 
are nothing more than the cofactors of A."

As far as I know, adjoint matrix is the TRANSPOSE of the matrix whose 
elements are replaced by its cofactors. 

The answer above, however, happens to be correct, since adj(A) is 
symmetric.

Regards,
Anshuman

Date: 05/08/00 at 14:52:45
From: Dr. Rob
Subject: Re: 3X3 Matrices - Solving Manually

Thanks for writing to Ask Dr. Math, Anshuman.

Your definition is correct, and more complete than the one in the
answer above. That one does not specify where in the adjoint the 
cofactors are to be found, whereas yours does. That one merely says that 
the entries (or elements) of the matrix are the cofactors. This is 
correct, but not complete.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/