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/
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