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Determinants of 4x4 Matrices


Date: 12/18/96 at 22:44:44
From: Abigal Orange
Subject: Determinants

Dear Dr. Math

Is there a way to solve 4 by 4 determinants other than taking up an 
entire page or using a graphing calculator?


Date: 12/19/96 at 03:27:42
From: Doctor Pete
Subject: Re: Determinants

The most efficient way to evaluate a 4 x 4 determinant is to use row
reduction to create zeros in a row or column, and then to use 
expansion by minors along that row/column.  

For example, let A be the matrix:

     3   2  -1   4
     2   1   5   7
     0   5   2  -6
    -1   2   1   0

Then what we would like to do is reduce rows or columns so that one
row/column has as many zeros in it as possible.  Remember that 
interchanging two rows or columns will negate det(A), as will negating 
any row or column of entries.  Multiplying a row or column by a 
constant c also multiplies det(A) by c.  Finally, adding a constant 
multiple of a row or column to another row or column will not affect 
det(A).

Looking at the above matrix, we notice that by using this last rule we 
can get the first column to be:

     0
     0
     0
    -1

We add 3 times row 4 to row 1, which I will write as R1 --> 3*R4 + R1.  
This changes row 1 to:

     0   8   2   4

and leaves everything else unchanged.  Then we add 2 times row 4 to 
row 2 (R2 --> 2R4 + R2), so this changes row 2 to:

     0   5   7   7

and again, everything else is unchanged.  Our new matrix is:

     0   8   2   4
     0   5   7   7
     0   5   2  -6
    -1   2   1   0.

This matrix has the same determinant as A.  Expanding by minors along 
the first column, we clearly see that the first three terms in column 
1 will contribute 0 to the determinant, and so we have:

     det(A) = -(-1) det B = det(B)

where B is the 3 x 3 determinant:

     8   2   4
     5   7   7
     5   2  -6.

(Notice that since the -1 appears in the 4th row of column 1, it has a
negative sign in front of it in det(A)).  Then det(B) is easily 
calculated to be:

     det(B) = 8*7*(-6) + 2*7*5 + 4*5*2 - 5*7*4 - 2*7*8 - 5*2*(-6)
            = -418.

In general, you will have to exercise some judgment to determine 
what rows or columns to reduce.  The idea is that for each additional 
0 you can get in a row, you eliminate the need to calculate another 
3 x 3 determinant.  Sometimes, though, it is easier to begin expanding 
by minors than to try to obtain another 0, especially if you must add 
a noninteger multiple to another row.

-Doctor Pete,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
College Linear Algebra
High School Linear Algebra

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