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Linear Equations and Matrices


Date: 07/22/98 at 11:50:53
From: Kimberly Lawson
Subject: Algebra II

I'm having problems solving this system of linear equations using 
matrices. Can you help?

    a  -2b    c   -d   -e =  4
   2a  -3b         d  -2e = -3
    a   -b  -2c   -d    e =  3
         b   -c         e =  0
    a    b        -d   -e = -1


Date: 07/22/98 at 14:26:02
From: Doctor Rob
Subject: Re: Algebra II

Put the coefficients of a from the five equations in the first column 
of a matrix, the coefficients of b in the second column, the 
coefficients of c in the third column, those of d in the fourth, and 
those of e in the fifth. In the sixth column, put the constants on the 
righthand side of each equation. That gives you:

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

To find the solutions, we need to row reduce this matrix. To do this, 
start by picking a nonzero entry in the first column. Subtract 
multiples of the row in which it appears from the other rows to make 
all the other entries in that column zero. 

I pick the 1 in the first column, fifth row. I subtract 1 times it from 
row 1, 2 times it from row 2, and 1 times it from row 3. That gives me:

   [ 0 -3  1  0  0  5 ]
   [ 0 -5  0  3  0 -1 ]
   [ 0 -2 -2  0  2  4 ]
   [ 0  1 -1  0  1  0 ]
   [ 1  1  0 -1 -1 -1 ]

Now look in the second column, and pick a nonzero entry in a row not 
already used. I pick the 1 in the fourth row. Subtract multiples of 
the row in which it appears from the other rows to make all other 
entries in that column zero. I get:

   [ 0  0 -2  0  3  5 ]
   [ 0  0 -5  3  5 -1 ]
   [ 0  0 -4  0  4  4 ]
   [ 0  1 -1  0  1  0 ]
   [ 1  0  1 -1 -2 -1 ]

Another thing you can do is multiply or divide a row by any nonzero 
number. I divide the third row by -4. Now move on to the third column, 
and do the same. I pick the third row:

   [ 0  0  0  0  1  3 ]
   [ 0  0  0  3  0 -6 ]
   [ 0  0  1  0 -1 -1 ]
   [ 0  1  0  0  0 -1 ]
   [ 1  0  0 -1 -1  0 ]

Now I divide the second row by 3, and I pick the second row next. Then:

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

Finally, I pick the first row to work with to get all the other entries 
in the fifth column to be zero:

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

Finally, we rearrange the rows so that the "1" elements in the first
five columns are in the same row as column:

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

Now the answers are found in the sixth column:

   a =  1
   b = -1
   c =  2
   d = -2
   e =  3

You can check that they satisfy the given system of equations.

You can do all problems of this type in this way, using three kinds of
operations: rearranging the rows, multiplying or dividing a row by a
constant, and adding a multiple of one row to another. Once the first
n columns of the n-by-(n+1) matrix each have one 1 and all the rest of 
the entries zero, and they are arranged in order so that those 1's are 
on the diagonal, the last column will contain the answers.

The three operations correspond to operation on the system of equations 
with these three operations: rearranging the equations, multiplying or 
dividing both sides of an equation by a constant, or adding a multiple 
of one equation to another.

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

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