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Solving a System of Simultaneous Linear Equations

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We solve these problems by first forming a matrix in which the ij-th entry is the jth coefficient of the ith equation and the last column contains the constant terms on the right side of each equation. Thus, for problem 1, the matrix is

1 1 0 1 3
0 1 1 0 4
1 2 1 1 8

Next we use Gauss-Jordan elimination on this matrix. First note that the leading coefficient of 1 in the third column can be elimination by subtracting the first row from the third, yielding

1 1 0 1 3
0 1 1 0 4
0 1 1 0 5

Next we subtract the second row from the third, yielding

1 1 0 1 3
0 1 1 0 4
0 0 0 0 1

Translating back into equations, we have the following:

x1 + x2 + x4 = 3
x2 + x3 = 4
0 = 1

Since the third equation has no solutions, the original system of equations has no solutions.

For problem 2, we start with the following matrix:

1 1 1 1 4
1 2 0 0 6

We start by subtracting the first row from the second, yielding

1 1 1 1 4
0 1 -1 -1 2

Next we subtract the second row from the first to eliminate the second 1 in this row. This yields

1 0 2 2 2
0 1 -1 -1 2

Now the matrix is completely reduced so we can proceed no further. The resulting equations are as follows:

x1 + 2 x3 + 2 x4 = 2
x2 - x3 - x4 = 2

Solving for x1 and x2 in terms of x3 and ...

Solution Summary

We illustrate with several examples how Gauss-Jordan elimination may be used to solve an arbitrary system of simultaneous linear equations.

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