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Solving using Matrices and Row Reduction

Systems with three equations and three variables can also be solved
using matrices and row
reduction. First, arrange the system in
the following form:

*a*_{1}*x* + *b*_{1}*y* + *c*_{1}*z* = *d*_{1}

*a*_{2}*x* + *b*_{2}*y* + *c*_{2}*z* = *d*_{2}

*a*_{3}*x* + *b*_{3}*y* + *c*_{3}*z* = *d*_{3}

where

*a*_{1, 2, 3}, *b*_{1, 2, 3}, and

*c*_{1, 2, 3} are the

*x*,

*y*,
and

*z* coefficients, respectively, and

*d*_{1, 2, 3} are constants.

Next, create a 3×4 matrix, placing the *x* coefficients in the
1st column, the *y* coefficients in the 2nd column, the *z* coefficients
in the 3rd column, and the constants in the 4th column, with a line
separating the 3rd column and the 4th column:

This is equivalent to writing

which is equivalent to the original three
equations (check the multiplication yourself).

Finally, row reduce the 3×4
matrix using the elementary row
operations. The result should be
the identity matrix on the left side of the line and a column of
constants on the right side of the line. These constants are the
solution to the system of equations:

*Note:* If the system row reduces to

then the system is
inconsistent.
If the system row reduced to

then the system has multiple solutions.

*Example*: Solve the following system:

5*x* + 3*y* = 2*z* - 4

2*x* + 2*z* + 2*y* = 0

3*x* + 2*y* + *z* = 1

First, arrange the equations:

5*x* + 3*y* - 2*z* = - 4

2*x* + 2*y* + 2*z* = 0

3*x* + 2*y* + 1*z* = 1

Next, form the

3×4 matrix:

Finally, row reduce the matrix:

Thus,

(*x*, *y*, *z*) = (3, - 5, 2).