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Home : Math & Science : Math Study Guides : Algebra II : Systems of Three Equations : Solving using Matrices and Row Reduction
Solving using Matrices and Row Reduction
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:
a1x + b1y + c1z = d1
where a1, 2, 3, b1, 2, 3, and c1, 2, 3 are the x, y,
and z coefficients, respectively, and d1, 2, 3 are constants.
a2x + b2y + c2z = d2 a3x + b3y + c3z = d3
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:
5x + 3y = 2z - 4
First, arrange the equations:2x + 2z + 2y = 0 3x + 2y + z = 1
5x + 3y - 2z = - 4
Next, form the 3×4 matrix:2x + 2y + 2z = 0 3x + 2y + 1z = 1
Finally, row reduce the matrix:
Thus, (x, y, z) = (3, - 5, 2). |
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