Equivalent Systems of Linear Equations
What do we mean when we say that two systems of linear equations are equivalent
to one another?
Definition Two systems of linear
equations are said to be equivalent to one another
if they have the same sets of solutions.
Example The systems below are all equivalent
to one another:
3x + y - z =
0
2x + y = 0
x - y + z = 0 |
x + y + z = 0
y + 2z =
0 |
3x + 2y + z = 0
3x + y = z = 0 |
Indeed, the solutions for all of these systems may be described in the
form: (x, y, z ) = (t,
-2t, t ). That is, for any value we choose
for t, we will get a triple
that satisfies all the equations. In each of the systems.
Why are we interested in the concept of equivalent
systems? The reason for this is that systems of equations are
often presented in such a way that the solutions are not immediately apparent.
Thus in order to find the solutions of the system we need to transform
it to another system for which the solutions are easier to find - while
taking care not to add or subtract solutions along the way! In the example
above, only the second system is presented in a way that makes it convenient
to state the solution set; the first and third are much less conveniently
written.
How do we make the transformation from one
system to an equivalent system? This is done via the following
basic operations and combinations of them:
-
Multiplying an equation by a non-zero constant.
-
Adding one equation (or a non-zero multiple of it) to another.
-
Interchanging between equations.
Example Consider the following system
of 3 equations in 3 unknowns:
x +
y = 2
2x + 3y + z
= 4
x + 2y + 2z =
6 |
Our goal is to apply the operations given above to transform this system
into an equivalent system from which
it is easy to find the solutions. We now do this step by step.
-
Subtract 2*(Row1) from Row2 and place the result in the second row; subtract
Row1 from Row2 and place in the third row. Leave Row1 as is.
x +
y = 2
y + z = 0
y + 2z = 4 |
-
Subtract Row2 from Row3, and place the result in row3. Leave Row1
and Row2 as they are.
x +
y = 2
y + z = 0
z = 4 |
From the last form of the system we can deduce the following unique
solution to the system:
z = 4, y = -4, and x = 2-(-4) = 6
Equivalently, we say that the unique solution to this system is (x,
y, z) = (6, -4, 4).