# Systems of Linear Equations

Given a system of equations

we wish to find all solutions. The equations are said to be inhomogeneous
since the constants on the

right hand side are not 0. Associated with this system is a homogeneous system
( called the reduced

homogeneous system of the original system) given by

In matrix notion the inhomogeneous and the homogeneous systems are
represented by A**x = b** and

A**x = 0** respectively.

We note that set of solutions of the homogeneous system has a special form.

THEOREM. The set of solutions of the homogeneous system of equations A**x = 0**
is a subspace of

REMARK. The set of solutions of A**x = 0** is called the null space of A
or the kernel of A and is

denoted by Null A or ker A

PROOF. If y and z are in K = Null A, we must show that ay + bz is also in K
in order to show that K

is a subspace of
.
We just test this out by the defining property of K , viz., that A annihilates
each

vector of K. Here, to test this out we use some properties of matrix
multiplication . We have that

A(ay + bz) = aA(y) + bA(z) = a0 + b0 = 0.

So ay + bz is in K and thus K is a subspace. Q.E.D.

The theorem governing the relation of an inhomogeneous system and its
associated homogeneous

systems is the following:

THEOREM. Let A**x = b** be a system of equations. Let S be the set of all
solutions of the reduced

homogeneous system A**x = 0**. Let x_{p} be one particular solution of the original
system A**x = b**. Then

is the set of all solutions of the system A**x = b**.

PROOF. First let y ∈S. Then we have that y + x_{p}
is a solution of A**x = b** since

Now let z be a solution of A**x = b**. Then we have that

So we have that y = z - x_{p} is a solution of A**x = 0** and thus y
∈S. This means that z has the desired

form z = y + x_{p}. Q.E.D.

If we can find the one particular solution x_{p} of A**x = b** by some method even
by guessing, we

would just have to produce all solutions of the associated homogeneous system Ax
= 0 and add x _{p} to

all solutions of A**x = 0** to find all solutions A**x = b**. This is the content of the
preceding theorem.

However, our method of Gaussian elimination to upper triangular form from which
the solutions are

read off produces all solutions of the inhomogeneous system at one stroke

EXAMPLE. Solve for all solutions the system of equations

SOLUTION. We write down the Gaussian steps for the augmented matrix as

We read of the solution x_{4} = 0, x_{3} = s (a variable),.

We write this in vector form as

Since we are interested in the last expression for all , we may replace s/2 by t to get

We see that

Instead of using a variable , we can generate a basis of the null space
numerically. Suppose that

the coefficient matrix A of the homogeneous system is taken to row reduced form
B. Suppose the

columns j_{1}, … , j_{p} are non pivot colunms. These columns take a variable in our
method of solution.

Instead of using a variable, we substitute alternately a single 1 and p - 1 0’s
for the coordinates, i.e., we

substitute (1, 0, 0, … , 0), (0, 1, 0, … , 0), (0, 0, 1, … , 0), … , (0, 0, 0, …
, 1) in the columns j_{1}, … , j_{p}

and read off the rest of the n - p coordinates for the pivot columns. We get p
vectors which will be a

basis for the kernel K. For example in our case, we use Gaussian elimination to
get

We then set the single nonpivot element x_{3} = 1 and read of the solutions as
x_{4} = 0, x_{3} = 1/2 and x_{1} =

5/2. So the vector v = (5/2, 1/2, 1, 0) is a basis of the Null Space of A.

EXAMPLE. Consider the system of equations

The augmented matrix is

and the solution can be read off as x_{4} = 2, x_{3} = 1 - x_{4} = -1, x_{2} = 2x_{3} + x_{4} =
0, and

x_{1} = (1/3)(x_{4} - x_{3} - 2x_{2}) = 1. Here

The solution space of the associated homogeneous system is K = {0}, and thus,
the associated

homogeneous system only has the trivial solution **0**. In general, when only the
solution **x _{p}** is showing

in the solution, the solution of the associated homogeneous solution is only the trivial solution.

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