What if a linear system has more equations than variables?

Practical problems in many fields of study—such as biology, business, chemistry, computer science, economics, electronics, engineering, physics and the social sciences—can often be reduced to solving a system of linear equations. Linear algebra arose from attempts to find systematic methods for solving these systems, so it is natural to begin this book by studying linear equations.

If , , and are real numbers, the graph of an equation of the form

is a straight line (if and are not both zero), so such an equation is called a linear equation in the variables and . However, it is often convenient to write the variables as

is called a linear equation in the variables

is a linear equation; the coefficients of

Given a linear equation

that is, if the equation is satisfied when the substitutions

A system may have no solution at all, or it may have a unique solution, or it may have an infinite family of solutions. For instance, the system

Show that, for arbitrary values of and ,

is a solution to the system

Simply substitute these values of

Because both equations are satisfied, it is a solution for all choices of and .

The quantities and in this example are called parameters, and the set of solutions, described in this way, is said to be given in parametric form and is called the general solution to the system. It turns out that the solutions to every system of equations (if there are solutions) can be given in parametric form (that is, the variables

When only two variables are involved, the solutions to systems of linear equations can be described geometrically because the graph of a linear equation

In particular, if the system consists of just one equation, there must be infinitely many solutions because there are infinitely many points on a line. If the system has two equations, there are three possibilities for the corresponding straight lines:

• The lines intersect at a single point. Then the system has a unique solution corresponding to that point.
• The lines are parallel (and distinct) and so do not intersect. Then the system has no solution.
• The lines are identical. Then the system has infinitely many solutions—one for each point on the (common) line.

With three variables, the graph of an equation

Before describing the method, we introduce a concept that simplifies the computations involved. Consider the following system

of three equations in four variables. The array of numbers

occurring in the system is called the augmented matrix of the system. Each row of the matrix consists of the coefficients of the variables (in order) from the corresponding equation, together with the constant term. For clarity, the constants are separated by a vertical line. The augmented matrix is just a different way of describing the system of equations. The array of coefficients of the variables

is called the coefficient matrix of the system and

Elementary Operations

The algebraic method for solving systems of linear equations is described as follows. Two such systems are said to be equivalent if they have the same set of solutions. A system is solved by writing a series of systems, one after the other, each equivalent to the previous system. Each of these systems has the same set of solutions as the original one; the aim is to end up with a system that is easy to solve. Each system in the series is obtained from the preceding system by a simple manipulation chosen so that it does not change the set of solutions.

As an illustration, we solve the system

First, subtract twice the first equation from the second. The resulting system is

which is equivalent to the original. At this stage we obtain

Finally, we subtract twice the second equation from the first to get another equivalent system.

Now this system is easy to solve! And because it is equivalent to the original system, it provides the solution to that system.

Observe that, at each stage, a certain operation is performed on the system (and thus on the augmented matrix) to produce an equivalent system.

The following operations, called elementary operations, can routinely be performed on systems of linear equations to produce equivalent systems.

1. Interchange two equations.
2.  Multiply one equation by a nonzero number.
3. Add a multiple of one equation to a different equation.

Suppose that a sequence of elementary operations is performed on a system of linear equations. Then the resulting system has the same set of solutions as the original, so the two systems are equivalent.

Elementary operations performed on a system of equations produce corresponding manipulations of the rows of the augmented matrix. Thus, multiplying a row of a matrix by a number means multiplying every entry of the row by . Adding one row to another row means adding each entry of that row to the corresponding entry of the other row. Subtracting two rows is done similarly. Note that we regard two rows as equal when corresponding entries are the same.

In hand calculations (and in computer programs) we manipulate the rows of the augmented matrix rather than the equations. For this reason we restate these elementary operations for matrices.

The following are called elementary row operations on a matrix.

1. Interchange two rows.
2. Multiply one row by a nonzero number.
3. Add a multiple of one row to a different row.

In the illustration above, a series of such operations led to a matrix of the form

where the asterisks represent arbitrary numbers. In the case of three equations in three variables, the goal is to produce a matrix of the form

This does not always happen, as we will see in the next section. Here is an example in which it does happen.

Solution:
The augmented matrix of the original system is

To create a in the upper left corner we could multiply row 1 through by . However, the can be obtained without introducing fractions by subtracting row 2 from row 1. The result is

The upper left is now used to “clean up” the first column, that is create zeros in the other positions in that column. First subtract times row 1 from row 2 to obtain

Next subtract times row 1 from row 3. The result is

This completes the work on column 1. We now use the in the second position of the second row to clean up the second column by subtracting row 2 from row 1 and then adding row 2 to row 3. For convenience, both row operations are done in one step. The result is

Note that the last two manipulations did not affect the first column (the second row has a zero there), so our previous effort there has not been undermined. Finally we clean up the third column. Begin by multiplying row 3 by

Now subtract times row 3 from row 1, and then add times row 3 to row 2 to get

The corresponding equations are

The algebraic method introduced in the preceding section can be summarized as follows: Given a system of linear equations, use a sequence of elementary row operations to carry the augmented matrix to a “nice” matrix (meaning that the corresponding equations are easy to solve). In Example 1.1.3, this nice matrix took the form

The following definitions identify the nice matrices that arise in this process.

A matrix is said to be in row-echelon form (and will be called a row-echelon matrix if it satisfies the following three conditions:

1. All zero rows (consisting entirely of zeros) are at the bottom.
2. The first nonzero entry from the left in each nonzero row is a , called the leading for that row.
3. Each leading is to the right of all leading s in the rows above it.

A row-echelon matrix is said to be in reduced row-echelon form (and will be called a reduced row-echelon matrix if, in addition, it satisfies the following condition:

4.     Each leading is the only nonzero entry in its column.

The row-echelon matrices have a “staircase” form, as indicated by the following example (the asterisks indicate arbitrary numbers).

The leading s proceed “down and to the right” through the matrix. Entries above and to the right of the leading s are arbitrary, but all entries below and to the left of them are zero. Hence, a matrix in row-echelon form is in reduced form if, in addition, the entries directly above each leading are all zero. Note that a matrix in row-echelon form can, with a few more row operations, be carried to reduced form (use row operations to create zeros above each leading one in succession, beginning from the right).

The importance of row-echelon matrices comes from the following theorem.

Every matrix can be brought to (reduced) row-echelon form by a sequence of elementary row operations.

In fact we can give a step-by-step procedure for actually finding a row-echelon matrix. Observe that while there are many sequences of row operations that will bring a matrix to row-echelon form, the one we use is systematic and is easy to program on a computer. Note that the algorithm deals with matrices in general, possibly with columns of zeros.

Step 1. If the matrix consists entirely of zeros, stop—it is already in row-echelon form.

Step 2. Otherwise, find the first column from the left containing a nonzero entry (call it ), and move the row containing that entry to the top position.

Step 3. Now multiply the new top row by

Step 4. By subtracting multiples of that row from rows below it, make each entry below the leading zero. This completes the first row, and all further row operations are carried out on the remaining rows.

Step 5. Repeat steps 1–4 on the matrix consisting of the remaining rows.

The process stops when either no rows remain at step 5 or the remaining rows consist entirely of zeros.

Observe that the gaussian algorithm is recursive: When the first leading has been obtained, the procedure is repeated on the remaining rows of the matrix. This makes the algorithm easy to use on a computer. Note that the solution to Example 1.1.3 did not use the gaussian algorithm as written because the first leading was not created by dividing row 1 by . The reason for this is that it avoids fractions. However, the general pattern is clear: Create the leading s from left to right, using each of them in turn to create zeros below it. Here is one example.

Solution:

The corresponding augmented matrix is

Create the first leading one by interchanging rows 1 and 2

Now subtract times row 1 from row 2, and subtract times row 1 from row 3. The result is

Now subtract row 2 from row 3 to obtain

This means that the following reduced system of equations

is equivalent to the original system. In other words, the two have the same solutions. But this last system clearly has no solution (the last equation requires that , and satisfy

To solve a linear system, the augmented matrix is carried to reduced row-echelon form, and the variables corresponding to the leading ones are called leading variables. Because the matrix is in reduced form, each leading variable occurs in exactly one equation, so that equation can be solved to give a formula for the leading variable in terms of the nonleading variables. It is customary to call the nonleading variables “free” variables, and to label them by new variables

To solve a system of linear equations proceed as follows:

1.  Carry the augmented matrix\index{augmented matrix}\index{matrix!augmented matrix} to a reduced row-echelon matrix using elementary row operations.
2.  If a row
   
   occurs, the system is inconsistent.
3.  Otherwise, assign the nonleading variables (if any) as parameters, and use the equations corresponding to the reduced row-echelon matrix to solve for the leading variables in terms of the parameters.

There is a variant of this procedure, wherein the augmented matrix is carried only to row-echelon form. The nonleading variables are assigned as parameters as before. Then the last equation (corresponding to the row-echelon form) is used to solve for the last leading variable in terms of the parameters. This last leading variable is then substituted into all the preceding equations. Then, the second last equation yields the second last leading variable, which is also substituted back. The process continues to give the general solution. This procedure is called back-substitution. This procedure can be shown to be numerically more efficient and so is important when solving very large systems.

Rank

It can be proven that the reduced row-echelon form of a matrix

The rank of matrix

Compute the rank of

Solution:

The reduction of

Because this row-echelon matrix has two leading s, rank

Suppose that rank

Suppose a system of equations in variables is consistent, and that the rank of the augmented matrix is .

1. The set of solutions involves exactly parameters.
2. If
   
   , the system has infinitely many solutions.
3. If , the system has a unique solution.

Proof:

The fact that the rank of the augmented matrix is means there are exactly leading variables, and hence exactly nonleading variables. These nonleading variables are all assigned as parameters in the gaussian algorithm, so the set of solutions involves exactly parameters. Hence if

Theorem 1.2.2 shows that, for any system of linear equations, exactly three possibilities exist:

1. No solution. This occurs when a row
   
   occurs in the row-echelon form. This is the case where the system is inconsistent.
2.  Unique solution. This occurs when every variable is a leading variable.
3.  Infinitely many solutions. This occurs when the system is consistent and there is at least one nonleading variable, so at least one parameter is involved.

 https://www.geogebra.org/m/cwQ9uYCZ Please answer these questions after you open the webpage:

1.  For the given linear system, what does each one of them represent?