If you're seeing this message, it means we're having trouble loading external resources on our website. Show
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. 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,, andare real numbers, the graph of an equation of the form is a straight line (ifandare not both zero), so such an equation is called a linear equation in the variablesand. However, it is often convenient to write the variables as, particularly when more than two variables are involved. An equation of the form is called a linear equation in thevariables. Heredenote real numbers (called the coefficients of, respectively) andis also a number (called the constant term of the equation). A finite collection of linear equations in the variablesis called a system of linear equations in these variables. Hence, is a linear equation; the coefficients of,, andare,, and, and the constant term is. Note that each variable in a linear equation occurs to the first power only. Given a linear equation, a sequenceofnumbers is called a solution to the equation if that is, if the equation is satisfied when the substitutionsare made. A sequence of numbers is called a solution to a system of equations if it is a solution to every equation in the system. 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,has no solution because the sum of two numbers cannot be 2 and 3 simultaneously. A system that has no solution is called inconsistent; a system with at least one solution is called consistent. Show that, for arbitrary values ofand,
 is a solution to the system
Simply substitute these values of,,, andin each equation.
Because both equations are satisfied, it is a solution for all choices ofand. The quantitiesandin 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,,are given in terms of new independent variables,, etc.). When only two variables are involved, the solutions to systems of linear equations can be described geometrically because the graph of a linear equationis a straight line ifandare not both zero. Moreover, a pointwith coordinatesandlies on the line if and only if—that is when,is a solution to the equation. Hence the solutions to a system of linear equations correspond to the pointsthat lie on all the lines in question. 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:
With three variables, the graph of an equationcan be shown to be a plane and so again provides a “picture” of the set of solutions. However, this graphical method has its limitations: When more than three variables are involved, no physical image of the graphs (called hyperplanes) is possible. It is necessary to turn to a more “algebraic” method of solution. 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 OperationsThe 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,in this manner. At each stage, the corresponding augmented matrix is displayed. The original system is
First, subtract twice the first equation from the second. The resulting system is
which is equivalent to the original. At this stage we obtainby multiplying the second equation by. The result is the equivalent system
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.
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 numbermeans 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.
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:
To create ain the upper left corner we could multiply row 1 through by. However, thecan be obtained without introducing fractions by subtracting row 2 from row 1. The result is
The upper leftis now used to “clean up” the first column, that is create zeros in the other positions in that column. First subtracttimes row 1 from row 2 to obtain
Next subtracttimes row 1 from row 3. The result is
This completes the work on column 1. We now use thein 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 byto obtain
Now subtracttimes row 3 from row 1, and then addtimes row 3 to row 2 to get
The corresponding equations are,, and, which give the (unique) solution. 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:
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 leadingis 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 leadings proceed “down and to the right” through the matrix. Entries above and to the right of the leadings 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 leadingare 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 byto create a leading. Step 4. By subtracting multiples of that row from rows below it, make each entry below the leadingzero. 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 leadinghas 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 leadingwas not created by dividing row 1 by. The reason for this is that it avoids fractions. However, the general pattern is clear: Create the leadings 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 subtracttimes row 1 from row 2, and subtracttimes 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,andsatisfy, and no such numbers exist). Hence the original system has no solution. 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, called parameters. Every choice of these parameters leads to a solution to the system, and every solution arises in this way. This procedure works in general, and has come to be called To solve a system of linear equations proceed as follows:
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. RankIt can be proven that the reduced row-echelon form of a matrixis uniquely determined by. That is, no matter which series of row operations is used to carryto a reduced row-echelon matrix, the result will always be the same matrix. By contrast, this is not true for row-echelon matrices: Different series of row operations can carry the same matrixto different row-echelon matrices. Indeed, the matrix The rank of matrixis the number of leadings in any row-echelon matrix to whichcan be carried by row operations. Compute the rank of Solution: The reduction ofto row-echelon form is
Because this row-echelon matrix has two leadings, rank. Suppose that rank, whereis a matrix withrows andcolumns. Thenbecause the leadings lie in different rows, andbecause the leadings lie in different columns. Moreover, the rank has a useful application to equations. Recall that a system of linear equations is called consistent if it has at least one solution. Suppose a system ofequations invariables is consistent, and that the rank of the augmented matrix is.
Proof: The fact that the rank of the augmented matrix ismeans there are exactlyleading variables, and hence exactlynonleading variables. These nonleading variables are all assigned as parameters in the gaussian algorithm, so the set of solutions involves exactlyparameters. Hence if, there is at least one parameter, and so infinitely many solutions. If, there are no parameters and so a unique solution. Theorem 1.2.2 shows that, for any system of linear equations, exactly three possibilities exist:
https://www.geogebra.org/m/cwQ9uYCZ 2. Based on the graph, what can we say about the solutions? Does the system have one solution, no solution or infinitely many solutions? Why 3. Change the constant term in every equation to 0, what changed in the graph? 4. For the following linear system: Can you solve it using Gaussian elimination? When you look at the graph, what do you observe? Many important problems involve linear inequalities rather than linear equations For example, a condition on the variablesandmight take the form of an inequalityrather than an equality. There is a technique (called the simplex algorithm) for finding solutions to a system of such inequalities that maximizes a function of the formwhereandare fixed constants. A system of equations in the variablesis called homogeneous if all the constant terms are zero—that is, if each equation of the system has the form Clearlyis a solution to such a system; it is called the trivial solution. Any solution in which at least one variable has a nonzero value is called a nontrivial solution. Show that the following homogeneous system has nontrivial solutions.
Solution: The reduction of the augmented matrix to reduced row-echelon form is outlined below.
The leading variables are,, and, sois assigned as a parameter—say. Then the general solution is,,,. Hence, taking(say), we get a nontrivial solution:,,,. The existence of a nontrivial solution in Example 1.3.1 is ensured by the presence of a parameter in the solution. This is due to the fact that there is a nonleading variable (in this case). But there must be a nonleading variable here because there are four variables and only three equations (and hence at most three leading variables). This discussion generalizes to a proof of the following fundamental theorem. If a homogeneous system of linear equations has more variables than equations, then it has a nontrivial solution (in fact, infinitely many). Proof: Suppose there areequations invariables where, and letdenote the reduced row-echelon form of the augmented matrix. If there areleading variables, there arenonleading variables, and soparameters. Hence, it suffices to show that. Butbecausehasleading 1s androws, andby hypothesis. So, which gives. Note that the converse of Theorem 1.3.1 is not true: if a homogeneous system has nontrivial solutions, it need not have more variables than equations (the system,has nontrivial solutions but.) Theorem 1.3.1 is very useful in applications. The next example provides an illustration from geometry. We call the graph of an equationa conic if the numbers,, andare not all zero. Show that there is at least one conic through any five points in the plane that are not all on a line. Solution: Let the coordinates of the five points be,,,, and. The graph ofpasses throughif This gives five equations, one for each, linear in the six variables,,,,, and. Hence, there is a nontrivial solution by Theorem 1.1.3. If, the five points all lie on the line with equation, contrary to assumption. Hence, one of,,is nonzero. Linear Combinations and Basic SolutionsAs for rows, two columns are regarded as equal if they have the same number of entries and corresponding entries are the same. Letandbe columns with the same number of entries. As for elementary row operations, their sumis obtained by adding corresponding entries and, ifis a number, the scalar productis defined by multiplying each entry ofby. More precisely:
A sum of scalar multiples of several columns is called a linear combination of these columns. For example,is a linear combination ofandfor any choice of numbersand. Let and and determine whetherandare linear combinations of,and. Solution: For, we must determine whether numbers,, andexist such that, that is, whether
Equating corresponding entries gives a system of linear equations,, andfor,, and. By gaussian elimination, the solution is,, andwhereis a parameter. Taking, we see thatis a linear combination of,, and. Turning to, we again look for,, andsuch that; that is,
leading to equations,, andfor real numbers,, and. But this time there is no solution as the reader can verify, sois not a linear combination of,, and. Our interest in linear combinations comes from the fact that they provide one of the best ways to describe the general solution of a homogeneous system of linear equations. When Example 1.3.1 is,,, and, whereis a parameter, and we would now express this by saying that the general solution is Now letandbe two solutions to a homogeneous system withvariables. Then any linear combinationof these solutions turns out to be again a solution to the system. More generally: In fact, suppose that a typical equation in the system is, and suppose that . Hence
A similar argument shows that Statement 1.1 is true for linear combinations of more than two solutions. The remarkable thing is that every solution to a homogeneous system is a linear combination of certain particular solutions and, in fact, these solutions are easily computed using the gaussian algorithm. Here is an example. Solve the homogeneous system with coefficient matrix
Solution: The reduction of the augmented matrix to reduced form is
so the solutions are,,, andby gaussian elimination. Hence we can write the general solutionin the matrix form
Here The solutionsandin Example 1.3.5 are denoted as follows: The gaussian algorithm systematically produces solutions to any homogeneous linear system, called basic solutions, one for every parameter. Moreover, the algorithm gives a routine way to express every solution as a linear combination of basic solutions as in Example 1.3.5, where the general solutionbecomes
Hence by introducing a new parameterwe can multiply the original basic solutionby 5 and so eliminate fractions. For this reason: Any nonzero scalar multiple of a basic solution will still be called a basic solution. In the same way, the gaussian algorithm produces basic solutions to every homogeneous system, one for each parameter (there are no basic solutions if the system has only the trivial solution). Moreover every solution is given by the algorithm as a linear combination of Letbe anmatrix of rank, and consider the homogeneous system invariables withas coefficient matrix. Then: Is 5x 3y 5 a linear equation in one variable?No, the given equation 5x−3y=5 is a linear equation in two variable having x and y as the two variables. Was this answer helpful?
For what value of k will the system of equations 2x 3y 5 and 4x KY 10 have infinite number of solutions?∴ The value of k is 6.
Which of the following pairs of linear equations are consistent or inconsistent 3x 2y 5 2x 3y 7?3x + 2y = 5; 2x - 3y = 7. Hence, the given lines are intersecting. So, the given pair of linear equations has exactly one solution and therefore it is consistent.
What is the solution of 5x 3y 5?Answer: first answer is x=0 and y= -5/3. second answer is x=1 and y=0.
|
