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3 Linear Systems and Matrices

3.1 Introduction to Linear Systems

The subject of linear algebra centers around the problem of solving systems of (algebraic) linear equations. In high school algebra the method of elimination is used to solve systems of two and three (simultaneous) linear equations, with attention ordinarily confined to systems that have one and only one solution, and with applications to “word problems” that have one and only one “answer.” In this section we introduce some of the basic terminology of linear algebra by reviewing the elementary technique of elimination from a slightly more general viewpoint. In subsequent sections we apply this same technique to the solution of systems involving many linear equations in many unknowns. The applications of the method of elimination are diverse and important, because so many mathematical problems involve the solution of systems of linear equations.

Recall that if a, b, and c are constants with a and b not both zero, then the graph of the equation

a x + b y = c

$a x + b y = c$ (1)

is a (straight) line in the xy-plane. For this reason, an equation of the form in (1) is called a linear equation in the variables x and y. Similarly, an equation that can be written in the form

a x + b y + c z = d

$a x + b y + c z = d$ (2)

is called linear in the three variables x, y, and z (even though its graph in xyz-space is a plane rather than a line). Thus the equations

3 x - 2 y = 5 and x - 7 y + 5 z = - 11

$3 x - 2 y = 5 and x - 7 y + 5 z = - 11$

are linear, because they involve only the first powers of the variables. By contrast, the equations

x + y + x y = 5 and x^{2} + y^{3} + \sqrt{z} = 1

$x + y + x y = 5 and x^{2} + y^{3} + \sqrt{z} = 1$

are not linear, because they cannot be rewritten to eliminate the higher powers, roots, and products of the variables.

Two Equations in Two Unknowns

A system of linear equations (also called a linear system) is simply a finite collection of linear equations involving certain variables. Sometimes we refer to the variables as the “unknowns” in the system. Thus a system of two linear equations in two unknowns x and y may be written in the form

\begin{array}{rcl} a_{1} x + b_{1} y & = & c_{1} \\ a_{2} x + b_{2} y & = & c_{2} . \end{array}

$\begin{array}{rcl} a_{1} x + b_{1} y & = & c_{1} \\ a_{2} x + b_{2} y & = & c_{2} . \end{array}$ (3)

By a solution of the system in (3) is meant a pair (x, y) of values—normally real numbers—that satisfy both equations simultaneously.

Example 1

The values $x = 2, y = - 1$ $x = 2, y = - 1$ constitute a solution of the system

\begin{array}{rcrcl} 2 x & - & y & = & 5 \\ x & + & 2 y & = & 0 \end{array}

$\begin{array}{rcrcl} 2 x & - & y & = & 5 \\ x & + & 2 y & = & 0 \end{array}$ (4)

because both $2 (2) - (- 1) = 5$ $2 (2) - (- 1) = 5$ and $(2) + 2 (- 1) = 0 .$ $(2) + 2 (- 1) = 0 .$ The values $x = 3, y = 1$ $x = 3, y = 1$ satisfy the first equation in (4) but do not satisfy the second. Hence (3, 1) is not a solution of the system in (4).

Example 2

The linear system

\begin{array}{rcrcl} x & + & y & = & 1 \\ 2 x & + & 2 y & = & 3 \end{array}

$\begin{array}{rcrcl} x & + & y & = & 1 \\ 2 x & + & 2 y & = & 3 \end{array}$ (5)

has no solution at all, because if $x + y = 1,$ $x + y = 1,$ then $2 x + 2 y = 2,$ $2 x + 2 y = 2,$ and so $2 x + 2 y \neq 3 .$ $2 x + 2 y \neq 3 .$ Thus, two numbers that satisfy the first equation in (5) cannot simultaneously satisfy the second.

A linear system is said to be consistent if it has at least one solution and inconsistent if it has none. Thus the system in Example 1 is consistent, whereas the system in Example 2 is inconsistent.

Three Possibilities

Given a system of linear equations, we ask: What is the set of all solutions of the system? In brief, what is the solution set of the system?

In the case of a linear system

\begin{array}{rcl} a_{1} x + b_{1} y & = & c_{1} \\ a_{2} x + b_{2} y & = & c_{2} \end{array}

$\begin{array}{rcl} a_{1} x + b_{1} y & = & c_{1} \\ a_{2} x + b_{2} y & = & c_{2} \end{array}$ (6)

of two equations in two unknowns, we can use our knowledge of elementary geometry to sort out the possibilities for its solution set. If neither equation in (6) has both left-hand coefficients zero, then their graphs in the xy-plane are two straight lines $L_{1}$ $L_{1}$ and $L_{2} .$ $L_{2} .$ Then exactly one of the following situations must hold.

The lines $L_{1}$ $L_{1}$ and $L_{2}$ $L_{2}$ intersect at a single point (as in Figure 3.1.1).
The lines $L_{1}$ $L_{1}$ and $L_{2}$ $L_{2}$ are parallel nonintersecting lines (as in Figure 3.1.2).
The lines $L_{1}$ $L_{1}$ and $L_{2}$ $L_{2}$ coincide—they actually are the same line (see Figure 3.1.3).

A pair (x, y) of real numbers constitutes a solution of the system in (6) if and only if the point (x, y) in the coordinate plane lies on both the two lines $L_{1}$ $L_{1}$ and $L_{2} .$ $L_{2} .$ In the case shown in Figure 3.1.1, there is exactly one such point. In the case shown in Figure 3.1.2, there is no such point, and in the case in Figure 3.1.3, there are infinitely many such points—every point on the line $L_{1} = L_{2}$ $L_{1} = L_{2}$ is such a point. We therefore see that there are just three possibilities for a linear system of two equations in two unknowns: It has either

FIGURE 3.1.1.

Two intersecting lines: a unique solution.

FIGURE 3.1.2.

Two parallel lines: no solution.

FIGURE 3.1.3.

Two coincident lines: infinitely many solutions.

exactly one solution;
no solution; or
infinitely many solutions.

It is a fundamental fact of linear algebra (which we will establish in Section 3.3) that, however many equations and variables may appear in a linear system, precisely these same three possibilities occur: A system of m linear equations in n variables either has a unique solution (that is, exactly one solution), or it has no solution, or it has infinitely many solutions. Thus it is impossible, for instance, for a linear system to have exactly 2 solutions or to have exactly 17 solutions.

The Method of Elimination

The next three examples illustrate how we can use the elementary method of elimination to solve a system of two equations in two unknowns. To solve a system means to determine what its solution set is. The basic idea of the method is this:

First, we add an appropriate constant multiple of the first equation to the second equation. The idea is to choose the constant in such a way as to eliminate the variable x from the second equation.
Next, the new second equation contains only the variable y, so we readily solve it for the value of y.
Finally, we determine the value of x by “back substitution” of this value of y in the first equation.

Examples 3 through 5 illustrate this method in the three cases corresponding to Figures 3.1.1 through 3.1.3 (respectively).

Example 3

In order to solve the system

\begin{array}{rcl} 5 x + 3 y & = & 1 \\ x - 2 y & = & 8, \end{array}

$\begin{array}{rcl} 5 x + 3 y & = & 1 \\ x - 2 y & = & 8, \end{array}$ (7)

we first interchange the two equations:

\begin{array}{rcl} x - 2 y & = & 8 \\ 5 x + 3 y & = & 1. \end{array}

$\begin{array}{rcl} x - 2 y & = & 8 \\ 5 x + 3 y & = & 1. \end{array}$

Then we multiply the first equation by $- 5$ $- 5$ and add the resulting terms to the second (without changing the first equation). The result is

\begin{array}{rcrcr} x & - & 2 y & = & 8 \\ 13 y & = & - 39. \end{array}

$\begin{array}{rcrcr} x & - & 2 y & = & 8 \\ 13 y & = & - 39. \end{array}$ (8)

Now the second equation immediately yields the value $y = - 3,$ $y = - 3,$ and back substitution of this value in the first equation yields

x = 2 y + 8 = 2 (- 3) + 8 = 2.

$x = 2 y + 8 = 2 (- 3) + 8 = 2.$

Taking it as obvious (for the moment) that the systems in (7) and (8) have the same solution set, we conclude that the original system in (7) has the unique solution $x = 2, y = - 3$ $x = 2, y = - 3$ .

Example 4

To solve the system

\begin{array}{rcr} 2 x + 6 y & = & 4 \\ 3 x + 9 y & = & 11, \end{array}

$\begin{array}{rcr} 2 x + 6 y & = & 4 \\ 3 x + 9 y & = & 11, \end{array}$ (9)

we first multiply the first equation by $\frac{1}{2}$ $\frac{1}{2}$ and get

\begin{array}{rcr} x + 3 y & = & 2 \\ 3 x + 9 y & = & 11. \end{array}

$\begin{array}{rcr} x + 3 y & = & 2 \\ 3 x + 9 y & = & 11. \end{array}$

We next multiply the first equation by $- 3$ $- 3$ and add each term to the corresponding term in the second equation. The result is

\begin{array}{rcl} x + 3 y & = & 2 \\ 0 & = & 5. \end{array}

$\begin{array}{rcl} x + 3 y & = & 2 \\ 0 & = & 5. \end{array}$ (10)

What, however, are we to make of the new second equation, $0 = 5$ $0 = 5$ ? The system in (10) actually is

\begin{array}{rcl} x + 3 y & = & 2 \\ 0 x + 0 y & = & 5. \end{array}

$\begin{array}{rcl} x + 3 y & = & 2 \\ 0 x + 0 y & = & 5. \end{array}$

Because 0 is simply not equal to 5, there are no values of x and y that satisfy the second equation. Hence there certainly can be no values that satisfy both simultaneously. We conclude that the original system in (9) has no solution.

Example 5

If, instead of the system in (9), we had begun in Example 4 with the system

\begin{array}{rcl} 2 x + 6 y & = & 4 \\ 3 x + 9 y & = & 6 \end{array}

$\begin{array}{rcl} 2 x + 6 y & = & 4 \\ 3 x + 9 y & = & 6 \end{array}$ (11)

and performed the same operations, we would have obtained, instead of (10), the system

\begin{array}{rcl} x + 3 y & = & 2 \\ 0 & = & 0. \end{array}

$\begin{array}{rcl} x + 3 y & = & 2 \\ 0 & = & 0. \end{array}$ (12)

Here, $0 = 0$ $0 = 0$ is shorthand for the equation

0 x + 0 y = 0,

$0 x + 0 y = 0,$

which is satisfied by all values of x and y. In terms of restrictions or conditions on x and y, one of our original two equations has in effect disappeared, leaving us with the single equation

x + 3 y = 2.

$x + 3 y = 2.$ (13)

Of course this is hardly surprising, because each equation in (11) is a multiple of the one in (13); in some sense we really had only one equation to begin with. At any rate, we can substitute any value of y we please in (13) and then solve for x. Thus our system in (11) has infinitely many solutions. To describe them explicitly, let us write $y = t,$ $y = t,$ where the parameter t is a new independent variable that we will use to generate solution pairs (x, y). Then Equation (13) yields $x = 2 - 3 t,$ $x = 2 - 3 t,$ so our infinite solution set of the system in (11) may be described as follows:

x = 2 - 3 t, y = t

$x = 2 - 3 t, y = t$

as the arbitrary parameter t ranges over the set of all real numbers. For instance, $t = 2$ $t = 2$ yields the solution $(- 4, 2),$ $(- 4, 2),$ and $t = - 3$ $t = - 3$ yields the solution $(11, - 3)$ $(11, - 3)$ .

Remark

There is some latitude in how we parameterize the solutions of a system having infinitely many solutions, including the names of any parameters we introduce for this purpose. For example, we could parameterize the solutions of the system in (11) alternatively by writing $x = s$ $x = s$ in (13), thereby getting the different parameterization

x = s, y = \frac{1}{3} (2 - s)

$x = s, y = \frac{1}{3} (2 - s)$

that yields the same solutions. For instance, the parameter values $s = - 4$ $s = - 4$ and $s = 11$ $s = 11$ here give the previously noted solutions $(- 4, 2)$ $(- 4, 2)$ and $(11, - 3)$ $(11, - 3)$ of the original system in (13).

Comment

These three examples illustrate the basic features of the method of elimination, which involves “transforming” a given linear system by means of a sequence of successive steps that do not change the solutions of the system. Each of these steps consists of performing one of the following three elementary operations:

Multiply one equation by a nonzero constant.
Interchange two equations.
Add a constant multiple of (the terms of) one equation to (corresponding terms of) another equation.

In subsequent sections of this chapter we discuss the systematic use of these elementary operations to eliminate variables successively in any linear system, whatever the number of equations and variables. In this way we will see that every linear system corresponds to precisely one of the three situations illustrated in Examples 3–5. That is, either

we discover (as in Example 3) a unique solution of the system; or
we eventually arrive at an inconsistent equation (as in Example 4), so that the system has no solution; or
we find that the original system has infinitely many solutions.

Three Equations in Three Unknowns

In the case of a system of three linear equations in three variables x, y, and z, we may proceed as follows: Assuming that the first equation involves x, use the first equation to eliminate x from the second and third equations by adding appropriate multiples of the first equation. Then, assuming that the new second equation involves y, use the new second equation to eliminate y from the new third equation. Solve the new third equation for z, back-substitute in the second equation to determine y, and, finally, back-substitute y and z in the first equation to find x. The following two examples illustrate this procedure.

Example 6

Solve the linear system

\begin{array}{rcrcrcr} x & + & 2 y & + & z & = & 4 \\ 3 x & + & 8 y & + & 7 z & = & 20 \\ 2 x & + & 7 y & + & 9 z & = & 23. \end{array}

$\begin{array}{rcrcrcr} x & + & 2 y & + & z & = & 4 \\ 3 x & + & 8 y & + & 7 z & = & 20 \\ 2 x & + & 7 y & + & 9 z & = & 23. \end{array}$ (14)

Solution

First, we add $- 3$ $- 3$ times the first equation to the second equation; the result is

\begin{array}{rcrcrcr} x & + & 2 y & + & z & = & 4 \\ 2 y & + & 4 z & = & 8 \\ 2 x & + & 7 y & + & 9 z & = & 23. \end{array}

$\begin{array}{rcrcrcr} x & + & 2 y & + & z & = & 4 \\ 2 y & + & 4 z & = & 8 \\ 2 x & + & 7 y & + & 9 z & = & 23. \end{array}$

Then addition of $- 2$ $- 2$ times the first equation to the third equation yields

\begin{array}{rcrcr} x + 2 y & + & z & = & 4 \\ 2 y & + & 4 z & = & 8 \\ 3 y & + & 7 z & = & 15. \end{array}

$\begin{array}{rcrcr} x + 2 y & + & z & = & 4 \\ 2 y & + & 4 z & = & 8 \\ 3 y & + & 7 z & = & 15. \end{array}$

We have now eliminated x from the second and third equations. To simplify the process of eliminating y from the third equation, we multiply the second equation by $\frac{1}{2},$ $\frac{1}{2},$ to obtain

\begin{array}{rcrcr} x + 2 y & + & z & = & 4 \\ y & + & 2 z & = & 4 \\ 3 y & + & 7 z & = & 15. \end{array}

$\begin{array}{rcrcr} x + 2 y & + & z & = & 4 \\ y & + & 2 z & = & 4 \\ 3 y & + & 7 z & = & 15. \end{array}$

Finally, addition of $- 3$ $- 3$ times the second equation to the third equation gives

\begin{array}{rcrcl} x + 2 y & + & z & = & 4 \\ y & + & 2 z & = & 4 \\ z & = & 3. \end{array}

$\begin{array}{rcrcl} x + 2 y & + & z & = & 4 \\ y & + & 2 z & = & 4 \\ z & = & 3. \end{array}$ (15)

This system has a triangular form that makes its solution easy. By back substitution of $z = 3$ $z = 3$ in the second equation in (15), we find that

y = 4 - 2 z = 4 - 2 (3) = - 2;

$y = 4 - 2 z = 4 - 2 (3) = - 2;$

then the first equation yields

\begin{array}{rcl} x & = & 4 - 2 y - z \\ = & 4 - 2 (- 2) - (3) = 5. \end{array}

$\begin{array}{rcl} x & = & 4 - 2 y - z \\ = & 4 - 2 (- 2) - (3) = 5. \end{array}$

Thus our original system in (14) has the unique solution $x = 5, y = - 2, z = 3$ $x = 5, y = - 2, z = 3$ .

Comment

The steps by which we transformed (14) into (15) show that every solution of the system in (14) is a solution of the system in (15). But these steps can be reversed to show similarly that every solution of the system in (15) is also a solution of the system in (14). Thus the two systems are equivalent in the sense that they have the same solution set. The computation at the end of Example 6 shows that (15) has the unique solution $(x, y, z) = (5, - 2, 3),$ $(x, y, z) = (5, - 2, 3),$ and it follows that this also is the unique solution of the original system in (14).

Example 7

Solve the system

\begin{array}{rcrcrcr} 3 x & - & 8 y & + & 10 z & = & 22 \\ x & - & 3 y & + & 2 z & = & 5 \\ 2 x & - & 9 y & - & 8 z & = & - 11. \end{array}

$\begin{array}{rcrcrcr} 3 x & - & 8 y & + & 10 z & = & 22 \\ x & - & 3 y & + & 2 z & = & 5 \\ 2 x & - & 9 y & - & 8 z & = & - 11. \end{array}$ (16)

Solution

In order to avoid fractions in the elimination of x, we first interchange the first two equations to get

\begin{array}{rcrcrcr} x & - & 3 y & + & 2 z & = & 5 \\ 3 x & - & 8 y & + & 10 z & = & 22 \\ 2 x & - & 9 y & - & 8 z & = & - 11. \end{array}

$\begin{array}{rcrcrcr} x & - & 3 y & + & 2 z & = & 5 \\ 3 x & - & 8 y & + & 10 z & = & 22 \\ 2 x & - & 9 y & - & 8 z & = & - 11. \end{array}$

Addition of $- 3$ $- 3$ times the first equation to the second gives

\begin{array}{rcrcrcr} x & - & 3 y & + & 2 z & = & 5 \\ y & + & 4 z & = & 7 \\ 2 x & - & 9 y & - & 8 z & = & - 11, \end{array}

$\begin{array}{rcrcrcr} x & - & 3 y & + & 2 z & = & 5 \\ y & + & 4 z & = & 7 \\ 2 x & - & 9 y & - & 8 z & = & - 11, \end{array}$

and then addition of $- 2$ $- 2$ times the first equation to the third equation gives

\begin{array}{rcrcrcr} x & - & 3 y & + & 2 z & = & 5 \\ y & + & 4 z & = & 7 \\ - & 3 y & - & 12 z & = & - 21. \end{array}

$\begin{array}{rcrcrcr} x & - & 3 y & + & 2 z & = & 5 \\ y & + & 4 z & = & 7 \\ - & 3 y & - & 12 z & = & - 21. \end{array}$

Finally, addition of 3 times the second equation to the third equation yields

\begin{array}{rcl} x - 3 y + 2 z & = & 5 \\ y + 4 z & = & 7 \\ 0 & = & 0. \end{array}

$\begin{array}{rcl} x - 3 y + 2 z & = & 5 \\ y + 4 z & = & 7 \\ 0 & = & 0. \end{array}$ (17)

Because the third equation has disappeared, we can choose $z = t$ $z = t$ arbitrarily and then solve for x and y:

\begin{array}{rcl} y & = & 7 - 4 z = 7 - 4 t, \\ x & = & 5 + 3 y - 2 z \\ = & 5 + 3 (7 - 4 t) - 2 t = 26 - 14 t . \end{array}

$\begin{array}{rcl} y & = & 7 - 4 z = 7 - 4 t, \\ x & = & 5 + 3 y - 2 z \\ = & 5 + 3 (7 - 4 t) - 2 t = 26 - 14 t . \end{array}$

Thus our original system in (16) has infinitely many solutions. Moreover, one convenient way of describing them is this:

x = 26 - 14 t, y = 7 - 4 t, z = t .

$x = 26 - 14 t, y = 7 - 4 t, z = t .$ (18)

The arbitrary parameter t can take on all real number values, and in so doing generates all the (infinitely many) solutions of the original system in (16). Figure 3.1.4 shows the three planes corresponding to the three equations in (16). These planes intersect in a straight line that is parameterized by the equations in (18).

FIGURE 3.1.4.

The three planes of Example 7.

A Differential Equations Application

In Chapter 1 we saw that a general solution of a first-order differential equation involves an arbitrary constant that must be determined in order to satisfy a given initial condition. The next example illustrates the fact that a general solution of a second-order differential equation typically involves two arbitrary constants. An initial value problem for such a differential equation ordinarily involves two initial conditions, whose imposition leads to a system of two equations with the two arbitrary constants as unknowns.

Example 8

If A and B are constants and

y (x) = A e^{3 x} + B e^{- 3 x},

$y (x) = A e^{3 x} + B e^{- 3 x},$ (19)

then differentiation yields

y ′ (x) = 3 A e^{3 x} - 3 B e^{- 3 x}

$y ′ (x) = 3 A e^{3 x} - 3 B e^{- 3 x}$ (20)

and

y ″ (x) = 9 A e^{3 x} + 9 B e^{- 3 x} = 9 y (x) .

$y ″ (x) = 9 A e^{3 x} + 9 B e^{- 3 x} = 9 y (x) .$

Thus the function y(x) defined by (19) satisfies the second-order differential equation

y ″ - 9 y = 0.

$y ″ - 9 y = 0.$ (21)

Now, suppose that we want to solve the initial value problem consisting of this equation and the two initial conditions

y (0) = 7, y^{′} (0) = 9.

$y (0) = 7, y^{′} (0) = 9.$ (22)

Then substitution of $x = 0$ $x = 0$ in (19) and (20) yields the linear system

\begin{array}{rcrcl} A & + & B & = & 7 \\ 3 A & - & 3 B & = & 9 \end{array}

$\begin{array}{rcrcl} A & + & B & = & 7 \\ 3 A & - & 3 B & = & 9 \end{array}$

that we readily solve for $A = 5, B = 2 .$ $A = 5, B = 2 .$ It follows that the particular solution

y (x) = 5 e^{3 x} + 2 e^{- 3 x}

$y (x) = 5 e^{3 x} + 2 e^{- 3 x}$

satisfies both the differential equation (21) and the initial conditions in (22).

In Chapter 5 we will study higher-order differential equations and initial value problems in detail, after developing the necessary methods of linear algebra in the intervening chapters.

3.1 Problems

In each of Problems 1–22, use the method of elimination to determine whether the given linear system is consistent or inconsistent. For each consistent system, find the solution if it is unique; otherwise, describe the infinite solution set in terms of an arbitrary parameter t (as in Examples 5 and 7).

$\begin{array}{r} x & + & 3 y & = & 9 \\ 2 x & + & y & = & 8 \end{array}$ $\begin{array}{r} x & + & 3 y & = & 9 \\ 2 x & + & y & = & 8 \end{array}$
$\begin{array}{rcrcl} 3 x & + & 2 y & = & 9 \\ x & - & y & = & 8 \end{array}$ $\begin{array}{rcrcl} 3 x & + & 2 y & = & 9 \\ x & - & y & = & 8 \end{array}$
$\begin{array}{rcl} 2 x + 3 y & = & 1 \\ 3 x + 5 y & = & 3 \end{array}$ $\begin{array}{rcl} 2 x + 3 y & = & 1 \\ 3 x + 5 y & = & 3 \end{array}$
$\begin{array}{rcr} 5 x - 6 y & = & 1 \\ 6 x - 5 y & = & 10 \end{array}$ $\begin{array}{rcr} 5 x - 6 y & = & 1 \\ 6 x - 5 y & = & 10 \end{array}$
$\begin{array}{rcl} x + 2 y & = & 4 \\ 2 x + 4 y & = & 9 \end{array}$ $\begin{array}{rcl} x + 2 y & = & 4 \\ 2 x + 4 y & = & 9 \end{array}$
$\begin{array}{rcl} 4 x - 2 y & = & 4 \\ 6 x - 3 y & = & 7 \end{array}$ $\begin{array}{rcl} 4 x - 2 y & = & 4 \\ 6 x - 3 y & = & 7 \end{array}$
$\begin{array}{rcr} x - 4 y & = & - 10 \\ - 2 x + 8 y & = & 20 \end{array}$ $\begin{array}{rcr} x - 4 y & = & - 10 \\ - 2 x + 8 y & = & 20 \end{array}$
$\begin{array}{rcr} 3 x - 6 y & = & 12 \\ 2 x - 4 y & = & 8 \end{array}$ $\begin{array}{rcr} 3 x - 6 y & = & 12 \\ 2 x - 4 y & = & 8 \end{array}$
$\begin{array}{rcrcrcl} x & + & 5 y & + & z & = & 2 \\ 2 x & + & y & - & 2 z & = & 1 \\ x & + & 7 y & + & 2 z & = & 3 \end{array}$ $\begin{array}{rcrcrcl} x & + & 5 y & + & z & = & 2 \\ 2 x & + & y & - & 2 z & = & 1 \\ x & + & 7 y & + & 2 z & = & 3 \end{array}$
$\begin{array}{rcrcrcr} x & + & 3 y & + & 2 z & = & 2 \\ 2 x & + & 7 y & + & 7 z & = & - 1 \\ 2 x & + & 5 y & + & 2 z & = & 7 \end{array}$ $\begin{array}{rcrcrcr} x & + & 3 y & + & 2 z & = & 2 \\ 2 x & + & 7 y & + & 7 z & = & - 1 \\ 2 x & + & 5 y & + & 2 z & = & 7 \end{array}$
$\begin{array}{rcrcrcr} 2 x & + & 7 y & + & 3 z & = & 11 \\ x & + & 3 y & + & 2 z & = & 2 \\ 3 x & + & 7 y & + & 9 z & = & - 12 \end{array}$ $\begin{array}{rcrcrcr} 2 x & + & 7 y & + & 3 z & = & 11 \\ x & + & 3 y & + & 2 z & = & 2 \\ 3 x & + & 7 y & + & 9 z & = & - 12 \end{array}$
$\begin{array}{rcrcrcr} 3 x & + & 5 y & - & z & = & 13 \\ 2 x & + & 7 y & + & z & = & 28 \\ x & + & 7 y & + & 2 z & = & 32 \end{array}$ $\begin{array}{rcrcrcr} 3 x & + & 5 y & - & z & = & 13 \\ 2 x & + & 7 y & + & z & = & 28 \\ x & + & 7 y & + & 2 z & = & 32 \end{array}$
$\begin{array}{rcrcrcl} 3 x & + & 9 y & + & 7 z & = & 0 \\ 2 x & + & 7 y & + & 4 z & = & 0 \\ 2 x & + & 6 y & + & 5 z & = & 0 \end{array}$ $\begin{array}{rcrcrcl} 3 x & + & 9 y & + & 7 z & = & 0 \\ 2 x & + & 7 y & + & 4 z & = & 0 \\ 2 x & + & 6 y & + & 5 z & = & 0 \end{array}$
$\begin{array}{rcrcrcr} 4 x & + & 9 y & + & 12 z & = & - 1 \\ 3 x & + & y & + & 16 z & = & - 46 \\ 2 x & + & 7 y & + & 3 z & = & 19 \end{array}$ $\begin{array}{rcrcrcr} 4 x & + & 9 y & + & 12 z & = & - 1 \\ 3 x & + & y & + & 16 z & = & - 46 \\ 2 x & + & 7 y & + & 3 z & = & 19 \end{array}$
$\begin{array}{rcrcrcr} x & + & 3 y & + & 2 z & = & 5 \\ x & - & y & + & 3 z & = & 3 \\ 3 x & + & y & + & 8 z & = & 10 \end{array}$ $\begin{array}{rcrcrcr} x & + & 3 y & + & 2 z & = & 5 \\ x & - & y & + & 3 z & = & 3 \\ 3 x & + & y & + & 8 z & = & 10 \end{array}$
$\begin{array}{rcrcrcr} x & - & 3 y & + & 2 z & = & 6 \\ x & + & 4 y & - & z & = & 4 \\ 5 x & + & 6 y & + & z & = & 20 \end{array}$ $\begin{array}{rcrcrcr} x & - & 3 y & + & 2 z & = & 6 \\ x & + & 4 y & - & z & = & 4 \\ 5 x & + & 6 y & + & z & = & 20 \end{array}$
$\begin{array}{rcrcrcr} 2 x & - & y & + & 4 z & = & 7 \\ 3 x & + & 2 y & - & 2 z & = & 3 \\ 5 x & + & y & + & 2 z & = & 15 \end{array}$ $\begin{array}{rcrcrcr} 2 x & - & y & + & 4 z & = & 7 \\ 3 x & + & 2 y & - & 2 z & = & 3 \\ 5 x & + & y & + & 2 z & = & 15 \end{array}$
$\begin{array}{rcrcrcl} x & + & 5 y & + & 6 z & = & 3 \\ 5 x & + & 2 y & - & 10 z & = & 1 \\ 8 x & + & 17 y & + & 8 z & = & 5 \end{array}$ $\begin{array}{rcrcrcl} x & + & 5 y & + & 6 z & = & 3 \\ 5 x & + & 2 y & - & 10 z & = & 1 \\ 8 x & + & 17 y & + & 8 z & = & 5 \end{array}$
$\begin{array}{rcrcrcr} x & - & 2 y & + & z & = & 2 \\ 2 x & - & y & - & 4 z & = & 13 \\ x & - & y & - & z & = & 5 \end{array}$ $\begin{array}{rcrcrcr} x & - & 2 y & + & z & = & 2 \\ 2 x & - & y & - & 4 z & = & 13 \\ x & - & y & - & z & = & 5 \end{array}$
$\begin{array}{rcrcrcl} 2 x & + & 3 y & + & 7 z & = & 15 \\ x & + & 4 y & + & z & = & 20 \\ x & + & 2 y & + & 3 z & = & 10 \end{array}$ $\begin{array}{rcrcrcl} 2 x & + & 3 y & + & 7 z & = & 15 \\ x & + & 4 y & + & z & = & 20 \\ x & + & 2 y & + & 3 z & = & 10 \end{array}$
$\begin{array}{rcrcrcr} x & + & y & - & z & = & 5 \\ 3 x & + & y & + & 3 z & = & 11 \\ 4 x & + & y & + & 5 z & = & 14 \end{array}$ $\begin{array}{rcrcrcr} x & + & y & - & z & = & 5 \\ 3 x & + & y & + & 3 z & = & 11 \\ 4 x & + & y & + & 5 z & = & 14 \end{array}$
$\begin{array}{rcrcrcl} 4 x & - & 2 y & + & 6 z & = & 0 \\ x & - & y & - & z & = & 0 \\ 2 x & - & y & + & 3 z & = & 0 \end{array}$ $\begin{array}{rcrcrcl} 4 x & - & 2 y & + & 6 z & = & 0 \\ x & - & y & - & z & = & 0 \\ 2 x & - & y & + & 3 z & = & 0 \end{array}$

In each of Problems 23–30, a second-order differential equation and its general solution y(x) are given. Determine the constants A and B so as to find a solution of the differential equation that satisfies the given initial conditions involving y(0) and $y ′ (0)$ $y ′ (0)$ .

$y ″ + 4 y = 0, y (x) = A \cos 2 x + B \sin 2 x, y (0) = 3, y ′ (0) = 8$ $y ″ + 4 y = 0, y (x) = A \cos 2 x + B \sin 2 x, y (0) = 3, y ′ (0) = 8$
$y ″ - 9 y = 0, y (x) = A \cosh 3 x + B \sinh 3 x, y (0) = 5, y ′ (0) = 12$ $y ″ - 9 y = 0, y (x) = A \cosh 3 x + B \sinh 3 x, y (0) = 5, y ′ (0) = 12$
$y ″ - 25 y = 0, y (x) = {A e}^{5 x} + {B e}^{- 5 x}, y (0) = 10, y ′ (0) = 20$ $y ″ - 25 y = 0, y (x) = {A e}^{5 x} + {B e}^{- 5 x}, y (0) = 10, y ′ (0) = 20$
$y ″ - 121 y = 0, y (x) = {A e}^{11 x} + {B e}^{- 11 x}, y (0) = 44, y ′ (0) = 22$ $y ″ - 121 y = 0, y (x) = {A e}^{11 x} + {B e}^{- 11 x}, y (0) = 44, y ′ (0) = 22$
$y ″ + 2 y ′ - 15 y = 0, y (x) = {A e}^{3 x} + {B e}^{- 5 x}, y (0) = 40, y ′ (0) = - 16$ $y ″ + 2 y ′ - 15 y = 0, y (x) = {A e}^{3 x} + {B e}^{- 5 x}, y (0) = 40, y ′ (0) = - 16$
$y ″ - 10 y ′ + 21 y = 0, y (x) = {A e}^{3 x} + {B e}^{7 x}, y (0) = 15, y ′ (0) = 13$ $y ″ - 10 y ′ + 21 y = 0, y (x) = {A e}^{3 x} + {B e}^{7 x}, y (0) = 15, y ′ (0) = 13$
$6 y ″ - 5 y ′ + y = 0, y (x) = {A e}^{x / 2} + {B e}^{x / 3}, y (0) = 7, y ′ (0) = 11$ $6 y ″ - 5 y ′ + y = 0, y (x) = {A e}^{x / 2} + {B e}^{x / 3}, y (0) = 7, y ′ (0) = 11$
$15 y ″ + y ′ - 28 y = 0, y (x) = {A e}^{4 x / 3} + {B e}^{- 7 x / 5}, y (0) = 41, y ′ (0) = 164$ $15 y ″ + y ′ - 28 y = 0, y (x) = {A e}^{4 x / 3} + {B e}^{- 7 x / 5}, y (0) = 41, y ′ (0) = 164$
A system of the form

$\begin{array}{rcl} a_{1} x + b_{1} y & = & 0 \\ a_{2} x + b_{2} y & = & 0, \end{array}$ $\begin{array}{rcl} a_{1} x + b_{1} y & = & 0 \\ a_{2} x + b_{2} y & = & 0, \end{array}$

in which the constants on the right-hand side are all zero, is said to be homogeneous. Explain by geometric reasoning why such a system has either a unique solution or infinitely many solutions. In the former case, what is the unique solution?
Consider the system

$\begin{array}{rcl} a_{1} x + b_{1} y + c_{1} z & = & d_{1} \\ a_{2} x + b_{2} y + c_{2} z & = & d_{2} \end{array}$ $\begin{array}{rcl} a_{1} x + b_{1} y + c_{1} z & = & d_{1} \\ a_{2} x + b_{2} y + c_{2} z & = & d_{2} \end{array}$

of two equations in three unknowns.
1. Use the fact that the graph of each such equation is a plane in xyz-space to explain why such a system always has either no solution or infinitely many solutions.
2. Explain why the system must have infinitely many solutions if $d_{1} = 0 = d_{2}$ $d_{1} = 0 = d_{2}$ .
The linear system

$\begin{array}{rcl} a_{1} x + b_{1} y & = & c_{1} \\ a_{2} x + b_{2} y & = & c_{2} \\ a_{3} x + b_{3} y & = & c_{3} \end{array}$ $\begin{array}{rcl} a_{1} x + b_{1} y & = & c_{1} \\ a_{2} x + b_{2} y & = & c_{2} \\ a_{3} x + b_{3} y & = & c_{3} \end{array}$

of three equations in two unknowns represents three lines $L_{1}, L_{2},$ $L_{1}, L_{2},$ and $L_{3}$ $L_{3}$ in the xy-plane. Figure 3.1.5 shows six possible configurations of these three lines. In each case describe the solution set of the system.
Consider the linear system

$\begin{array}{rcl} a_{1} x + b_{1} y + c_{1} z & = & d_{1} \\ a_{2} x + b_{2} y + c_{2} z & = & d_{2} \\ a_{3} x + b_{3} y + c_{3} z & = & d_{3} \end{array}$ $\begin{array}{rcl} a_{1} x + b_{1} y + c_{1} z & = & d_{1} \\ a_{2} x + b_{2} y + c_{2} z & = & d_{2} \\ a_{3} x + b_{3} y + c_{3} z & = & d_{3} \end{array}$

of three equations in three unknowns to represent three planes $P_{1}, P_{2},$ $P_{1}, P_{2},$ and $P_{3}$ $P_{3}$ in xyz-space. Describe the solution set of the system in each of the following cases.
1. The three planes are parallel and distinct.
2. The three planes coincide— $P_{1} = P_{2} = P_{3}$ $P_{1} = P_{2} = P_{3}$ .
3. $P_{1}$ $P_{1}$ and $P_{2}$ $P_{2}$ coincide and are parallel to $P_{3}$ $P_{3}$ .
4. $P_{1}$ $P_{1}$ and $P_{2}$ $P_{2}$ intersect in a line L that is parallel to $P_{3}$ $P_{3}$ .
5. $P_{1}$ $P_{1}$ and $P_{2}$ $P_{2}$ intersect in a line L that lies in $P_{3}$ $P_{3}$ .
6. $P_{1}$ $P_{1}$ and $P_{2}$ $P_{2}$ intersect in a line L that intersects $P_{3}$ $P_{3}$ in a single point.

FIGURE 3.1.5.

Three lines in the plane (Problem 33).

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Table of Contents for 3 Linear Systems and Matrices

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Table of Contents for
3 Linear Systems and Matrices