Chapter 2 Supplementary Exercises
3.4 The Geometry of Linear Systems
3.4 The Geometry of Linear Systems 165
THEOREM3.4.2 LetW be the plane inR3 that contains the pointx0and is parallel to the noncollinear vectorsv1andv2. Then an equation of the plane throughx0that is parallel tov1andv2is given by
x=x0+t1v1+t2v2 (3) Ifx0=0,then the plane passes through the origin and the equation has the form
x=t1v1+t2v2 (4)
Remark Observe that the line throughx0represented by Equation (1) is the translation byx0of x0
x W
t2v2 t1v1 z
y
x
Figure 3.4.3
the line through the origin represented by Equation (2) and that the plane throughx0represented by Equation (3) is the translation byx0of the plane through the origin represented by Equation (4) (Figure 3.4.4).
Figure 3.4.4
x0
v x
y
x = x0 + tv
x = x0 + t1v1 + t2v2
x = t1v1 + t2v2 x = tv
x0
v2 v1 z
y
x
Motivated by the forms of Formulas (1) to (4), we can extend the notions of line and plane toRnby making the following definitions.
DEFINITION1 Ifx0andvare vectors inRn, and ifvis nonzero, then the equation
x=x0+tv (5)
defines theline throughx0that is parallel tov. In the special case wherex0=0, the line is said topass through the origin.
DEFINITION2 Ifx0,v1,andv2are vectors inRn, and ifv1andv2are not collinear, then the equation
x=x0+t1v1+t2v2 (6) defines theplane throughx0that is parallel tov1andv2. In the special case where x0=0, the plane is said topass through the origin.
Equations (5) and (6) are calledvector formsof a line and plane inRn. If the vectors in these equations are expressed in terms of their components and the corresponding components on each side are equated, then the resulting equations are calledparametric equationsof the line and plane. Here are some examples.
E X A M P L E 1 Vector and Parametric Equations of Lines inR2andR3 (a) Find a vector equation and parametric equations of the line in R2 that passes
through the origin and is parallel to the vectorv=(−2,3).
(b) Find a vector equation and parametric equations of the line in R3 that passes through the pointP0(1,2,−3)and is parallel to the vectorv=(4,−5,1).
(c) Use the vector equation obtained in part (b) to find two points on the line that are different fromP0.
Solution(a) It follows from (5) withx0=0that a vector equation of the line isx=tv.
If we letx=(x, y), then this equation can be expressed in vector form as (x, y)=t(−2,3)
Equating corresponding components on the two sides of this equation yields the para- metric equations
x= −2t, y=3t
Solution(b) It follows from (5) that a vector equation of the line isx=x0+tv. If we letx=(x, y, z), and if we takex0=(1,2,−3), then this equation can be expressed in vector form as
(x, y, z)=(1,2,−3)+t(4,−5,1) (7) Equating corresponding components on the two sides of this equation yields the para- metric equations
x =1+4t, y=2−5t, z= −3+t
Solution(c) A point on the line represented by Equation (7) can be obtained by sub- stituting a specific numerical value for the parametert. However, sincet=0 produces (x, y, z)=(1,2,−3), which is the pointP0, this value oftdoes not serve our purpose.
Taking t=1 produces the point (5,−3,−2)and takingt = −1 produces the point (−3,7,−4). Any other distinct values fort(exceptt =0) would work just as well.
E X A M P L E 2 Vector and Parametric Equations of a Plane inR3 Find vector and parametric equations of the planex−y+2z=5.
Solution We will find the parametric equations first. We can do this by solving the equation for any one of the variables in terms of the other two and then using those two variables as parameters. For example, solving forxin terms ofyandzyields
x =5+y−2z (8)
and then usingyandzas parameterst1andt2, respectively, yields the parametric equa- tions
x =5+t1−2t2, y=t1, z=t2
To obtain a vector equation of the plane we rewrite these parametric equations as We would have obtained dif-
ferent parametric and vector equations in Example 2 had we solved (8) foryorzrather than x. However, one can show the same plane results in all three cases as the parameters vary from−⬁to⬁.
(x, y, z)=(5+t1−2t2, t1, t2) or, equivalently, as
(x, y, z)=(5,0,0)+t1(1,1,0)+t2(−2,0,1)
E X A M P L E 3 Vector and Parametric Equations of Lines and Planes inR4 (a) Find vector and parametric equations of the line through the origin ofR4that is
parallel to the vectorv=(5,−3,6,1).
(b) Find vector and parametric equations of the plane inR4that passes through the point x0=(2,−1,0,3)and is parallel to bothv1=(1,5,2,−4)andv2=(0,7,−8,6).
Solution(a) If we letx=(x1, x2, x3, x4), then the vector equationx=tvcan be ex- pressed as
(x1, x2, x3, x4)=t(5,−3,6,1)
Equating corresponding components yields the parametric equations x1=5t, x2= −3t, x3=6t, x4=t
3.4 The Geometry of Linear Systems 167 Solution(b) The vector equationx=x0+t1v1+t2v2can be expressed as
(x1, x2, x3, x4)=(2,−1,0,3)+t1(1,5,2,−4)+t2(0,7,−8,6) which yields the parametric equations
x1=2+t1
x2= −1+5t1+7t2
x3=2t1−8t2
x4=3−4t1+6t2
LinesThroughTwo Points in Rn
Ifx0andx1are distinct points inRn, then the line determined by these points is parallel to the vectorv=x1−x0(Figure 3.4.5), so it follows from (5) that the line can be expressed
x0
x1 v
Figure 3.4.5
in vector form as
x=x0+t(x1−x0) (9)
or, equivalently, as
x=(1−t)x0+tx1 (10)
These are called thetwo-point vector equationsof a line inRn.
E X A M P L E 4 A Line Through Two Points inR2
Find vector and parametric equations for the line inR2that passes through the points P (0,7)andQ(5,0).
Solution We will see below that it does not matter which point we take to bex0and which we take to bex1, so let us choosex0=(0,7)andx1=(5,0). It follows that x1−x0=(5,−7)and hence that
(x, y)=(0,7)+t(5,−7) (11)
which we can rewrite in parametric form as
x=5t, y=7−7t
Had we reversed our choices and takenx0=(5,0)andx1=(0,7), then the resulting vector equation would have been
(x, y)=(5,0)+t(−5,7) (12)
and the parametric equations would have been x=5−5t, y=7t
(verify). Although (11) and (12) look different, they both represent the line whose equa- tion in rectangular coordinates is
7x+5y =35
(Figure 3.4.6). This can be seen by eliminating the parametert from the parametric
x y
2 3 4 5 6
1 1 2 3 4 5 6 7
7x + 5y = 35
Figure 3.4.6
equations (verify).
The pointx=(x, y) in Equations (9) and (10) traces an entire line inR2 as the parametertvaries over the interval(−⬁,⬁). If, however, we restrict the parameter to vary fromt=0 tot=1, thenxwill not trace the entire line but rather just theline segmentjoining the pointsx0andx1. The pointxwill start atx0whent =0 and end at x1whent =1. Accordingly, we make the following definition.
DEFINITION3 Ifx0andx1are vectors inRn, then the equation
x=x0+t(x1−x0) (0≤t≤1) (13) defines theline segment fromx0tox1. When convenient, Equation (13) can be written as
x=(1−t)x0+tx1 (0≤t≤1) (14)
E X A M P L E 5 A Line Segment from One Point to Another inR2
It follows from (13) and (14) that the line segment inR2fromx0=(1,−3)tox1=(5,6) can be represented either by the equation
x=(1,−3)+t (4,9) (0≤t ≤1) or by the equation
x=(1−t)(1,−3)+t(5,6) (0≤t ≤1) Dot Product Form of a
Linear System
Our next objective is to show how to express linear equations and linear systems in dot product notation. This will lead us to some important results about orthogonality and linear systems.
Recall that alinear equationin the variablesx1, x2, . . . , xnhas the form
a1x1+a2x2+ · · · +anxn=b (a1, a2, . . . , annot all zero) (15) and that the correspondinghomogeneousequation is
a1x1+a2x2+ · · · +anxn=0 (a1, a2, . . . , annot all zero) (16) These equations can be rewritten in vector form by letting
a=(a1, a2, . . . , an) and x=(x1, x2, . . . , xn) in which case Formula (15) can be written as
a·x=b (17)
and Formula (16) as
a·x=0 (18)
Except for a notational change fromntoa, Formula (18) is the extension toRnof Formula (6) in Section 3.3. This equation reveals thateach solution vectorxof a homogeneous equation is orthogonal to the coefficient vectora. To take this geometric observation a step further, consider the homogeneous system
a11x1 +a12x2 + · · · +a1nxn =0 a21x1 +a22x2 + · · · +a2nxn =0 ... ... ... ... am1x1+am2x2+ · · · +amnxn=0
If we denote the successive row vectors of the coefficient matrix byr1,r2, . . . ,rm, then we can rewrite this system in dot product form as
r1·x =0 r2·x =0 ... ... rm·x=0
(19)
from which we see that every solution vectorxis orthogonal to every row vector of the coefficient matrix. In summary, we have the following result.
3.4 The Geometry of Linear Systems 169
THEOREM3.4.3 IfAis anm×nmatrix,then the solution set of the homogeneous linear system Ax=0consists of all vectors inRn that are orthogonal to everyrow vector ofA.
E X A M P L E 6 Orthogonality of Row Vectors and Solution Vectors
We showed in Example 6 of Section 1.2 that the general solution of the homogeneous linear system
⎡
⎢⎢
⎣
1 3 −2 0 2 0
2 6 −5 −2 4 −3
0 0 5 10 0 15
2 6 0 8 4 18
⎤
⎥⎥
⎦
⎡
⎢⎢
⎢⎢
⎢⎢
⎢⎣ x1 x2
x3
x4
x5
x6
⎤
⎥⎥
⎥⎥
⎥⎥
⎥⎦
=
⎡
⎢⎢
⎣ 0 0 0 0
⎤
⎥⎥
⎦
is
x1= −3r−4s−2t, x2=r, x3= −2s, x4=s, x5=t, x6=0 which we can rewrite in vector form as
x=(−3r−4s−2t, r,−2s, s, t,0)
According to Theorem 3.4.3, the vectorxmust be orthogonal to each of the row vectors r1=(1,3,−2,0,2,0)
r2=(2,6,−5,−2,4,−3) r3=(0,0,5,10,0,15) r4=(2,6,0,8,4,18)
We will confirm thatxis orthogonal tor1, and leave it for you to verify thatxis orthogonal to the other three row vectors as well. The dot product ofr1andxis
r1·x=1(−3r−4s−2t)+3(r)+(−2)(−2s)+0(s)+2(t)+0(0)=0 which establishes the orthogonality.
The Relationship Between Ax=0and Ax=b
We will conclude this section by exploring the relationship between the solutions of a homogeneous linear systemAx=0and the solutions (if any) of a nonhomogeneous linear systemAx=bthat has the same coefficient matrix. These are calledcorresponding linear systems.
To motivate the result we are seeking, let us compare the solutions of the correspond- ing linear systems
⎡
⎢⎢
⎣
1 3 −2 0 2 0
2 6 −5 −2 4 −3
0 0 5 10 0 15
2 6 0 8 4 18
⎤
⎥⎥
⎦
⎡
⎢⎢
⎢⎢
⎢⎢
⎢⎣ x1
x2 x3
x4
x5
x6
⎤
⎥⎥
⎥⎥
⎥⎥
⎥⎦
=
⎡
⎢⎢
⎣ 0 0 0 0
⎤
⎥⎥
⎦ and
⎡
⎢⎢
⎣
1 3 −2 0 2 0
2 6 −5 −2 4 −3
0 0 5 10 0 15
2 6 0 8 4 18
⎤
⎥⎥
⎦
⎡
⎢⎢
⎢⎢
⎢⎢
⎢⎣ x1
x2 x3
x4
x5
x6
⎤
⎥⎥
⎥⎥
⎥⎥
⎥⎦
=
⎡
⎢⎢
⎣ 0
−1 5 6
⎤
⎥⎥
⎦
We showed in Examples 5 and 6 of Section 1.2 that the general solutions of these linear systems can be written in parametric form as
homogeneous−→x1= −3r−4s−2t, x2=r, x3= −2s, x4=s, x5=t, x6=0
nonhomogeneous−→x1= −3r−4s−2t, x2=r, x3= −2s, x4=s, x5=t, x6= 13
which we can then rewrite in vector form as
homogeneous−→(x1, x2, x3, x4, x5, x6)=(−3r−4s−2t, r,−2s, s, t,0)
nonhomogeneous−→(x1, x2, x3, x4, x5, x6)=
−3r−4s−2t, r,−2s, s, t,13 By splitting the vectors on the right apart and collecting terms with like parameters, we can rewrite these equations as
homogeneous−→(x1, x2, x3, x4, x5)=r(−3,1,0,0,0)+s(−4,0,−2,1,0,0)+t(−2,0,0,0,1,0) (20)
nonhomogeneous−→(x1, x2, x3, x4, x5)=r(−3,1,0,0,0)+s(−4,0,−2,1,0,0) +t(−2,0,0,0,1,0)+
0,0,0,0,0,13 (21) Formulas (20) and (21) reveal that each solution of the nonhomogeneous system can be obtained by adding the fixed vector
0,0,0,0,0,13
to the corresponding solution of the homogeneous system. This is a special case of the following general result.
THEOREM3.4.4 The general solution of a consistent linear systemAx=b can be obtained byadding anyspecific solution ofAx=bto the general solution ofAx=0.
Proof Letx0be any specific solution ofAx=b, letWdenote the solution set ofAx=0, and letx0+W denote the set of all vectors that result by addingx0to each vector in W. We must show that ifxis a vector inx0+W, thenxis a solution ofAx=b, and conversely that every solution ofAx=bis in the setx0+W.
Assume first thatxis a vector inx0+W.This implies thatxis expressible in the formx=x0+w,whereAx0=bandAw=0.Thus,
Ax=A(x0+w)=Ax0+Aw=b+0=b which shows thatxis a solution ofAx=b.
Conversely, letxbe any solution ofAx=b.To show thatxis in the setx0+Wwe must show thatxis expressible in the form
x=x0+w (22)
wherewis inW(i.e.,Aw=0).We can do this by takingw=x−x0.This vector obvi- ously satisfies (22), and it is inWsince
Aw=A(x−x0)=Ax−Ax0=b−b=0
Remark Theorem 3.4.4 has a useful geometric interpretation that is illustrated in Figure 3.4.7.
Ax = b
Ax = 0
0 x0
Figure 3.4.7 The solution set ofAx=bis a translation of the solution space ofAx=0.
If, as discussed in Section 3.1, we interpret vector addition as translation, then the theorem states that ifx0isanyspecific solution ofAx=b, then theentiresolution set ofAx=bcan be obtained by translating the solution space ofAx=0by the vectorx0.
Exercise Set 3.4
In Exercises1–4, find vector and parametric equations of the line containing the point and parallel to the vector.
1.Point:(−4,1); vector:v=(0,−8) 2.Point:(2,−1); vector:v=(−4,−2) 3.Point:(0,0,0); vector:v=(−3,0,1) 4.Point:(−9,3,4); vector:v=(−1,6,0)
In Exercises5–8, use the given equation of a line to find a point on the line and a vector parallel to the line.
5. x=(3−5t,−6−t) 6.(x, y, z)=(4t,7,4+3t) 7. x=(1−t)(4,6)+t(−2,0) 8. x=(1−t)(0,−5,1)
3.4 The Geometry of Linear Systems 171 In Exercises9–12, find vector and parametric equations of
the plane that contains the given point and is parallel to the two vectors.
9.Point:(−3,1,0); vectors:v1=(0,−3,6)and v2=(−5,1,2)
10.Point:(0,6,−2); vectors:v1=(0,9,−1)and v2=(0,−3,0)
11.Point:(−1,1,4); vectors:v1=(6,−1,0)and v2=(−1,3,1)
12.Point:(0,5,−4); vectors:v1=(0,0,−5)and v2=(1,−3,−2)
In Exercises13–14, find vector and parametric equations of the line inR2 that passes through the origin and is orthogonal tov.
13. v=(−2,3) 14. v=(1,−4)
In Exercises15–16, find vector and parametric equations of the plane inR3that passes through the origin and is orthogonal tov.
15. v=(4,0,−5)[Hint: Construct two nonparallel vectors or- thogonal tovinR3].
16. v=(3,1,−6)
In Exercises17–20, find the general solution to the linear sys- tem and confirm that the row vectors of the coefficient matrix are orthogonal to the solution vectors.
17. x1+ x2+ x3=0 2x1+2x2+2x3=0 3x1+3x2+3x3=0
18. x1+3x2−4x3=0 2x1+6x2−8x3=0
19.x1+5x2+x3+2x4− x5=0 x1−2x2−x3+3x4+2x5=0 20.x1+3x2−4x3=0
x1+2x2+3x3=0
21.(a) The equationx+y+z=1 can be viewed as a linear sys- tem of one equation in three unknowns. Express a general solution of this equation as a particular solution plus a general solution of the associated homogeneous equation.
(b) Give a geometric interpretation of the result in part (a).
22.(a) The equationx+y=1 can be viewed as a linear system of one equation in two unknowns. Express a general solu- tion of this equation as a particular solution plus a general solution of the associated homogeneous system.
(b) Give a geometric interpretation of the result in part (a).
23.(a) Find a homogeneous linear system of two equations in three unknowns whose solution space consists of those vectors in R3 that are orthogonal to a=(1,1,1) and b=(−2,3,0).
(b) What kind of geometric object is the solution space?
(c) Find a general solution of the system obtained in part (a), and confirm that Theorem 3.4.3 holds.
24. (a) Find a homogeneous linear system of two equations in three unknowns whose solution space consists of those vectors inR3that are orthogonal toa=(−3,2,−1)and b=(0,−2,−2).
(b) What kind of geometric object is the solution space?
(c) Find a general solution of the system obtained in part (a), and confirm that Theorem 3.4.3 holds.
25. Consider the linear systems
⎡
⎣ 3 2 −1
6 4 −2
−3 −2 1
⎤
⎦
⎡
⎣x1
x2
x3
⎤
⎦=
⎡
⎣00 0
⎤
⎦
and ⎡
⎣ 3 2 −1
6 4 −2
−3 −2 1
⎤
⎦
⎡
⎣x1
x2
x3
⎤
⎦=
⎡
⎣ 24
−2
⎤
⎦
(a) Find a general solution of the homogeneous system.
(b) Confirm thatx1=1, x2=0, x3=1 is a solution of the nonhomogeneous system.
(c) Use the results in parts (a) and (b) to find a general solution of the nonhomogeneous system.
(d) Check your result in part (c) by solving the nonhomoge- neous system directly.
26. Consider the linear systems
⎡
⎣1 −2 3
2 1 4
1 −7 5
⎤
⎦
⎡
⎣x1
x2
x3
⎤
⎦=
⎡
⎣00 0
⎤
⎦
and ⎡
⎣1 −2 3
2 1 4
1 −7 5
⎤
⎦
⎡
⎣x1
x2
x3
⎤
⎦=
⎡
⎣ 27
−1
⎤
⎦
(a) Find a general solution of the homogeneous system.
(b) Confirm thatx1=1, x2=1, x3=1 is a solution of the nonhomogeneous system.
(c) Use the results in parts (a) and (b) to find a general solution of the nonhomogeneous system.
(d) Check your result in part (c) by solving the nonhomoge- neous system directly.
In Exercises27–28, find a general solution of the system, and use that solution to find a general solution of the associated homo- geneous system and a particular solution of the given system.
27.
⎡
⎣36 48 12 25
9 12 3 10
⎤
⎦
⎡
⎢⎢
⎣ x1
x2
x3
x4
⎤
⎥⎥
⎦=
⎡
⎣ 37 13
⎤
⎦
28.
⎡
⎣9 −3 5 6
6 −2 3 1
3 −1 3 14
⎤
⎦
⎡
⎢⎢
⎣ x1
x2
x3
x4
⎤
⎥⎥
⎦=
⎡
⎣ 45
−8
⎤
⎦
29.Letx=x0+tvbe a line inRn, and letT:Rn→Rnbe a ma- trix operator on Rn. What kind of geometric object is the image of this line under the operatorT? Explain your reason- ing.
True-False Exercises
TF.In parts (a)–(f ) determine whether the statement is true or false, and justify your answer.
(a) The vector equation of a line can be determined from any point lying on the line and a nonzero vector parallel to the line.
(b) The vector equation of a plane can be determined from any point lying in the plane and a nonzero vector parallel to the plane.
(c) The points lying on a line through the origin inR2orR3are all scalar multiples of any nonzero vector on the line.
(d) All solution vectors of the linear systemAx=bare orthogo- nal to the row vectors of the matrixAif and only ifb=0.
(e) The general solution of the nonhomogeneous linear system Ax=bcan be obtained by addingbto the general solution of the homogeneous linear systemAx=0.
(f ) Ifx1andx2are two solutions of the nonhomogeneous linear systemAx=b, thenx1−x2is a solution of the corresponding homogeneous linear system.
Working withTechnology
T1.Find the general solution of the homogeneous linear system
⎡
⎢⎢
⎢⎣
2 6 −4 0 4 0
0 0 1 2 0 3
6 18 −15 −6 12 −9
1 3 0 4 2 9
⎤
⎥⎥
⎥⎦
⎡
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎣ x1
x2
x3
x4
x5
x6
⎤
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎦
=
⎡
⎢⎢
⎢⎣ 0 0 0 0
⎤
⎥⎥
⎥⎦
and confirm that each solution vector is orthogonal to every row vector of the coefficient matrix in accordance with Theorem 3.4.3.