Answer Key for Introduction to Linear Algebra (6th Edition) by Strang | Complete Solutions to Exercises

ALL 10 CHAPTERS COVERED
ss ss ss

, 2 Solutions ssto
ssExercises


Problem Set 1.1, page 8 ss ss ss ss




1 The sscombinations ssgive ss(a) ss a ssline ssin ssR3 (b) ss a ssplane ssin ssR3 (c) ss all ssof ssR3.

2 v ss+ ssw ss = s s (2, ss3) ssand ssv ss− ssw s s = s s (6, ss−1) sswill ssbe ssthe ssdiagonals ssof ssthe ssparallelogram
ss with
v ssand ssw ssas sstwo sssides ssgoing ssout ssfrom ss(0, ss0).

3 This ssproblem ss gives ssthe ss diagonals ssv ss + ss w ss and ss v ss − ss w ss of ssthe
ss parallelogram ssand ss asks ss for ssthe sssides: ssThe ssopposite ssof ssProblem ss2. ssIn
ssthis ssexample ssv ss= ss(3, ss3) ssand ssw ss= ss(2, ss−2).

4 3v ss+ ssw ss= ss(7, ss5) ssand sscv ss+ ssdw ss= ss(2c ss+ ssd, ssc ss+ ss2d).

5 u+v ss= ss (−2, ss3, ss1) ssand ssu+v+w ss = ss(0, ss0, ss0) ssand ss2u+2v+w ss= ss( ssadd
ss first ssanswers) ss= ss (−2, ss3, ss1). ssThe ssvectors ssu, ssv, ssw ssare ssin ssthe sssame
ss plane ssbecause ssa sscombination ssgives ss(0, ss0, ss0). ssStated ssanother ssway: ssu ss=
ss−v ss− ssw ssis ssin ssthe ssplane ssof ssv ssand ssw.

6 The sscomponents ssof ssevery sscv ss+ ssdw ssadd ssto sszero ssbecause ssthe sscomponents ssof ssv ssand ssof ssw
add ssto sszero. ssc ss= ss3 ssand ssd ss= ss9 ssgive ss(3, ss3, ss−6). ssThere ssis ssno sssolution ssto sscv+dw ss= ss(3, ss3, ss6)
because ss3 ss+ ss3 ss+ ss6 ssis ssnot sszero.

7 The ssnine sscombinations ssc(2, ss1) ss+ ssd(0, ss1) sswith ssc ss= ss0, ss1, ss2 ssand ssd ss= ss(0, ss1,

ss2) sswill sslie sson ssa sslattice. ss If sswe sstook ssall sswhole ssnumbers ssc ssand ssd, ssthe
ss lattice sswould sslie ssover ssthe sswhole ssplane.

8 The ssother ssdiagonal ssis ssv ss− ssw ss(or sselse ssw ss− ssv). ss Adding ssdiagonals ssgives ss2v ss(or ss2w).

9 The s s fourth ss corner s s can s s be s s (4, ss4) ss or s s (4, ss0) ss or s s (−2, ss2). ss Three ss possible s s parallelograms!

10 i ss− ssj ss= ss(1, ss1, ss0) ssis ssin ssthe ssbase ss(x-y ssplane). ssi ss+ ssj ss+ ssk ss= ss(1, ss1, ss1) ssis ssthe

ssopposite sscorner ssfrom ss(0, ss0, ss0). ssPoints ssin ssthe sscube sshave ss0 ss≤ ss x ss≤ ss1, ss0
ss ≤ ssy ss≤ ss1, ss0 ss≤ ssz ss≤ ss1.
11 Four ssmore sscorners ss(1, ss1, ss0), ss(1, ss0, ss1), ss(0, ss1, ss1), ss(1, ss1, ss1). ss The sscenter ss point ssis ss(ss1ss, ss1ss, ss1ss).
2 ss ss 2 ss ss 2
Centers ssof ssfaces ssare ss(ss1ss, ss1sss, ss0),
s
1 1 1 1
s ss(sss ss, ss ss, ss1) ssand ss(0, sss ss,sss sss), ss
1 1 1 1 1 1
s(1, ss ss, ss ss) ssand ss(ss ss, ss0, ss ss), ss(ss ss, ss1, ss ss).
2 ss ss 2 2 ss ss 2 2 ss ss 2 2 2 2 2 2
ss ss 2

12 The sscombinations ssof ssi ss= s s (1, ss0, ss0) ssand ssi ss+ ssj ss = ss (1, ss1, ss0) ssfill ssthe ssxy ss plane ssin ssxyz
ss space.

13 Sum ss= sszero ssvector. ss Sum ss= ss−2:00 ssvector ss= ss8:00 ssvector. s s 2:00 ssis ss30◦ ss from sshorizontal
√
= ss(cos ssπ ss, sssin ssπ ss) ss= ss( 3/2, ss1/2).
6 6

14 Moving ssthe ssorigin ssto ss6:00 ssadds ssj ss = ss (0, ss1) ssto ssevery ssvector. ss So ssthe sssum
ss of sstwelve ssvectors sschanges ssfrom ss0 ssto ss12j ss= ss(0, ss12).

, Solutions ssto 3
ssExercises
3ss 1ss
15 The ss point s s v ss+ ss w ss is ss three-fourths ss of ss the s s way ss to s s v ss starting ss from ss w.
s s The
vector ss
4 4
1 1 1 1
v ss+ ss ssw ssis sshalfway ssto ssu ss= ss v ss+
ss ss ss w. ssThe ssvector ssv ss+ ssw ssis ss2u ss(the ssfar sscorner ssof ssthe
4 4 2 2
parallelogram).

16 All ss combinations s s with s s c ss+ ssd s s = s s 1 ss are ss on ss the ss line ss that
ss passes ss through ss v ss and ss w. ssThe sspoint ssV s s = ss−v ss+ ss2w ssis sson ssthat
line ssbut ssit ssis ssbeyond ssw.
ss


17 All ssvectors sscv ss+ sscw ssare sson ssthe ssline sspassing ssthrough ss(0, ss0) ssand ssu ss= 1
s s ss v ss+ ss1ssw. s s That
2 2

line sscontinues ssout ssbeyond ssv ss+ ssw ssand ssback ssbeyond ss(0, ss0). ss With ssc ss≥ ss 0,
ss half ssof ssthis ssline ssis ssremoved, ssleaving ssa ssray ssthat ssstarts ssat ss(0, ss0).

18 The sscombinations sscv ss+ ssdw sswith ss0 ss≤ ss c ss≤ ss1 ssand ss0 ss≤ ssd ss≤ ss1 ssfill ssthe
parallelogram sswith sssides ssv ssand ssw. ssFor ssexample, ssif ssv ss= ss(1, ss0) ssand ssw ss=
ss


ss (0, ss1) ssthen sscv ss+ ssdw ssfills ssthe ssunit sssquare. ssBut sswhen ssv ss= ss(a, ss0) ssand ssw
ss = ss(b, ss0) ssthese sscombinations ssonly ssfill ssa sssegment ssof ssa ssline.

19 With ssc ss≥ 0 ssand ssd ss≥
ss ss0 sswe ssget ssthe ssinfinite ss“cone” ssor ss“wedge” ssbetween ssv
ss and ssw. s s For ssexample, ssif ssv ss= ss(1, ss0) ssand ssw ss= ss(0, ss1), ssthen ssthe sscone ssis ssthe
whole ssquadrant ssx ss≥ ss0, ssy ss≥
ss


0. ss Question: ss What s s if ss w ss = ss −v? ss The ss cone s s opens ssto s s a ss half-space.
ss But ss the s s combinations ssof ssv ss= ss(1, ss0) ssand ssw ss= ss(−1, ss0) ssonly ssfill ssa
ss line.
20 (a) s s ss
1
u ss+ ss1ssv ss+ 1
ss ss w ssis ssthe sscenter ssof ssthe sstriangle ssbetween ssu, ssv ssand ssw; 1
s s ssu ss+ 1
ss ss w
ss lies
3 3 3 2 2

between ssu ssand ssw (b) ss To ssfill ssthe sstriangle sskeep sscss≥ss0, ssdss≥ss0, ssess≥ss0, ssand sscss+ssdss+sse ss= ss1.

21 The sssum ssis ss(v ss− ssu)ss+(w ss− ssv)ss+(uss− ssw) ss = sszero ssvector. s s Those ssthree sssides
ss of ssa sstriangle ssare ssin ssthe sssame ssplane!
22 The ssvector ss1ss(u ss+ ssv ss+ ssw) ssis ssoutside ssthe sspyramid ssbecause ssc ss+ ssd ss+ sse ss= ss
1
+ ss1 ss+ ss1
ss ss > ss1.
2 2 2 2

23 All ssvectors ssare sscombinations ssof ssu, ssv, ssw ssas ssdrawn ss(not ssin ssthe sssame

ss plane). ss Start ssby ssseeing ssthat sscu ss+ ssdv ssfills ssa ssplane, ssthen ssadding ssew
ss fills ssall ssof ssR3.

24 The sscombinations ssof ssu ssand ssv ss fill ssone ssplane. ss The sscombinations ssof ssv ss and
ss w ss fill ssanother ssplane. ss Those ssplanes ssmeet ssin ssa ssline: ss only ssthe ssvectors sscv
ss are ssin ssboth ssplanes.

25 (a) ss For ssa ssline, sschoose ssu ss= ssv ss= ssw ss= ssany ssnonzero ssvector (b) ss For ssa ssplane, sschoose
u s s and s s v ss in s s different s s directions. ss A s s combination ss like s s w ss = ss u ss + s s v ss is s s in s s the s s same s s plane.

, 4 Solutions ssto
ssExercises


26 Two ssequations sscome ssfrom ssthe sstwo sscomponents: ssc ss+ ss3d ss= ss14 ssand ss2c ss+ ssd
= ss8. ssThe sssolution ssis ssc ss= ss2 ssand ssd ss= ss4. ssThen ss2(1, ss2) ss+ ss4(3, ss1) ss=
ss


ss (14, ss8).

27 A ssfour-dimensional sscube sshas ss24 s s = ss 16 ss corners ss and ss2 ss· ss4 ss = ss 8
ss three-dimensional ssfaces ssand ss24 two-dimensional ssfaces ssand ss32
ss ss edges ssin
ss Worked ssExample ss2.4 ssA.

28 There ssare ss6 ssunknown ssnumbers ssv1, ssv2, ssv3, ssw1, ssw2, ssw3. ssThe sssix ssequations
ss come ssfrom ssthe sscomponents ssof ssv ss+ ssw ss= ss(4, ss5, ss6) ssand ssv ss− ssw ss= ss(2, ss5, ss8).
Add ssto ssfind ss2v ss= ss(6, ss10, ss14)
ss


so ssv ss= ss(3, ss5, ss7) ssand ssw ss= ss(1, ss0, ss−1).

29 Fact ss: ssFor ssany ssthree ssvectors ssu, ssv, ssw ssin ssthe ssplane, sssome sscombination sscu ss+

dv ss+ ssew ssis ssthe sszero ssvector ss(beyond ssthe ssobvious ssc ss= ssd ss= sse ss= ss0). s s So
ss


if ssthere ssis ssone sscombination ssCu ss+ ssDv ss+ ssEw
ss ss that s s produces ss b, ss there s s will
ss be ss many ss more—just ss add ss c, ssd, sse ss or ss 2c, ss2d, ss2e ssto ssthe ssparticular
ss solution ssC, ssD, ssE.

The ssexample sshas ss3u ss− ss2v ss+ ssw ss = ss 3(1, ss3) ss− ss2(2, ss7) ss+ ss1(1, ss5) ss = ss (0, ss0). s s It ssalso
ss has


−2u ss+ ss1v ss+ ss0w ss= ssb ss= ss(0, ss1). ssAdding ssgives ssu ss− ssv ss+ ssw ss= ss(0, ss1). ssIn ssthis sscase ssc, ssd, sse
equal ss3, ss−2, ss1 ssand ssC, ssD, ssE ss = ss−2, ss1, ss0.

Could ssanother ssexample sshave ssu, ssv, ssw ssthat sscould ssNOT sscombine ssto ssproduce ssb
? ssYes.
ss ss The ssvectors ss(1, ss1), ss(2, ss2), ss(3, ss3) ssare sson ssa ssline ssand ssno sscombination
ss produces ssb. ss We sscan sseasily sssolve sscu ss+ ss dv ss+ ss ew ss = ss 0 ss but ssnot ssCu
ss + ssDv ss+ ssEw ss= ssb.

30 The sscombinations ssof ssv ssand ssw ssfill ssthe ssplane ssunless ssv ssand ssw sslie sson ssthe sssame
line ssthrough ss(0, ss0).
ss ss Four ssvectors sswhose sscombinations ssfill ss4-dimensional
space: s s one ssexample ssis ssthe ss“standard ssbasis” ss(1, ss0, ss0, ss0), ss(0, ss1, ss0, ss0), ss(0,
ss


ss 0, ss1, ss0), ssand ss(0, ss0, ss0, ss1).

31 The ssequations sscu ss+ ssdv ss+ ssew ss= ssb ssare


2c s s −d = ss 1 So ssd ss= c ss = ss3/4
−c ss+2d ss ss −e ss= ss0 ss 2e ssthen d ss= ss2/4
−d ss+2e ss= ss0 ss c ss= ss3e e ss= ss1/4
ss then ss4e
ss = ss1