082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

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276 Chapter 8 Matrix algebra( )a bIf A is the matrix , we write its determinant asc d∣ a bc d∣ . Note that the straightlines || indicate that we are discussing the determinant, which is a scalar, rather thanthe matrix itself. If the matrixAis such that |A| = 0, then it has no inverse and is saidtobe singular. If |A| ≠ 0 thenA −1 exists andAissaidtobe non-singular.Asingular matrixAhas |A| = 0.Anon-singular matrixAhas |A| ≠ 0.Example8.23 IfA =( ) 1 2andB=5 0( ) −1 2find |A|, |B|and |AB|.−3 1Solution |A| =∣ 1 25 0∣ = (1)(0) − (2)(5) = −10|B| =∣ −1 2−3 1∣ = (−1)(1) − (2)(−3) =5( )( ) ( )1 2 −1 2 −7 4AB ==5 0 −3 1 −5 10|AB| = (−7)(10) − (4)(−5) = −50We note that |A||B| = |AB|.The resultobtained inExample 8.23 istruemore generally:IfAandBaresquare matrices of the same order, |A||B| = |AB|.8.6.2 OrthogonalmatricesA non-singular square matrixAsuch thatA T = A −1 is said to be orthogonal. Consequently,ifAisorthogonalAA T =A T A =I.Example8.24 Find the inverse ofA =( ) 0 −1. Deduce thatAisan orthogonal matrix.1 0Solution From the formula forthe inverse of a 2 ×2 matrix wefindA −1 = 1 ( ) ( ) 0 1 0 1=1 −1 0 −1 0This isclearly equal tothe transpose ofA. HenceAisan orthogonal matrix.To findthe inversesoflargermatricesweshallneedtostudydeterminantsfurther.Thisisdone inSection8.7.

8.6 The inverse of a 2 × 2 matrix 277EXERCISES8.61 IfA=( ) 5 6findA−4 8−1 .2 Find the inverse, ifitexists,ofeach ofthe followingmatrices:( ) ( ) ( )1 0 −1 0 2 3(a) (b) (c)0 1 0 −1 4 1( ) ( ) ( )−1 0 6 2 −6 2(d) (e) (f)−1 7 9 3 9 3⎛ ⎞1 1(g) ⎜22⎝0 1 ⎟⎠2( ) ( ) 3 0 7 83 IfA= andB=−1 4 4 3find |AB|, |BA|.( ) a b4 IfA= ,B =c dfindAB, |A|, |B|, |AB|.( ) e fg hVerifythat |AB| = |A||B|.( ) 1 25 IfA= findA3 4−1 .Find values ofthe constantsaandbsuch thatA +aA −1 =bI.( ) ( )1 1 2 16 IfA= andB=0 3 −1 3findAB, (AB) −1 ,B −1 ,A −1 andB −1 A −1 .Deduce that (AB) −1 =B −1 A −1 .7 Given that the matrix⎛⎞cos ωt −sin ωt 0M = ⎝sin ωt cos ωt 0⎠0 0 1is orthogonal, findM −1 .( ) a b8 (a) IfA= andkis ascalar constant,c dshowthat the inverse ofthe matrixkAis 1 k A−1 .( ) 1 1(b) Findthe inverse of andhence write1 0⎛ ⎞1 1down the inverse of ⎜33⎟⎝13 0 ⎠ .Solutions164( ) 8 −64 51⎛( ) ( ) 1 0 −1 0− 1 ⎞32 (a) (b) (c) ⎜ 10 100 1 0 −1 ⎝ 2− 1 ⎟⎠5 5⎛( ) −1 0(d)− 1 − 1 ⎞11 (e) No inverse (f) ⎜ 12 18⎟⎝ 1 1 ⎠7 7( )4 62 −2(g)0 23 −132, −1324( ) ae +bg af+bh,ce+dg cf+dhad−bc,eh−fg,(ad −bc)(eh − fg)( ) −2 15 32 −1 a=−2,b=52( ) 1 46 AB= , (AB)−3 9−1 = 1 ( ) 9 −4,21 3 1B −1 = 1 ( ) 3 −1,A7 1 2−1 = 1 ( ) 3 −13 0 1⎛⎞cos ωt sinωt 07 ⎝−sin ωt cos ωt 0⎠0 0 1( ) ( ) 0 1 0 38 (b) ,1 −1 3 −3

8.6 The inverse of a 2 × 2 matrix 277

EXERCISES8.6

1 IfA=

( ) 5 6

findA

−4 8

−1 .

2 Find the inverse, ifitexists,ofeach ofthe following

matrices:

( ) ( ) ( )

1 0 −1 0 2 3

(a) (b) (c)

0 1 0 −1 4 1

( ) ( ) ( )

−1 0 6 2 −6 2

(d) (e) (f)

−1 7 9 3 9 3

⎛ ⎞

1 1

(g) ⎜2

2

0 1 ⎟

2

( ) ( ) 3 0 7 8

3 IfA= andB=

−1 4 4 3

find |AB|, |BA|.

( ) a b

4 IfA= ,B =

c d

findAB, |A|, |B|, |AB|.

( ) e f

g h

Verifythat |AB| = |A||B|.

( ) 1 2

5 IfA= findA

3 4

−1 .

Find values ofthe constantsaandb

such thatA +aA −1 =bI.

( ) ( )

1 1 2 1

6 IfA= andB=

0 3 −1 3

findAB, (AB) −1 ,B −1 ,A −1 andB −1 A −1 .

Deduce that (AB) −1 =B −1 A −1 .

7 Given that the matrix

cos ωt −sin ωt 0

M = ⎝sin ωt cos ωt 0⎠

0 0 1

is orthogonal, findM −1 .

( ) a b

8 (a) IfA= andkis ascalar constant,

c d

showthat the inverse ofthe matrixkA

is 1 k A−1 .

( ) 1 1

(b) Findthe inverse of andhence write

1 0

⎛ ⎞

1 1

down the inverse of ⎜3

3

⎝1

3 0 ⎠ .

Solutions

1

64

( ) 8 −6

4 5

1

( ) ( ) 1 0 −1 0

− 1 ⎞

3

2 (a) (b) (c) ⎜ 10 10

0 1 0 −1 ⎝ 2

− 1 ⎟

5 5

( ) −1 0

(d)

− 1 − 1 ⎞

1

1 (e) No inverse (f) ⎜ 12 18

⎝ 1 1 ⎠

7 7

( )

4 6

2 −2

(g)

0 2

3 −132, −132

4

( ) ae +bg af+bh

,

ce+dg cf+dh

ad−bc,eh−fg,

(ad −bc)(eh − fg)

( ) −2 1

5 3

2 −1 a=−2,b=5

2

( ) 1 4

6 AB= , (AB)

−3 9

−1 = 1 ( ) 9 −4

,

21 3 1

B −1 = 1 ( ) 3 −1

,A

7 1 2

−1 = 1 ( ) 3 −1

3 0 1

cos ωt sinωt 0

7 ⎝−sin ωt cos ωt 0⎠

0 0 1

( ) ( ) 0 1 0 3

8 (b) ,

1 −1 3 −3

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