Schaum's Outline of Discrete Mathematics, Third Edition (Schaum's Outlines)

APP. A] VECTORS AND MATRICES 425

A.11. Find the transpose of each matrix:

A=

[

1 − 23

78 − 9

]

; B=

⎡

⎣

123

245

356

⎤

⎦; C=[ 1 ,− 3 , 5 ,− 7 ]; D=

⎡

⎣

2

− 4

6

⎤

⎦

Rewrite the rows of each matrix as columns to obtain the transposes of the matrices:

AT=

⎡

⎣

17

− 28

3 − 9

⎤

⎦,BT=

⎡

⎣

123

245

356

⎤

⎦,CT=

⎡

⎢⎢

⎣

1

− 3

5

− 7

⎤

⎥⎥

⎦,D

T=[ 2 ,− 4 , 6 ]

(Note thatBT=B; such a matrix is said to besymmetric. Note also that the transpose of the row vectorCis a column

vector, and the transpose of the column vectorDis a row vector.)

A.12. Prove Theorem A.2(i):A(BC)=A(BC).

LetA=[aij],B=[bjk], andC=[ckl]. Furthermore, letAB=S=[sik]andBC=T=[tjl].

Then

sik=ai 1 b 1 k+ai 2 b 2 k+···+aimbmk=

∑m

j= 1

aijbjk

tjl=bj 1 c 1 i+bj 2 c 2 i+···+bjncnl=

∑n

k=l

bjkckl

Now, multiplyingSbyC, i.e., (AB)byC, the element in theith row andlth column of the matrix(AB)Cis

si 1 c 1 l+si 2 c 2 l+···+sincnl=

∑n

k= 1

sikckl=

∑n

k= 1

∑m

j= 1

(aijbjk)ckl

On the other hand, multiplyingAbyT, i.e.,AbyBC, the element in theith row andlth column of the matrix

A(BC)is

ai 1 t 1 l+ai 2 t 2 l+···+aimtml=

∑m

j= 1

aijtjl=

∑m

k= 1

∑n

j= 1

aij(bjkckl)

Since the above sums are equal, the theorem is proven.

SQUARE MATRICES, DETERMINANTS, INVERSES

A.13. Find the diagonal and trace of each matrix:

(a)A=

⎡

⎣

136

2 − 58

4 − 27

⎤

⎦;(b)B=

[

t− 23

− 4 t+ 5

]

;(c)C=

[

12 − 3

4 − 56

]

.

(a) The diagonal consists of the elementsa 11 ,a 22 ,a 33 , that is, the scalars 1,− 5 ,7. The trace is the sum of the diagonal

elements; hence tr(A)= 1 − 5 + 7 =3.

(b) The diagonal consists of the pair{t− 2 ,t+ 5 }. Thus tr(B)=t− 2 +t+ 5 = 2 t+3.

(c) The diagonal and trace are defined only for square matrices.

A.14. Let A =

[

12

4 − 3

]

. Find: (a)A^2 ;(b)A^3 ;(c)f (A) where f(x) = 2 x^3 − 4 x +5;

(d) g(A)whereg(x)=x^2 + 2 x−11.

(a) A^2 =AA=

[

12

4 − 3

][

12

4 − 3

]

=

[

1 + 82 − 6

4 −12 8+ 9

]

=

[

9 − 4

− 817

]

.

(b) A^3 =AA^2 =

[

12

4 − 3

][

9 − 4

− 817

]

=

[

9 − 16 − 4 + 34

36 + 24 − 16 − 51

]

=

[

− 730

60 − 67

]

.