¹x×ç Íù Übç Homework 1-Solution 1 (~l*3 2(a) 4 æ£ (T.6), wìÑ 9/16 (û) í,{ª)
1. (a) C + E = E + C =
5 −5 8 4 2 9 5 3 4
(b) ÄÑ A ¸ B í×ü.°, FJ A + B .æÊ
(c) D − F = 3 − (−4) −2 − 5 2 − 2 4 − 3
= 7 −7 0 1
.
(d) −3C + 5O =
3 · 3 3 · (−2) 3 · 3 3 · 4 3 · 1 3 · 5 3 · 2 3 · 1 3 · 3
−
0 0 0 0 0 0 0 0 0
=
−9 3 −9
−12 −3 −15
−6 −3 −9
.
(e) 2C − 3E =
6 −2 6 8 2 10 4 2 6
−
6 −12 15 0 3 12
9 6 3
=
6 − 6 −2 − (−12) 6 − 15 8 − 0 2 − 3 10 − 12 4 − 9 2 − 6 6 − 3
=
0 10 −9 8 −1 −2
−5 −4 3
.
(f) ÄÑ B ¸ F ×ü.°, ̶ó‹, FJ 2B + F .æÊ
2. (a) AT =
1 2 2 1 3 4
, and (AT)T =
1 2 2 1 3 4
T
= 1 2 3 2 1 4
= A.
(b) (C +E)T =
3 −1 3 4 1 5 2 1 3
+
2 −4 5 0 1 4 3 2 1
T
=
5 −5 8 4 2 9 5 3 4
T
=
5 4 5
−5 2 3 8 9 4
CT+
ET =
3 −1 3 4 1 5 2 1 3
T
+
2 −4 5 0 1 4 3 2 1
T
=
3 4 2
−1 1 1 3 5 3
+
2 0 3
−4 1 2 5 4 1
=
5 4 5
−5 2 3 8 9 4
.
(c) (2D + 3F )T = 6 −4 4 8
+ −12 15 6 9
T
= −6 11 10 17
T
= −6 10 11 17
. (d) D − DT = 3 −2
2 4
−
3 2
−2 4
= 0 −4 4 0
.
(e) 2AT + B = 2
1 2 2 1 3 4
+
1 0 2 1 3 2
=
2 4 4 2 6 8
+
1 0 2 1 3 2
=
3 4 6 3 9 10
.
(f) (3D − 2F )T = 9 −6 6 12
− −8 10 4 6
T
= 17 −16
2 6
T
=
17 2
−16 6
.
3. (a) (2A)T = 2 4 6 4 2 8
T
=
2 4 4 2 6 8
.
(b) ÄÑ A ¸ B ×ü.°Ì¶óÁ, A − B .æÊ, FJ (A − B)T ?.æÊ
(c) (3BT − 2A)T =
3 1 2 3 0 1 2
−2 1 2 3 2 1 4
T
= 3 6 9 0 3 6
− 2 4 6 4 2 8
T
=
1 2 3
−4 1 −2
T
=
1 −4 2 1 3 −2
.
¹x×ç Íù Übç Homework 1-Solution 2 (d) ÄÑ A ¸ B ×ü.°, FJ AT ¸ BT í×ü?.ó°Ì¶óÁ, FJ (3AT −5BT)T .æÊ
(e) (−A)T = −1 −2 −3
−2 −1 −4
T
=
−1 −2
−2 −1
−3 −4
; −(AT) = −
1 2 2 1 3 4
=
−1 −2
−2 −1
−3 −4
(f) ÄÑ CE ¸ FT í×ü.°, FJ (C + E + FT)T .æÊ
4. (a) a12= −3, a22= −5, a23= 4 (b) b11 = 4, b31 = 5
(c) c13= 2, c31= 6, c33= −1
5. (a) A + B =
1 0 1 1 1 0 0 1 1
+
0 1 1 1 0 1 1 1 0
=
1 1 2 2 1 1 1 2 1
.
(b) B + C =
0 1 1 1 0 1 1 1 0
+
1 1 0 0 1 1 1 0 1
=
1 2 1 1 1 2 2 1 1
.
(c) A + B + C =
1 0 1 1 1 0 0 1 1
+
0 1 1 1 0 1 1 1 0
+
1 1 0 0 1 1 1 0 1
=
2 2 2 2 2 2 2 2 2
.
(d) A + CT =
1 0 1 1 1 0 0 1 1
+
1 0 1 1 1 0 0 1 1
=
2 0 2 2 2 0 0 2 2
.
(e) B − C =
0 1 1 1 0 1 1 1 0
−
1 1 0 0 1 1 1 0 1
=
−1 0 1 1 −1 0 0 1 −1
.
6. (a) B = −A = −1 0 0 0
. (b) C = 1 1
1 1
−A= 0 1 1 1
. Theoretical Exercises
T.3 I
A=
a b c c d e e e f
.
(a) A − AT =
a b c c d e e e f
−
a c e b d e c e f
=
0 b − c c − e c − b 0 0 e − c 0 0
.
(b) A + AT
a b c c d e e e f
+
a c e b d e c e f
=
2a b+ c c + e c+ b 2d 2e e+ c 2e 2f
.
(c) (A + AT)T =
2a b+ c c + e c+ b 2d 2e e+ c 2e 2f
T
=
2a c+ b e + c b+ c 2d 2e c+ e 2e 2f
.
(·<: (A + AT)T = A + AT.)
¹x×ç Íù Übç Homework 1-Solution 3 T.6 J A Ñ,úiä³ (upper triangular matrix), † aij = 0, ç i > jv, ] AT = [aTij], w2 aTij = aji/ aTij = aji = 0 ç j > i v (·<¤ví i ¸ j [Wb¸b), FJÊ AT 25jÖFÊ5 Wb×kbv(à:a23), wMÑÉ (¹AT23úi(¬,j5jÖîÑ0), ] AT Ñ-úiä³(lower triangular matrix) FJJä³ A Ñ,úiä³ (upper triangular matrix), †wā0ä³ AT Ñ- úiä³(lower triangular matrix)
T.7 A − AT 23úi((main diagonal) 5jÖ (entry) îÑÉ I A = [aij] Ñø_ n ¼j³ (square matrix of order n), / AT = [aTij] Ñ A íā0ä³ (the transpose of A), w2 aTij = aji. I B = [bij] = A − AT, † B 253úi(5jÖÑ
bii= aii−aTii = aii−aii= 0. (1 ≤ i ≤ n).
C6, 5ªâJ-)ø: I
A=
a11 a12 · · · a1n
a21 a22 · · · a2n
... ... ...
an1 an2 · · · ann
.
Ñ n ¼j³, †
A − AT =
a11 a12 · · · a1n
a21 a22 · · · a2n
... ... ... an1 an2 · · · ann
−
a11 a21 · · · an1 a12 a22 · · · an2 ... ... ... a1n a2n · · · ann
=
0 a12−a21 a13−a31 · · · a1n−an1 a21−a12 0 a23−a32 · · · a2n−an2 a31−a13 a32−a23 0 · · · a3n−an3
... ... ... . .. ... an1−a1n an2−a2n an3−a3n · · · 0
w23úi( (main diagonal) 5MÑÉ (zeros).
