(46)-线代2矩阵1笔记版_08.2026考研数学高途王喆全程班_赠送2025课程_25考研数学（三）全年智达班_{2}--资料

2025第二章
矩阵一、矩阵的运算
 a + b a + b a + b 
11 11 12 12 1n 1n
 
a + b a + b a + b
1.加法A + B = (a + b )
=
 21 21 22 22 2n 2n 
mn mn ij ij mn
 
 
a + b a + b a + b
 
m1 m1 m2 m2 mn mn
mn
 a a a 
↳
11 12 1n
 
a a a
 21 22 2n 
2.数乘A = (a ) =
ij mn  
 
a a a
 
m1 m2 mn
mn
SE 0 * *E : XA An #E : XA = 0
= =
=> X = 0 A = 0二、矩阵的乘法
设A ,B , 则AB = C = (c ) ,其中
ms sn mn ij mn
s

c = a b + a b + + a b = a b ,(i = 1,2, ,m; j = 1,2, ,n).
ij i1 1 j i2 2 j is sj ik kj
k=1
: ① r(Al
r(B)
= S
& B3/7
+
BE - AX 07.
=
EIAE A B # A B
& . = 0 = o = 0
YEAAB
B
= 0 = =0
③ EEE ABC CAB) C A (BL
= . =
三、方阵的幂 Ak = A  A  ... A称为方阵 A的k次幂.
k个
(AB)" # A 4. B" (A + Bl = A + 2 AB + B2
12
A
AB BA
= + + +特殊矩阵——成比例矩阵(秩 1 矩阵)
/]x35 EXEXER(4/EBE
1 3 : =
.
( =( ⑳b Gibz Gibs
- dbb - I , I
Grh ⑫ Grbs
,
asb
anb b
,
(
) 3) , )(213)
# A
=
=
63 9
<JAX5= * 2THAE +r(2-BT)
2. = =
a B = (G , G2as)(b) = b + + -
- -特殊矩阵——成比例矩阵(秩 1 矩阵)
trA)
B XEX9 EBE /AV A
A = ·
3
,
.
nyn
A x93
it ::
BT BT G tULA)
A 2 =
. = . .
,
n-
= A" = J . B . d . BT ... f . B = d . tr(A) . B T
-
nu
1
n- n- 1
BT
+r(A) 2
= . . = +(A) A四、矩阵的转置
 a a a   a a a 
11 12 1n 11 21 m1
   
a a a a a a
设A =  21 22 2n  ,则AT =  12 22 m2 
mn   nm  
   
a a a a a a
   
m1 m2 mn 1n 2n mn
mn nm
A A + A Anxn (AT) (A)
=
-
,
JERBE
# A A EJA
=
,运算规律
(1) (AT )T = A；
(2) (A + B)T = AT + BT ； 1555 . *RE, MurETT
(3)
(A)T
=
AT
；
(4)
(AB)T
=
BT AT
,
推广：(A A A )T = A TA T A T ,
1 2 k k k−1 1
(Ak )T
=
(AT )k
. |A4 ||" (A + 14 (A*
+
=
=
五、对称矩阵 AT = A.
(* ( (+ 14
=
六、反对称矩阵 AT = −A.第二部分 题型解析
题型一、求方阵的 n
(d
次方(★★) :
解题思路：求An
有如下几种思路：
O
(i)
思路 1——找规律法.
G
思路 2——如果A为成比例矩阵(或秩为 1 矩阵)，则A = αβT ，其中α,β
为列向量，且An
=
[β,α]n−1A
=
tr(A)n−1A.
思路 3——如果A = kE + B，其中B为幂零矩阵，可用二项式展开定理
" KEY CrDER ! CrEl? B2
计算An
.
A
= (kE + B) =
Cr
· +
B
+ =...
思路
4——如果A可相似对角化，即P−1AP
=
Λ，则An
=
PΛnP−1
.
思路
5——如果A是分块对角阵，可用分块对角阵性质计算An
. 1 2 3
 
【例2.1】 设 A = 2 4 6 ,求 An .
 
 
3 6 9
 
23
ti . 14 % ) I
= A A
= = 24s
369
123
I I (2)(123)
a
FRA 2
= 2 + 0 = = .
369
(123)(z)
<T
2 14
· =
. =
diy CT d2i 2 1422T 14 (ii)
C 2 <
: A = = . ... . = = 1 2 3
 
【例2.2】 设 A = 0 1 4 ，求 An .
 
 
0 0 1
 
( 1)
B
A = + = E +
/ =
B =
& B B ):08)(=
= = 8
= 1 2 3
 
【例2.2】 设 A = 0 1 4 ，求 An .
 
 
0 0 1
 
? B2
B) Cr ? ChE B Cn En
A (E + = · E + + ·
: =
+) B
n(
B
E + n- + .
=
I I O O I " "() + I
I
+
o · :
O 8 I 00 o
4
I 24 - n
I
I
=
O
I
44
O
6
I 2 1 1
 
【例2.3】 设 A = 1 2 1 ,求An.
 
 
1 1 2
 
#
: #(A -XE) = 0 = x = = x z = 1x3 = 4
)i)
(i) dr
Ex xn 1 At (A - E(X = 0 = d. = =
= = .
Ex3 4A5 (A-4E)X = 0 + * = ) , )
=
,
(P (6 dats)
=I
,) " )
= . = P
0
44
/ i
=
mp + Ap A P Xpt A Pap 441
x = - =
=
,
,
4 14"
1472
-题型二、矩阵的逆(★★★)
一、可逆矩阵 对于 n 阶方阵 A ，如果有一个 n 阶方阵 B ，使
AB = BA = E，则称矩阵 A 是可逆的，并把 B 称为 A 的逆矩阵.
二、矩阵可逆的条件 n 阶方阵 A 可逆的充分必要条件是 A  0,即 A
. - 1
a = =
是
1
非奇异矩阵，且A−1
= A*.
A
<ElABI /BAKE)
=> A FEB = => (A1 1B1 = 1
.
=> (A) + 0三、可逆矩阵的性质
1.(A−1)−1 = A;
1
2.
A−1
= ;
A
1
3.(A)−1
=
A−1
;

4.(AT )−1
=
(A−1)T
;
B)
5.(AB)−1
=
B−1A−1
;
(A
= + A
+ BT
=
推广(1) (A A A )−1 = A −1A −1 A −1 ；
1 2 m m m−1 1
(2)
(An )−1
=
(A−1)n
.四、分块对角阵的逆
−1
 A    A−1  
1 1
   
A A−1
1.  2  =  2  .
   
   
 A  A−1
   
s s
−1
 A   A −1 
1 n
   
A
 2   
2. = .
   A −1 
2
   
A A −1
   
n 1思路 1——判断A可逆的方法主要有
1.定义法；2.| A | 0，这是主要方法.
思路 2——求逆矩阵的方法有：
1.定义法：如果有一个 n 阶方阵 B ，使AB = BA = E，则B = A−1 .一般适
用于抽象矩阵.
1
2.伴随法：A−1
=
A*
，一般适用二阶或者三阶数值型矩阵；
A
( )
3.初等行变化法：
(
A | E
)
⎯
行
⎯→ E | A
−1
，一般适用于三阶及
n n n n
以上的数值型矩阵.−1 −1
A O A−1 O  O A  O B−1 
4.分块矩阵法： = ， = ，其
       
O B O B−1 B O A−1 O
       
中A,B均可逆.
思路 3——求逆矩阵的运算要充分利用逆矩阵的性质与“加变乘”思
想来进行计算.【例2.4】
下列命题正确的是( ).
(A) 若AB = E，则 A
D
可逆，且A−1 = B
(B) 若A,B均为n阶可逆矩阵，则A + B必可逆
(C) 若A,B均为n阶不可逆矩阵，则A − B必不可逆
(D) 若A,B均为n阶不可逆矩阵，则AB必不可逆.
-A
BBB#. R(1
:
A (1 1) B
=
(i) AB = C = E,
(A) =
.
.
(2( A ) o B = ) = j)A + B = 7 % %
(B) = =
01
(c) (B A 1 % ) B = (89) A - B = (0 -)
= =
(b) A B / * = (A) =0 (B1 =0 111 (ABI = (A) 1B1 = 0 = AB7
. . ,【例2.5】 设 A为n阶非奇异矩阵，为 n 维列向量， b 为常数.记分块矩
 E O   A 
阵P = ,Q = ，其中 A 是矩阵 A的伴随矩阵,E 为
   
− T A A  T b
   
n
阶单位矩阵. (1) 计算并化简PQ；
my)
(in))) (
pa
= =
- - a
+
# A 1A)E #* 1AI At
= = .
*
A 2 )o A2)
I
PQ b)
:
=
=
#
O - + 1
.(2) 证明：矩阵Q可逆的充分必要条件是 T A−1  b.
T iR35
7 F
[ii] (Q)
, ,
IPQI
IQ)
=
IP/
E 8
LERA (Pl (E) /All (A)
: = = · =
*
GTA IAI
-
A
2
IPQ1 = (A) 1 kI(b-2TA2) P (b-LTA)
= . =
.
IAI(b-2A+
d)
0
=
+
(b 2TA 2)
(a) =
: . -
AFF Q7 1Q1 LA b
-: : /Alto : + 0 E【例2.6】 设n阶矩阵 A ， A2 + A − 4E = 0，证明 A + 2E可逆，并求
( )−1
A + 2E .
A
LEDA : : + A - 4E = 0
E)
(A zE
(A 2E) - 0
=
.: + -
(A 2E) (A-El 2 E
. + =
<A-E)
(A 2E)
: + E
=
z
ATLE LET A-E
:
=【例2.7】 设n阶矩阵 A ( )2 对称，可逆，且 A + B = E，证明:
( )−1 ( )T
E + A−1B E − BT A−1 = ( A + B )( A − B ) .
+
BIT. BTA" (A + B) (E (BA+ (T)
(E A (E = A + A -
+ -
(AtBIJ" (E- (A+P (BT/T) CAtBIT CE- (AT? B)
[AT A
= · = . .
Bit A B) CA B1" (A -B)
= (A + A)E - = + ·
=
= (A + 1) = (A + B) · (A + B) = E
+
(a B) (A + BI
= + =
F
· (A + B) (A - B) =
= 2
. 0 0 0 1 2
 
0 0 0 3 4
 
【例2.8】 设 A =  2 1 1 0 0，求 A−1 .
 
1 2 1 0 0
 
 
1 1 2 0 0
 
2 I
I
))
(ii)
=) n,) m A
=
iSAi
=
112
#Al = () (
=
=-11 1 20 I
I 2 I , 10 : ner) - I
(AzIE)
=
12 12 1 0 1 0
1200
1
211 10 0
Vi
r - 12 o ↓ 12 E E I
re-): I
O 1 + 01 +
00 - 4 11 -3
2 2
I 10 II -II
12 o o r
r (4) 131
01 + 10 -I -
o
o
13
I
·
001
-k
-4
I
3 - 2
-
g 8 F
I
I
I
r - 2 - -I I I I -
. ! · o Al)0 All O O -I
I
3 : I
3
- I
I - 08
- - 5
o 10 -
I 3 Ai 8 - 2 10 =
-k
01
· - 4
-
8 j题型三、伴随矩阵(★★★)
 a a a 
11 12 1n
 
a a a
 21 22 2n 
一、伴随矩阵 设A = 则其伴随矩阵
 
 
a a a
 
n1 n2 nn
 A A A 
11 21 n1
 
A A A
 12 22 n2 
A* =
 
 
A A A
 
1n 2n nn*
it A. A A A IAI E
&E = = .
:
an
I
LEA A. A
=
:
i
Uni
O
- ( I m)0 I O ---
I
=
0 = E
.... 0
"
"
....
.....二、伴随矩阵的性质 设 A 为n阶可逆矩阵，则
1.A* =| A | A−1 ; (A7) n r(A)
, = m
I
*
r(A )
=
1 r(A)
=
n+
,
2. AA* = A*A = A E（恒成立） r(Akn-
o
,
+
3. A* = A n−1 （恒成立） D A ** A]
,
* 1 = (IA) · Al = 1AP-IA") = 1AP
② AX7 (E) MA* E **, Al (A*
= o = 0
,
, , .
1
"+
4. (A*)−1 = (A−1)* = A : = 1A1
A
" )" = A =
(A) A+
=
* )
(A) (Y) ( + Y A
=
. =( )T ( )*
5.
A*
=
AT
.
*
( )
6. kA
⑨*
= k
n−1
A
*
7. (A*)* =| A |n−2 A(n  2)
8. ( AB )* = B*A* .
推广:(1) (A A A )* = A *A * A * .
1 2 m m m−1 1
(2)
(An )*
=
(A*)n解题思路：伴随矩阵是线性代数的常考点之一，解此类问题的思路：
思路 1——熟记伴随矩阵定义两大要素：代数余子式、转置排列；
思路 2——掌握基本公式AA* = A*A = A E 及在 A 可逆时
A* =| A | A−1 ，并能够将其作各种恒等变形推导出伴随矩阵和逆矩阵的
各种关系式.【例2.9】 设n阶矩阵 A 非奇异( n  2 )， A  是矩阵 A 的伴随矩阵，则
C
( ).
( ) n−1 ( ) n+1
(A) A = A A (B) A = A A
( ) n−2 ( ) n+2
(C) A = A A (D) A = A A
* "
(** ) 11 L* 1AM. AM ? A
= . = =【例2.10】 设 A, B均为 2 阶矩阵， A*, B* 分别为 A , B 的伴随矩阵，若
O A
A = 2, B = 3，则分块矩阵 的伴随矩阵为（ B ）.
 
B O
 
 O 3B*   O 2B*   O 3A*   O 2A* 
(A) (B) (C) (D) .
       
2A* O 3A* O 2B* O 3B* O
       
A
O ** (
(Al1B1
(l = 2x3 = 6 +0 .
=
Bo
(
)
(A 7 = =B
·
C
=
I
-