7. If A=[aij ]m×n , B=[bij ]m×n
Equal matrix:
Then A=B if and only if aij =bij ∀ 1≤i≤m, 1≤ j≤n
Ex 1: (Equal matrix)
A=[1 2
3 4 ] B=[a b
c d ]
If A=B
Then a=1, b=2, c=3, d=4
15. Three basic matrix operators:
(1) matrix addition
(2) scalar multiplication
(3) matrix multiplication
Zero matrix: 0m×n
Identity matrix of order n: I n
16. Then (1) A+B = B + A
(2) A + ( B + C ) = ( A + B ) + C
(3) ( cd ) A = c ( dA )
(4) 1A = A
(5) c( A+B ) = cA + cB
(6) ( c+d ) A = cA + dA
If A,B,C ∈M m×n , c,d :scalar
Properties of matrix addition and scalar multiplication:
17. If A∈Mm×n , c :scalar
Then (1) A+0m×n =A
(2) A+(− A)=0m×n
(3) cA=0m×n ⇒ c= 0 or A= 0m× n
Notes:
(1) 0m×n: the additive identity for the set of all m×n matrices
(2) –A: the additive inverse of A
Properties of zero matrices:
18. If A=
[
a11 a12 ⋯ a1n
a21 a22 ⋯ a2n
⋮ ⋮ ⋮ ⋮
am1 am2 ⋯ amn
]∈M m×n
Then AT
=
[
a11 a21 ⋯ am1
a12 a22 ⋯ am2
⋮ ⋮ ⋮ ⋮
a1n a2n ⋯ amn
]∈ M n×m
Transpose of a matrix :
Transpose is the interchange of rows and columns of a
given matrix.
19. A=[2
8] (b) A=
[1 2 3
4 5 6
7 8 9 ] (c
)
A=
[0 1
2 4
1 −1]
Sol: (a)
A=[2
8] ⇒ A
T
=[2 8 ]
(b)
A=
[1 2 3
4 5 6
7 8 9 ] ⇒ AT
=
[1 4 7
2 5 8
3 6 9 ](c
)
A=
[0 1
2 4
1 −1] ⇒ AT
=[0 2 1
1 4 −1]
(a)
Ex 8: (Find the transpose of the following matrix)
20. (1) ( A
T
)
T
=A
(2 ) ( A+B)
T
=A
T
+B
T
(3) (cA)T
=c ( AT
)
(4 ) ( AB )
T
=B
T
A
T
Properties of transposes
21. A square matrix A is symmetric if A = AT
Ex:
If A=
[1 2 3
a 4 5
b c 6 ] is symmetric, find a, b, c?
A square matrix A is skew-symmetric if AT
= –A
Skew-symmetric matrix:
Sol:
A=
[1 2 3
a 4 5
b c 6 ] AT
=
[1 a b
2 4 c
3 5 6 ] a= 2,b= 3,c= 5
A=A
T
⇒
Symmetric matrix:
22. If A=
[0 1 2
a 0 3
b c 0 ] is a skew-symmetric, find a, b, c?
Note:
AAT
is symmetric
Pf: (AAT
)T
=( AT
)T
AT
=AAT
AA∴
T
is symmetric
Sol:
A=
[0 1 2
a 0 3
b c 0 ] −AT
=
[0 −a −b
−1 0 −c
−2 −3 0 ]
A=−A
T
a=−1,b=−2,c=−3⇒
Ex:
23. ab = ba (Commutative law for multiplication)
(1)If m≠ p, then AB is defined ,BA is undefined .
(3)If m=p=n,then AB∈M m×m ,BA∈ M m×m
(2)If m=p, m≠n, then AB∈M m×m ,BA∈ M n×n (Sizes are not the
same)
(Sizes are the same, but matrices are not equal)
Real number:
Matrix:
AB≠BA
m×nn× p
Three
situations:
24. A=[1 3
2 −1] B=[2 −1
0 2 ]
Sol:
AB=[1 3
2 −1 ][2 −1
0 2 ]=[2 5
4 −4 ]
Note: AB≠BA
BA=[2 −1
0 2 ][1 3
2 −1]=[0 7
4 −2 ]
Ex 4:
Sow that AB and BA are not equal for the matrices.
and
25. (Cancellation is not
valid)
ac=bc, c≠0
⇒ a=b (Cancellation law)
Matrix:
AC=BC C ≠0
(1) If C is invertible, then A = B
Real number:
A≠B(2) If C is not invertible, then
26. 26
/61
A=[1 3
0 1 ], B=[2 4
2 3 ], C=[ 1 −2
−1 2 ]
Sol:
AC=[1 3
0 1 ][ 1 −2
−1 2 ]=[−2 4
−1 2 ]
So AC=BC
But A≠B
BC=[2 4
2 3 ][ 1 −2
−1 2 ]=[−2 4
−1 2 ]
Ex 5: (An example in which cancellation is not valid)
Show that AC=BC
Elementary Linear Algebra: Section 2.2, p.65
27. A∈M n×n
If there exists a matrix B∈ M n× n such that AB=BA=In ,
Note:
A matrix that does not have an inverse is called
noninvertible (or singular).
Consider
Then (1) A is invertible (or nonsingular)
(2) B is the inverse of A
Inverse matrix:
28. If B and C are both inverses of the matrix A, then B = C.
Pf: AB=I
C (AB)=CI
(CA)B=C
IB=C
B=C
Consequently, the inverse of a matrix is unique.
Notes:(1) The inverse of A is denoted by A
−1
(2) AA
−1
=A
−1
A=I
Thm 2.7: (The inverse of a matrix is unique)
29. [A ∣ I ]⃗Gauss-Jordan Elimination [I ∣ A−1
]
Ex 2: (Find the inverse of the matrix)
A=[ 1 4
−1 −3 ]
Sol:
AX=I
[ 1 4
−1 −3 ][x11 x12
x21 x22
]=[1 0
0 1 ]
[ x11+4x21 x12+4x22
−x11−3x21 −x12−3x22
]=[1 0
0 1 ]
Find the inverse of a matrix by Gauss-Jordan
Elimination:
34. (1) A
0
=I
(2) A
k
=AA⋯A⏟
k factors
(k> 0)
(3) Ar
⋅As
=Ar+s
r,s:integers
( A
r
)
s
=A
rs
(4) D=
[
d1 0 ⋯ 0
0 d2 ⋯ 0
⋮ ⋮ ⋱ ⋮
0 0 ⋯ dn
]⇒ Dk
=
[
d1
k
0 ⋯ 0
0 d2
k
⋯ 0
⋮ ⋮ ⋮
0 0 ⋯ dn
k ]
Power of a square matrix:
35. If A is an invertible matrix, k is a positive integer, and c is a scalar
not equal to zero, then
(1) A
−1
is invertible and ( A
−1
)
−1
=A
(2) Ak
is invertible and ( Ak
)−1
=A−1
A−1
⋯A−1
⏟
k factors
=( A−1
)k
=A−k
(3) c A is invertible and (cA)−1
=
1
c
A−1
,c≠0
(4)A
T
is invertible and ( A
T
)
−1
=(A
−1
)
T
Thm 2.8 : (Properties of inverse matrices)
36. Thm 2.9: (The inverse of a product)
If A and B are invertible matrices of size n, then AB is invertible and
(AB)
−1
=B
−1
A
−1
If AB is invertible, then its inverse is unique.
So (AB)−1
=B−1
A−1
Pf:
(AB)(B−1
A−1
)=A( BB−1
) A−1
=A(I )A−1
=( AI) A−1
=AA−1
=I
(B
−1
A
−1
)( AB)=B
−1
( A
−1
A)B=B
−1
(I )B=B
−1
( IB)=B
−1
B=I
Note:
(A1 A2 A3⋯An )
−1
=An
−1
⋯A3
−1
A2
−1
A1
−1
37. Thm 2.10 (Cancellation properties)
If C is an invertible matrix, then the following properties hold:
(1) If AC=BC, then A=B (Right cancellation property)
(2) If CA=CB, then A=B (Left cancellation property)
Pf:
AC=BC
(AC )C−1
=(BC)C−1
A(CC−1
)=B(CC−1
)
AI=BI
A=B
(C is invertible, so C
-1
exists)
Note:If C is not invertible, then cancellation is not valid.
38. Thm 2.11: (Systems of equations with unique solutions)
If A is an invertible matrix, then the system of linear equations
Ax = b has a unique solution given by
x=A
−1
b
Pf:
( A is nonsingular)
Ax=b
A−1
Ax=A−1
b
Ix=A−1
b
x=A−1
b
This solution is unique.
If x1 and x2 were two solutions of equation Ax=b .
then Ax1 =b=Ax2
⇒ x1=x2 (Left cancellation property)