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Matrices: §2.2 Properties of Matrices
Satya Mandal, KU
Satya Mandal, KU
Matrices: §2.2 Properties of Matrices
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Goals
We will describe properties of matrices with respect to
addition, scalar multiplications and matrix multiplication and
others. Among what we will see
1. Matrix multiplication does not commute. That means,
not always AB = BA.
2. We will define transpose AT of a matrix A and discuss
its properties.
Satya Mandal, KU
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Algebra of Matrices
Let A, B, C be m × n matrices and c, d be scalars. Then,
A+B =B +A
A + (B + C ) = (A + B) + C
(cd)A = c(dA)
c(A + B) = cA + cB
(c + d)A = cA + dA
Commutativity of addition
Associativity of addition
Associativity of scalar multiplication
a Distributive property
a Distributive property
These seem obvious, expected and are easy to prove.
Satya Mandal, KU
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Zero
The m × n matrix with all entries zero is denoted by Omn . For
a matrix A of size m × n and a scalar c, we have
◮ A + Omn = A (This property is stated as:Omn is the
additive identity in the set of all m × n matrices.)
◮ A + (−A) = Omn . (This property is stated as:−A is the
additive inverse of A.)
◮ cA = Omn
=⇒ c = 0 or A = Omn .
Remark. So far, it appears that matrices behave like real
numbers.
Reading Assignment: Example 2 from §2.2 of the textbook.
Satya Mandal, KU
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Properties of Matrix Multiplication
Let A, B, C be matrices and c is a constant. Assume all the
matrix products below are defined. Then
A(BC ) = (AB)C
Associativity Matrix Product
A(B + C ) = AB + AC
Distributive Property
(A + B)C = AC + BC
Distributive Property
c(AB) = (cA)B = A(cB)
Proofs would be routine checking, which we would skip.
Reading Assignment: Example 3, 4, 5 from §2.2 of the
textbook.
Satya Mandal, KU
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Definition
For a positive integer, In would denote the square matrix of
order n whose main diagonal (left to right) entries are 1 and
rest of the entries are zero. So,


1 0 0
1 0
, I3 =  0 1 0 
I1 = [1], I2 =
0 1
0 0 1
Satya Mandal, KU
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Properties of the Identity Matrix
Let A be a m × n matrix. Then
◮ AIn = A
◮ Im A = a
◮ If A is a square matrix of size n × n, then
AIn = In A = A.
◮
In is called the Identity matrix of order n. Because of
above, we say that In is the multiplicative identity for the
set of all square matrices of order n.
Satya Mandal, KU
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Proof.
Proof. We will prove for 3 × 3 matrices A. Write


a b c
So,
A= u v w 
x y z



 
a b c
a b c
1 0 0
AI3 =  u v w   0 1 0  =  u v w  = A
x y z
x y z
0 0 1
Similarly, I3 A = A. The proof is complete.
Satya Mandal, KU
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Use of Matrix Algebra to solve
systmes of linear equation
Now that we know some Matrix Algebra, we will use this to
give a proof of the following theorem that was stated before:
Theorem. For a system of linear equations in n variables,
precisely one of the following is true:
◮ The system has exactly one solution.
◮ The system has an infinite number of solutions.
◮ The system has no solution.
Satya Mandal, KU
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The Proof.
Write the equation in the matrix form Ax = b. If the first or
the last statements are true then there is nothing to prove.
So, assume both are false. So, the system has at least two
distint solutions x1 , x2 with x1 6= x2 . So,
Ax1 = b and
Ax2 = b.
With y = x1 − x2 6= 0 we have
Ay = A(x1 − x2 ) = Ax1 − Ax2 = b − b = 0.
Satya Mandal, KU
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For any scalar c, we have
A(x1 + cy) = Ax1 + cAy = b + 0 = b
So, x1 + cy is a solution of the given system Ax = b, for all
scalars c, which is infinitely many. The proof is complete.
Satya Mandal, KU
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Definition
Algebra of Transpose
Definition of Transpose of a Matrix
Definition. Given a m × n matrix A, the transpose of A,
denoted by AT , is formed by writing the the columns of A as
rows (equivalently, writing the rows as columns). So,
transpose AT of


a11 a12 a13 · · · a1n
 a21 a22 a13 · · · a2n 


 an m × n matrix
a
a
a
·
·
·
a
A=
31
32
33
3n


 ··· ··· ··· ··· ··· 
am1 am2 am3 · · · amn
is given by:
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Definition
Algebra of Transpose
Transpose of a Matrix: Continued



A =


T
a11
a12
a13
···
a1n
a21
a22
a23
···
a2n
a31
a32
a33
···
a3n
···
···
···
···
···
am1
am2
am3
···
amn






an n × m matrix
Reading Assignment: Example 8 from §2.2 of the textbook.
Satya Mandal, KU
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Definition
Algebra of Transpose
Properties of Transpose
Let A, B be matrices and c be a scalar. Assume all the
multiplications below are defined. Then,
(AT )T = A
(A + B)T = AT + B T
(cA)T = cAT
(AB)T = B T AT
Double transpose of A is itself
transpose of sum
transpose of scalar multiplication
transpose of product.
Again, to prove we check entry-wise equalities.
Reading Assignment: Example 9 from §2.2 of the textbook.
Satya Mandal, KU
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Example: Noncommutativity
Example: Non-Cancellation
Let me draw your attention, how algebra of matrices differ
from that of the algebra of real numbers:
◮ Matrix product is not commutative. That means
AB 6= BA, for some matrices A, B. See §2.2 Example 4.
◮ Cancellation property fails. That means there are
matrices A, B, C , with C 6= 0, such that
AC = BC
but
A 6= B.
See §2.2 Example 5.
Satya Mandal, KU
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Example: Noncommutativity
Example: Non-Cancellation
Example of noncommutativity AB 6= BA:
We have
−1 1
1 −1
0 1
=
1 1
1 1
1 0
0 1
1 −1
1 1
=
1 0
1 1
1 −1
Right hand sides of these two equations are not equal. So,
commitativity fails for these two matrices.
Satya Mandal, KU
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Example: Noncommutativity
Example: Non-Cancellation
Example: AC = BC but A 6= B:
We have
1 1
2 2
1
1
0 0
1
1
=
=
1 1
2 2
−1 −1
−1 −1
0 0
So, cancellation property fails for matrix product.
Satya Mandal, KU
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Solve for Matrix X
Solve for the matrix X when


−1 1
A =  3 −1 
2 −1
◮
◮
◮

2 3
B =  4 −1 
11 −1

(1) 5X + 3A = 2B
(2) 2B + 4A = 5X
(3) X + 2A − 2B = O
Satya Mandal, KU
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Solution
◮
For (1)
3
2
X = B− A
5
5




−1 1
2 3
= .4  4 −1  − .6  3 −1 
2 −1
11 −1


1.4 .6

= −.2 .2 
3.2 .2
5X + 3A = 2B =⇒
Satya Mandal, KU
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Solution: Continued
◮
◮
For (2), X = 52 B + 45 A =

 



0
2
−1 1
2 3
.4  4 −1  + .8  3 −1  =  4 −1.2 
6 −1.2
2 −1
11 −1
For (3) X = −2A + 2B =

 



6 4
2 3
−1 1
−2  3 −1  + 2  4 −1  =  2 0 
18 0
11 −1
2 −1
Satya Mandal, KU
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Example
Let


2 3
7
A =  4 −1 19  , B = 
11 −1 −19

2 −1 1
C = 0 0 0
0 0 0

Demonstrate AC = BC bur A 6= B.
Satya Mandal, KU

2 −1 1
4 3 −1  ,
11 2 −1


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Solution:
We have


2
2 3
7
AC =  4 −1 19   0
0
11 −1 −19


2
2 −1 1
BC =  4 3 −1   0
0
11 2 −1
So, AC = BC .
Satya Mandal, KU
 
4
−1 1
0 0 = 8
22
0 0
 
4
−1 1
0 0 = 8
22
0 0

−2
2
−4
4 ,
−11 −11

−2
2
−4
4 
−11 −11
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Example
Let f (x) = x 2 − 2x + 2

4 −2 0
A =  8 −4 4 
22 0 0

Compute f (A).
Solution: We have

 


0
0
−8
4 −2 0
4 −2 0
A2 =  8 −4 4   8 −4 4  =  88 0 −16 
88 −44 0
22 0 0
22 0 0
Satya Mandal, KU
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Solution: Continued
Let f (A) = A2 − 2A + 2I3 =



0
0
−8
 88 0 −16  − 2 
88 −44 0

−6

72
=
44
Satya Mandal, KU



1 0 0
4 −2 0
8 −4 4  + 2  0 1 0 
0 0 1
22 0 0

4
−8
10 −24 
−44 2
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§2.2: Exwecise 14a, 14c, 15, 19, 23, 25, 31, 32, 33, 34, 55
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Matrices: §2.2 Properties of Matrices