MATHS MATHS PRESENTATIONPRESENTATION

GROUP MEMBERS : DRISHTI (1838)

VANDANA (1831)

DIMPY(1833)

Operations with Matrices Properties of Matrix Operations The Inverse of a Matrix

MATRICES

Matrix:

A= [ a ij ]=[a11 a12 a13 ⋯ a1n

a21 a22 a 23 ⋯ a2n

a31 a32 a33 ⋯ a3n

⋮ ⋮ ⋮ ⋮am1 am2 am3 ⋯ amn

]m× n

∈ M m×n

(i, j)-th entry: a ij

row: m

column: n

size: m×n

i-th row vector

r i= [ ai1 a i2 ⋯ a in ]

j-th column vector

c j=[ c1j

c 2j

⋮cmj

]row matrix

column matrix

Square matrix: m = n

Diagonal matrix:

A=diag ( d 1 ,d 2 ,⋯ ,d n) =[d1 0 … 00 d 2 ⋯ 0⋮ ⋮ ⋱ ⋮0 0 ⋯ d n

]∈M n×n

Trace:

If A= [ aij ]n×n

Then Tr ( A )=a11 +a 22 +⋯ +ann

Ex:

A= [ 1 2 34 5 6 ] =[r1

r2 ]

= [c 1 c 2 c 3 ]

r 1= [1 2 3 ] , r 2= [ 4 5 6 ]

c1=[14 ] , c2=[25 ] , c3=[36 ]A= [ 1 2 3

4 5 6 ]⇒

⇒

If A= [ a ij ]m× n , B= [ b ij ]m×n

Equal matrix:

Then A=B if and only if aij =b ij ∀ 1≤ i≤ m, 1≤ j≤ n

Ex 1: (Equal matrix)

A= [1 23 4 ] B= [a b

c d ]If A=B

Then a= 1, b= 2, c= 3, d= 4

Matrix addition:

If A= [ a ij ]m× n , B= [ bij ]m×n

Then A+B= [ a ij ]m×n +[ b ij ]m×n=[ aij +b ij ]m×n

Ex 2: (Matrix addition)

[−1 20 1 ]+[ 1 3

−1 2 ]=[−1+1 2+30−1 1+2 ]=[ 0 5

−1 3 ]

[ 1−3−2 ]+[−1

32 ]= [ 1−1

−3+3−2+2 ] =[000 ]

Matrix subtraction:A−B=A+(−1)B

Scalar multiplication:If A= [ aij ]m× n , c : scalar

Ex 3: (Scalar multiplication and matrix subtraction)

A= [ 1 2 4−3 0 −12 1 2 ] B= [ 2 0 0

1 −4 3−1 3 2 ]

Find (a) 3A, (b) –B, (c) 3A – B

Then cA= [ caij ]m× n

(a)

3A=3[ 1 2 4−3 0 −12 1 2 ]

(b)

−B= (−1 )[ 2 0 01 −4 3

−1 3 2 ](c)

3A−B= [ 3 6 12−9 0 −36 3 6 ]−[ 2 0 0

1 −4 3−1 3 2 ]

Sol:

=[ 3 6 12−9 0 −36 3 6 ]=[ 3 (1 ) 3 ( 2 ) 3 ( 4 )

3 (−3 ) 3 (0 ) 3 (−1 )3 ( 2 ) 3 (1 ) 3 ( 2 ) ]=[−2 0 0

−1 4 −31 −3 −2 ]

=[ 1 6 12−10 4 −6

7 0 4 ]

Matrix multiplication:If A= [ aij ]m× n , B= [ bij ]n× p

Then AB= [ aij ]m×n [ b ij ]n× p=[ c ij ]m× p

cij=∑k=1

n

aik bkj =ai1 b1j+ai2 b2j+⋯+ain bnjwhere

Notes: (1) A+B = B+A, (2) AB ≠ BA

Size of AB

[a11 a12 ⋯ a1n

⋮ ⋮ ⋮a i1 a i2 ⋯ a in

⋮an1

⋮an2

⋯⋮

ann] [b11 ⋯ b1j ⋯ b1n

b21 ⋮ b2j ⋯ b2n

⋮ ⋮ ⋮ ⋮bn1 ⋯ bnj ⋯ bnn

]=[ c i1 c i2 ⋯ c ij ⋯ cin ]

A= [−1 34 −25 0 ] B= [−3 2

−4 1 ] Ex 4: (Find AB)

Sol:

AB= [(−1)(−3)+(3)(−4 ) (−1 )(2 )+(3)(1 )( 4 )(−3)+(−2 )(−4) (4 )( 2)+(−2 )(1)(5)(−3 )+(0 )(−4 ) (5 )(2 )+( 0)(1 ) ]

=[ −9 1−4 6−15 10 ]

Matrix form of a system of linear equations:

{ a11 x1 +a12 x2+⋯+a1n xn =b1

a21 x1 +a22 x2+⋯+a2n x n =b2

⋮am1 x1 +am2 x 2+⋯+amn xn =bm

}

= = =

A x b

m linear equations

Single m atrix equation

Ax =bm ×nn×1 m ×1[ a11 a12 ⋯ a1n

a21 a22 ⋯ a2n

⋮ ⋮ ⋮ ⋮am1 am2 ⋯ amn

] [ x1

x 2

⋮x n

]=[b1

b2

⋮bm

]⇓

Partitioned matrices:

A= [ a11 a12 a13 a14

a21 a22 a23 a24

a31 a32 a33 a34]=[A11 A12

A21 A22 ]submatrix

A= [ a11 a12 a13 a14

a21 a22 a23 a24

a31 a32 a33 a34]=[r 1

r 2

r 3]

A= [ a11 a12 a13 a14

a21 a22 a23 a24

a31 a32 a33 a34]=[ c1 c 2 c3 c4 ]

Three basic matrix operators: (1) matrix addition (2) scalar multiplication (3) matrix multiplication

Zero matrix: 0m×n

Identity matrix of order n: I n

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:

If A ∈ M m×n , c : scalar

Then ( 1 ) A+ 0 m × n =A

( 2 ) A+ (− A )= 0m× n

( 3 ) cA= 0 m× 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:

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.

A= [28 ] (b) A= [ 1 2 3

4 5 67 8 9 ] (c

)A= [0 1

2 41 −1 ]

Sol: (a)A= [ 2

8 ] ⇒ AT = [2 8 ]

(b)A= [ 1 2 3

4 5 67 8 9 ] ⇒ AT =[ 1 4 7

2 5 83 6 9 ]

(c)

A= [0 12 41 −1 ] ⇒ AT=[ 0 2 1

1 4 −1 ]

(a)

Ex 8: (Find the transpose of the following matrix)

(1 ) ( AT )T =A

(2 ) ( A+B )T =AT +B T

(3 ) ( cA )T =c ( AT )

(4 ) ( AB )T =BT AT

Properties of transposes

A square matrix A is symmetric if A = AT

Ex:

If A= [1 2 3a 4 5b 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 3a 4 5b c 6 ] AT =[1 a b

2 4 c3 5 6 ] a= 2, b= 3, c= 5

A=AT

⇒

Symmetric matrix:

If A= [0 1 2a 0 3b 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 2a 0 3b c 0 ] −AT =[ 0 −a −b

−1 0 −c−2 −3 0 ]

A=−AT a= − 1, b=− 2, c=− 3⇒

Ex:

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:

A=[1 32 −1 ] B=[2 −1

0 2 ]Sol:

AB= [1 32 −1 ] [2 −1

0 2 ]=[ 2 54 −4 ]

Note: AB ≠ BA

BA= [2 −10 2 ][1 3

2 −1 ]=[0 74 −2 ]

Ex 4: Sow that AB and BA are not equal for the matrices.

and

(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

/61

A= [1 30 1 ] , B= [2 4

2 3 ] , C= [ 1 −2−1 2 ]

Sol:

AC= [1 30 1 ] [ 1 −2

−1 2 ]=[−2 4−1 2 ]

So AC=BC

But A ≠ B

BC= [2 42 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

A∈ M n×n

If there exists a matrix B∈ M n× n such that AB=BA=I n ,

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:

If B and C are both inverses of the matrix A, then B = C.

Pf: AB=IC (AB )=CI(CA)B=CIB=CB=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)

[ 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 00 1 ]

[ x11+4x21 x12+4x22

−x11−3x21 −x12−3x22 ]=[ 1 00 1 ]

Find the inverse of a matrix by Gauss-Jordan Elimination:

(1)⇒[ 1 4 ⋮ 1−1 −3 ⋮ 0 ] r⃗12

(1) , r 21(−4 ) [1 0 ⋮ −3

0 1 ⋮ 1 ](2 )⇒ [ 1 4 ⋮ 0

−1 −3 ⋮ 1 ] r⃗12(1 ) , r 21

(−4 )[1 0 ⋮ −40 1 ⋮ 1 ]

⇒ x11=−3, x21=1

⇒ x 12=− 4, x 22=1

X=A−1=[ x11 x12

x21 x22 ]=[−3 −41 1 ] ( AX=I=AA-1)

Thus

⇒ x11 + 4x21 1−x11 − 3x21 0

(1 )

x12 + 4x22 0−x12 − 3x22 1

( 2)

r⃗ 12(1) ,r21

(−4 ) [1 0 ⋮ −3 −40 1 ⋮ 1 1 ] ¿ A I I A−1 ¿ ¿¿

If A can’t be row reduced to I, then A is singular.

Note:

A= [ 1 −1 01 0 −1

−6 2 3 ]

R⃗2−R1 [ 1 −1 0 ⋮ 1 0 00 1 −1 ⋮ −1 1 0

−6 2 3 ⋮ 0 0 1 ]

Sol:

[ A ⋮ I ] = [ 1 −1 01 0 −1

−6 2 3

⋮⋮⋮

1 0 00 1 00 0 1 ]

R⃗3+6R1 [1 −1 00 1 −10 −4 3

⋮⋮⋮

1 0 0−1 1 06 0 1 ]

−⃗R3 [1 −1 0 ⋮ 1 0 00 1 −1 ⋮ −1 1 00 0 1 ⋮ −2 −4 −1 ]R⃗3+ 4R2 [1 −1 0 ⋮ 1 0 0

0 1 −1 ⋮ −1 1 00 0 −1 ⋮ 2 4 1 ]

Ex 3: (Find the inverse of the following matrix)

R⃗2+R3 [1 −1 0 ⋮ 1 0 00 1 0 ⋮ −3 −3 −10 0 1 ⋮ −2 −4 −1 ] R⃗1+R2 [1 0 0 ⋮ −2 −3 −1

0 1 0 ⋮ −3 −3 −10 0 1 ⋮ −1 −4 −1 ]

So the matrix A is invertible, and its inverse is

A−1=[−2 −3 −1−3 −3 −1−2 −4 −1 ]

= [ I ⋮ A−1 ]

Check: AA−1 =A−1 A=I

(1 ) A0 =I

(2 ) Ak =AA⋯A⏟k factors

(k> 0 )

(3 ) A r⋅A s =A r+s r,s :integers

( Ar )s =Ars

(4 ) D= [d 1 0 ⋯ 00 d 2 ⋯ 0⋮ ⋮ ⋱ ⋮0 0 ⋯ d n

]⇒ D k=[ d1k 0 ⋯ 0

0 d 2k ⋯ 0

⋮ ⋮ ⋮0 0 ⋯ d n

k ]

Power of a square matrix:

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=1c A−1 ,c≠0

(4 ) AT is invertible and ( AT )−1=( A−1)T

Thm 2.8 ： (Properties of inverse matrices)

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 A 2−1 A1

−1

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=BIA=B

(C is invertible, so C -1 exists )

Note:If C is not invertible, then cancellation is not valid.

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=bA−1 Ax=A−1 b Ix=A−1b x=A−1b

This solution is unique.

If x 1 and x 2 were two solutions of equation Ax=b .

then Ax 1 =b=Ax2 ⇒ x 1 =x 2 (Left cancellation property)