3.6 Matrix operations:

1) Multiplication of a matrix by a constant:

 A = k= Multiply each element of matrix by the constant k k(A) = (k*A’s all elements)

2) Equality of 2 Matrices

 A = B = If (A) =(B) then (x1=z1 , x2=z2 , x3=z3) , (y1=t1 , y2=t2 , y3=t3). Two matrices are ‘equal’ if their corresponding elements are equal

1)  if A = A1 then A is a symmetric matrix

Proof

 A = A1   = If,A =A1 then a2=b1, a3=c1, b3=c2 A = Thus A is a symmetric matrix ( elements are equal with respect to the principal diagonal:[a1-->b2-->c3])

2) if A = - A1 then A is a skew symmetric matrix

Proof

 A = - A1   = If A =-A1 then a1= -a1,a2= -b1, a3=-c1, b1=-a2,b2= -b2, b3= -c2, c1 =  -a3,c2= -b3,c3=-c3 a1=0,b2=0,c3=0 A = Thus A is both symmetric and skew. ( elements are equal and negative with respect to the principal diagonal [0,0,0])

3.6.1 Addition & Subtraction of Matrices:

1. If A and B are two matrices of the same order then their sum (A+B) is the matrix obtained by adding the corresponding elements of A with the elements of B.

2. If A and B are two matrices of the same order then their difference (A-B) is the matrix obtained by subtracting the corresponding elements of B from the elements of A.

 A = B = A+B = a1+x1=x1+a1 Since addition of numbers is commutative, addition of matrices is also commutative. Thus A+B=B+A A = B = A-B = a1-x1 x1-a1. Since subtraction of numbers is not commutative, subtraction of matrices is also not commutative. Thus  A-BB-A. We notice that for addition and subtraction matrices need to be of same order. A = A1   = A+ A1 = This is a symmetric matrix ( elements are equal with respect to the principal diagonal). A = A1   = A- A1 = This is a skew matrix as elements across the principal diagonal are zero.

3.6.2 Multiplication of matrices:

A =   3 x 2

B =   2 x 3

Elements in Rows of A are

(a1,a2), (b1,b2) and (c1,c2).

Elements in Columns of B are

(x1,y1), (x2,y2) and(x3,y3)

Let us define multiplication operation of 2 pairs of elements [(a1,a2)   (x1,y1)] = a1*x1+a2*y1

What did we do? We multiplied row elements s of A with corresponding column elements of B.

Let us use the following abbreviations.

FR(First row of A)           FC(First column of B)

SR (Second row of A)      SC(Second Column of B)

TR (Third row of A)         TC(Third Column of B)

We define Multiplication of Matrices as follows:

 <-------------------------Columns of B------------------------------ > Rows of A FC=(x1,y1) SC=(x2,y2) TC=(x3,y3) FR=(a1,a2) FR*FC =(a1,a2)* (x1,y1) =a1*x1+a2*y1=P1 FR*SC=       =(a1,a2)* (x2,y2) =a1*x2+a2*y2=P2 FR*TC =(a1,a2)* (x3,y3) =a1*x3+a2*y3=P3 SR=(b1,b2) SR*FC =(b1,b2)*(x1,y1) =b1*x1+b2*y1=Q1 SR*SC =(b1,b2)* (x2,y2) =b1*x2+b2*y2=Q2 SR*TC =(b1,b2)* (x3,y3) =b1*x3+b2*y3=Q3 TR=(c1,c2) TR*FC =(c1,c2)*(x1,y1) =c1*x1+c2*y1=R1 TR*SC =(c1,c2)* (x2,y2) =c1*x2+c2*y2=R2 TR*TC =(c1,c2)* (x3,y3) =c1*x3+c2*y3=R3

AB =  =   matrix3 x 3

Note : A is a 3x2 matrix and B is a 2x3 matrix. AB became a 3 x 3 matrix.

A = , C = (3 x 3).

Can we multiply A with C? A’s row values (a1, a2) which are 2 in number, can not be matched with C’s column values (x1,y1,z1) which are 3 in number.

Generalization : For  matrix multiplication, we multiply each column values of each row of one matrix (A) with each row values of another matrix (B)for every column of B.

Hence these two numbers should be equal (Number of columns of A = Number of rows of B)

Because of this reason AC is not possible.

General Method:

 A= m x n matrix B=  n x p matrix FR(First row of A) SR (Second row of A) TR (Third row of A) …………………… m’thR(m’th row of A)   FC(First column of B) SC(Second Column B) ……………………… P’th(p’th Column of B) AB =  m x p matrix If A is a matrix of order m x n and B is a matrix of order n x p then AB is a matrix of order m x p.

If A = , B = ,C= , I=

Exercise :  Prove the following which is also true for any matrices.

 No. Operation Property 1 (AB)1= B1 A1 2 AI=IA=A This is true for any Square matrix. 3 A+B =B+A Commutative Property 4 A-B B-A 5 AB BA 4 A(BC) =(AB)C Associative Property 5 A(B+C)=AB+AC Distributive property w.r.t addition 6 A(B-C) =AB-AC Distributive property w.r.t subtraction

3.6.2 Problem 1: If  A =  and B =  show that ABBA

Solution :

AB =

BA =

The matrices AB and BA are not same

3.6.2 Problem 2: IF  A=  and B= verify that (A+B)1  = A1+B1

Solution :

A+B =              (A+B)1  =

A1=    B1=  A1+B1=

3.6.2 Problem 3:   If A=  show that A2-8A+13I =0

Solution :

A2= A*A = =

-8A =

13I =  =

A2-8A+13I =  =

3.6.2 Problem 4:  Find  x and y given =

Solution :

Multiplication of two matrices gives us

LHS =

RHS = =

So we have

x+3y = -7              --à(1)

5x-2y = -1             --à(2)

5x+15y = -35        ---à(3) (Multiply (1) by 5)

-17y =34 ( Subtract (3) from (2))

y = -2  By substituting this value in (1) we get

x-6 = -7

x = -1

Verfication:

Verify the equality of product of matrices by substituting values of x and y.

Application of Matrix Theory:

Let us work with a real life problem for solution using theory of matrix.

3.6.2 Problem 5: The 11 railway connections between cities of Karnataka, Maharashtra and Gujarat are as follows. Find the number of direct routes from  Mangaluru to Ahmadabad and from Belagavi to Vadodara.

 No Starting place Destination 1 Mangaluru Mumbai 2 Mangaluru Pune 3 Hubballi Pune 4 Belagavi Nagapura 5 Mumbai Ahmadabad 6 Mumbai Surat 7 Pune Ahmadabad 8 Pune Surat 9 Pune Vadodara 10 Nagapura Surat 11 Nagapura Vadodara

Solution :

For easy understanding let us represent the problem in a diagram:

If cities are represented in alphabets as above then

Routes from Karnataka to Maharashtra

Equivalent

Matrix

Routes from Maharashtra to Gujarat:

Equivalent

Matrix

Product of matrices

Equivalent Matrix

 To b1 b2 b3 a1 1 1 0 a2 0 1 1 a3 0 0 1

P =

 To c1 c2 c3 b1 1 1 0 b2 1 1 1 b3 0 1 1

Q =

PQ =  =

 To c1 c2 c3 a1 2 2 1 a2 1 2 2 a3 0 1 1

Thus we conclude that, there are 2 direct routes from   Mangaluru to  Ahmadabad and 1 route from  Belagavi to Vadodara

3.6. Summary of learning

 No. Points  learnt 1 (AB)1= B1 A1 2 AI=IA=A 3 A+B =B+A 4 A-B B-A 5 AB BA 4 A(BC) =(AB)C 5 A(B+C)=AB+AC 6 A(B-C) =AB-AC

Note the following properties by taking suitable matrices.

1.  If A  0 and AB=AC then it is not necessary that B=C

2.  If AB = 0 then it is not necessary that A=0 or B=0

3.  If A=0 or B=0 then AB = 0 = BA

4.  (A+B)(A-B)  A2-B2

3.6.2 Problem 4:  Find matrix M such that M =

Solution:

First we need to find the order of matrix M

Note the order of matrix  is 2 by 2

Since the product of matrices is 1 by 2

M has to be a 1 by 2 matrix so that

(1 by 2 Matrix)*(2 by 2 Matrix) = (1 by 2 Matrix)

Let M = , On multiplying the two matrices we get

=

Since it is given that the product of M and the given matrix is equal to

It follows that x = 1 and x+2y = 2

On Solving, we get x = 1 and  y = 1/2.

Matrix M =

Verification:

Verify that =

3.6.3 Determinants:

Let A = be a 2 x 2 square matrix. We define the ‘determinant’ of A denoted by |A| as a real number  |A|= ad-bc.

A square matrix is called ‘singular matrix’ if its determinant is zero.

A square matrix is called ‘non-singular matrix’ if its determinant is not zero.

For a given matrix A, its inverse A-1 is defined as a matrix such that AA-1 = I (identity matrix)

Let us find the properties of A-1

Let A =  and A-1 =

Then AA-1 =

Since by definition  AA-1=, it follows that we need to have

ae+bg = 1, af+bh = 0, ce+dg = 0 and cf+dh = 1.

By expressing e, f, g and h in terms of a, b, c and d we get

Thus

A-1 =  =

Thus if |A| = 0, then A-1  does not exist.

Hence, a singular matrix does not have its inverse.

Also note that | A-1| = (ad-bc)

3.6.3 Problem 1:  Find the inverse of A= and also its determinant

Solution:

Here |A| = (2*-3-0*5) = -6

Since |A| 0 A-1 exists

A-1 = -1/6=

|A-1|= -1/6

Also note |AA-1|= (-6)(-1/6) = 1

3.6.4 Solving of simultaneous linear equations

We have learnt to solve simultaneous linear equations by algebraic method and by graphical method.

Now we will learn two more methods using matrix theory.

3.6.4.1 Inverse Matrix method

Let ax+by = p and cx+dy = q be two linear equations

Let A = a 2 by 2 matrix, X=  a 2 by 1 matrix and P =  a 2 by 1 matrix.

Hence AX =  which is =  = P

Thus in matrix form AX = P. Let us multiply both sides by A-1 (It exists when ad-bc 0)

AA-1X = A-1P

i.e. IX = A-1P (since IX = X) we have,

X = = =  (By matrix multiplication)

3.6.4 Problem 1: Solve 2x-3y+6 = 0 and 6x+y+8 = 0 using matrix inversion method.

Solution:

The given equations are rewritten as:

2x-3y = -6 and

6x+y = -8

So we have

=

Let A = , then |A| = 2-(-3*6) = 20. Since |A|0, A-1 exists and

A-1 = 1/20 =

Since X = A-1P

X = = = =

Thus x = -3/2 and y = 1

3.6.4.2. Cramer’s method

As in 3.6.4.1 let ax+by = p and cx+dy = q be two linear equations. Let A, X and P be matrices as defined in 3.6.4.1.

Hence AX =  which is == P

Thus in matrix form AX= P and we note |A| = ad-bc.

Let P2A=   and A1P= be two other matrices formed by elements of matrices A and P.

3.6.4 Problem 2: Solve the problem 3.6.4 Problem 1 using Cramer’s method.

Solution:

The given equations are rewritten as:

2x-3y = -6 and

6x+y = -8

So we have

A= , X= and P  =

|A| = 2+18=20

Then P2A== and A1P=   =

|A| = 2+18=20, |P2A| = -6-24 = -30 ,  |A1P| =  -16+36 = 20

x = (-30/20) = -3/2 and y = 20/20 = 1

These are the same solutions that we got in problem 3.6.4 Problem 1.