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

 

Additional Points:

 

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

 

e = d/(ad-bc), f =  -b/(ad-bc), g =  -c/(ad-bc), h = a/(ad-bc)

 

Thus

A^-1 =  =

 

Note that ad-bc = |A|

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)

 

Thus x = (dp-bq)/(ad-bc) and y = (aq-pc)/(ad-bc).

 

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 P_2A=   and A_1P= be two other matrices formed by elements of matrices A and P.

Then x = |P_2A|/|A| = (pd-bq)/(ad-bc) and y = |A_1P|/|A|=(aq-cp)/(ad-bc).

 

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 P_2A== and A_1P=   = 

 

|A| = 2+18=20, |P_2A| = -6-24 = -30 ,  |A_1P| =  -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.