MATRICES MATRICES AND AND

DETERMINANTSDETERMINANTSDETERMINANTSDETERMINANTS

E i ti Examination corner

1 one mark question in part A1 - two mark question in part Bq p1 five mark question in part C1 four mark question in part D1 four mark question in part D1 two or four mark question in

part E part E

Important 5 mark questions area:area:

Fi di th i f t iFinding the inverse of a matrix.Solving the simultaneous g

equations.

P bl d t i t tProblems on determinants ; etc.

S l b t i th dSolve by matrix method:

x y z 6+ + =x y z 2 + =x y z 6+ + =

yx z 2 = Given system of equation can be written in the matrix equation as written in the matrix equation as

AX D=

Wh A i ffi i t t iWhere A is coefficient matrix,X is variable matrix andD is constant matrix.

1 1 1 x 61 1 1 y 2 = 1 1 1 y 21 0 1 z 2

=

Coefficient matrix

Variable matrix

Constant matrix

matrix matrix

N it th t iNow we can write the matrix equation as 1X A D=

Here A has its inverse onlyHere A has its inverse only when it is non singular matrix.

( ) ( ) ( )1 1 1

A 1 1 1 1 1 0 1 1 1 1 0 1 4= = + + + =( ) ( ) ( )A 1 1 1 1 1 0 1 1 1 1 0 1 41 0 1

+ + +

Hence A is non singular itsHence A is non-singular, its inverse exists and is given by

( ) =1 1A adj AA

T11 12 13C C C

A

11 12 13

21 22 23adjA= C C C 31 32 33C C C

H ( )1 i jC M+

1 1 1

Here ( )1 i jij ijC - M+=

1 1 1A 1 1 1

M11 M12 M13

A 1 1 11 0 1

= 1 0 1

C12 = - M12 = 2C11 = M11 = 1C13 = M13 = 1C13 = M13 = 1

11C 1= 12C 2= 13C 1=

( )21C 1 0 1= = 31C 1 1 2= + =( )21

22C 1 1 2= = 31C 1 1 2+

( )32C 1 1 0= =22( )23C 0 1 1= =

( )3233C 1 1 2= = ( )23 33

( )

1 1 2adj A 2 2 0

C11 C21

C C

C31

C( )adj A 2 2 01 1 2

=

C12

C13

C22

C23

C32

C331 1 2

1

C13 C23 C33

( )1 1X A D adj A DA

= = A

1 1 2 6 4 11 1X 2 2 0 2 8 2 X 2 2 0 2 8 2

4 41 1 2 2 12 3

= = = 1 1 2 2 12 3

Th f 1 There fore x = 1, y = 2, z = 3.

CayleyCayley Hamilton TheoremHamilton Theorem

State Cayley Hamilton Theorem

y yy y

State Cayley Hamilton Theorem

and verify for the matrix 1 2

y

hence find its inverse.

3 4

Every square matrix satisfies its h t i ti tiown characteristic equation.

CayleyCayley Hamilton TheoremHamilton Theorem1 2

A Let

y yy y

A3 4

=

Let

Its characteristic equation is Its characteristic equation is

A I 01- 2

0=

A I 0 = 03 4 -=

( ) ( ) 2( ) ( )1 4 6 0 = 2 5 2 0 =

CayleyCayley Hamilton TheoremHamilton Theorem2 1 2 1 2 7 10A = = A A I2 5 2

y yy y

A3 4 3 4 15 22

A A I5 2

7 10 1 2 1 0 =

7 10 1 2 1 05 2

15 22 3 4 0 10 00 0

= 0 0

Hence Caley Hamilton theorem is ifi dverified.

CayleyCayley Hamilton TheoremHamilton Theorem

Now let us find its inverse using C-H

y yy y

Now let us find its inverse using C-H theorem

B C l H ilt th t

Here you have to write I

By Cayley Hamilton theorem we get,2A 5A 2I O =

Multiplying both sides by A-1 we get

A 5A 2I O

1A 5I 2A O =

CayleyCayley Hamilton TheoremHamilton Theoremy yy y

12 A A 5 I =

1 4 21A

= A 3 12

DeterminantsDeterminants

U i ti f

DeterminantsDeterminants

Using properties of determinant, prove that

a b ab b2 21 2 2+ ab a b ab a a b

2 2

2 2

2 1 22 2 1

+ b a a b2 2 1

( )a b 32 21= + +( )a b1+ +

DeterminantsDeterminantsA l i

DeterminantsDeterminantsApplyingC1 C1 bC3 and C2 C2 + aC3If each element of a row (or column) If each element of a row (or column) of a determinant constant of a determinant constant of a determinant, constant of a determinant, constant

multiplies of corresponding multiplies of corresponding elements of other rows (or columns) elements of other rows (or columns) elements of other rows (or columns) elements of other rows (or columns) are added then the determinant is are added then the determinant is unalteredunalteredunaltered.unaltered.

DeterminantsDeterminants

C1 C1 bC3Now C2 C2 + aC3C1 C1 bC3Now C2 C2 + aC3

2 2+ 2 22 2

1 a b 2ab 2b+ +2 2 21 a b 2b2b+ 02 2 +2 2

2 2

2ab 1 a b 2a0 2 21 a b+ + ( )( )2 2 2 22b 2a 1 a b( )2 22b b 1 a b 2 21 a b+ + ( )2 22a a 1 a b + ( ) + +2 2a 1 a b

DeterminantsDeterminantsDeterminantsDeterminants

Take out 1 + a2 + b2 from C1and C22

DeterminantsDeterminants1 0 2b

2 2 2

2 2

(1 a b ) 0 1 2ab 1 b

= + +2 2b a 1 a b

ApplyApply

C3 C3 + 2b C1 2a C2( a b )2 2 21 0 0

1 0 1 0= + +C3 C3 + 2b C1 2a C2( a b )b a a b2 2

1 0 1 01

= + + + +

DeterminantsDeterminants

( )a b a b22 2 2 21 1 + + + +

DeterminantsDeterminants

( )a b a b1 1 = + + + +

( )3( )a b RHS32 21= + + =

One mark questionsOne mark questions

Fi d if i t i 1 x

qq

Find x, if is symmetric matrix 3 1

A matrix is said to be symmetric matrix if its transpose is equal to matrix if its transpose is equal to itself.

x1 1 3 xx

1 1 33 1 1

=

x 3 =x3 1 1

One mark questionsOne mark questions

y x

qq

Find x and y , if is skew-

symmetric matrix

y3 0

symmetric matrix.

We know that a matrix A is We know that, a matrix A is skew-symmetric matrix if A = - ASo, y = 0 and x = - 3.

Solve by Crammers rule:yx + y = 3 and x y = 1 x y 1 .

Determinant of coefficient matrix

1 11 1 2 = = = 1 1 2

1 1 = = =

1 11 1 2 1 1 2

1 1 = = =

1

3 13 1 4

1 1 = = =

1 3

1 1

2

1 31 3 2

1 1 = = =

41 4x 22

= = =

2 22 2and y 12

= = =

2

Two mark questionTwo mark questiona 6 b+c

qq

Prove that b 6 c+a =0c 6 a+b

without expanding at any stage

c 6 a+b

without expanding at any stage.

A l C C + CApply C3 C3 + C1

Two mark questionTwo mark question

a 6 a b c+ +

qq

a 6 a b cLHS b 6 a b c

+ += + +

c 6 a b c+ +

Let us take common a+b+cfrom 3rd column and 6 from from 3rd column and 6 from 2nd column.

Two mark questionTwo mark question

( )a 1 1

qq

( )LHS=6 a+b+c b 1 1c 1 1c 1 1

=0 RHS= RHS.

If two rows /columns of a If two rows /columns of a d t i td t i t id ti l th th id ti l th th determinantdeterminant are identical then the are identical then the

value of the value of the d t i t i h d t i t i h determinant vanishes. determinant vanishes.