MATRICES - · PDF filecrackiitjee.in MATRICES

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crackiitjee.in MATRICES Computers play a very important role in not only scientific advances but in our day to day activities such as railway reservation, banking and shopping. The origin of computers goe to the study of matrix algebra. To know about the matrices, and its operation, we define it as under : Matrix Theory 1. Matrix and its elements An ordered rectangular arrangement of numbers in rows and columns is defined as a MATRIX. Each number is called an element or entry of the matrix. A = 2i 3 1 1 0 4 ,B 4 2i 2i 3 3 1 2 1 i 2 3 C = 1 5 4D 3 are examples of matrices. A has two rows (horizontal lines) and 3 vertical lines called columns. Its order is defined as 2 × 3. Similarly B has 3 rows and 3 columns and its order is 3 × 3. Order of C is 1 × 3 and order of D is |x|. Notation For more Study Material and Latest Questions related to IIT-JEE visit www.crackiitjee.in For more Study Material and Question Bank visit www.crackiitjee.in 1

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MATRICES

Computers play a very important role in not only scientific advances but in our

day to day activities such as railway reservation, banking and shopping. The origin

of computers goe to the study of matrix algebra. To know about the matrices,

and its operation, we define it as under :—

Matrix Theory

1. Matrix and its elements

An ordered rectangular arrangement of numbers in rows and columns is defined

as a MATRIX. Each number is called an element or entry of the matrix.

A =

2i 3 11 0 4

, B 4 2i 2i 33 1 2

1 i 2 3

C = 1 5 4 D 3

are examples of matrices. A has two rows (horizontal lines) and 3 vertical lines

called columns. Its order is defined as 2 × 3.

Similarly B has 3 rows and 3 columns and its order is 3 × 3.

Order of C is 1 × 3

and order of D is |x|.

Notation

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Let A be a matrix of order ‘m × n’ i.e. it has m rows and n columns. We denote aij

as the element or entry common to its row and jth column.

aij = the entry common to ith row and jth column a11 is therefore the element

common to 1st row and 1st columns it is also called the leading element.

A =

11 12 13 ij 1n

21 22 23 2j 2n

i1 i2 i3 ij in

m1 m2 m3 mj mn

a a a a a

a a a a a

a a a a a

a a a a a

= (aij)m × n or (aij) 1 < i < m, 1 < j < n

Types of Matrices

(i) Zero matrix : A matrix whose every entry is 0 is called zero matrix and is

denoted by 0.

(0) or 0 0 0

0 0 or 0 0 0

0 0 0

(ii) Row matrix and Column matrix

A matrix which has only one row is called a row matrix. Its order is of the form

1 × n

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0 1 3 5 ,

A matrix which has only one column is called a column matrix. Its order is of the

form m × 1.

11

37 ,

78

9

(ii) Square Matrix

A matrix in which the number of rows is equal to the number of columns is called

a square matrix. Its order is of the form n × n.

i.e. it can be 1 × 1, 2 × 2, 3 × 3

3 4 03 4

2 , , 1 2 11 0

1 1 1

Note on Square matrix

Consider a 3 × 3 matrix

A =

11

22

33

a i j i j

i j a i j

i j i j a

a11, a22, a33 are elements which are present in the matrix

a11 leading element.

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a11, a22, a23 from the main diagonal.

elements aij where i = j

The elements above it are of the form i < j

The elements below the main diagonal are of the form where i > j.

(iv) Triangular matrix

A square matrix A = (aij)n × n is called an upper triangular matrix if

aij = 0 i j, all elements below the main diagonal are zeros.

A =

2 3 1

0 4 2

0 0 5

or B =

a b c d

0 e f g

0 0 h k

0 0 0 e

Similarly a square matrix A = (aij)n × n is called a tower triangular matrix if

a = 0 i j

[all the elements above the main diagonal are zeros]

C =

a 0 0 03 0 0

b c 0 04 5 0 , D

d e f 06 7 8

g h k l

(v) Diagonal matrix

A square matrix A = (aij)n × n is called a diagonal matrix, if aij = 0 i j

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i.e. all non-diagonal elements are zeros.

also it is both an upper triangular as well as a lower triangular matrix.

Also if d1, d2, …, dn are its diagonal elements, the remaining elements being

zeros, it is denoted as diagonal (d1, d2, …, dn)

i.e. diagonal (d1, d2, …, dn) =

1

2

3

n

d 0 0 0

0 d 0 0

0 0 d 0

0 0 0 d

(vi) Scalar matrix

If in a diagonal matrix, all diagonal entries are equal to d 0, the called a scalar

matrix.

Aij =

d 0 0 0

0 d 0 0

0 if i j 0 d 0 0

d if i j

0 0 0 d

(vii) Unit matrix or Identity matrix

A square matrix whose every element in the main diagonal is 1 and 0’s elsewhere

is called a unit matrix. It is daroted by I. or In if it of order n × n.

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I = 0 if i j

1 if i j

I2 = 3 4

1 0 0 01 0 0

1 0 0 1 0 0, I 0 1 0 , I

0 1 0 0 1 00 0 1

0 0 0 1

OPERATION ON MATRICES

(1) Equality of two matrices

Two matrices A = (aij)m × n, B = (bij)m × n are 5 said to be equal if they have the

same order and their corresponding entries are equal.

0(A) = 0(B) and aij = bij i & j A = B.

i.e.

3 2 2 3 2 11 1 2

0 6 100 3 5

4 2 2

also x 4 2 4

y 6 3 v

if and only if x 2 u 4

y 3 v 6

(2) Addition of two matrices

Def. : If A and B two matrices of the same order

i.e. A = (aij)m × n, B = (bij)m × n 0(A) = 0(B) = m × n

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then teir sum A + B is obtained by adding their corresponding elements. i.e.,

(A + B)ij = Aij + Bij = aij + bij i and j

Also note that 0(A + B) = m × n = 0(A) = 0(B)

Two matrices of different orders can never be added

of A = 2 3 1 1 2 1

, B5 6 7 2 3 1

A + B =

1 5 02 1 3 2 1 1

3 3 85 2 6 3 7 1

(3) Scalar multiplication

Let A = (aij)m × n be any matrix and k be any number, then k.A (scalar

multiplication of A by a scalar k is obtained by multiplying each entry of A by k.

i.e. (kA)ij = k.Aij = kaij

Also 0(kA) = 0(A) = m × n.

If A = 2 3 1 4 6 2

, then 2A5 6 7 10 12 14

and 1 3 2 1 21

A5 2 3 7 22

(4) Negative of a matrix

If A = (aij)m × n is any matrix, then

– A = (– 1) A is called negative of matrix A.

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i.e. – A is obtained by multiplying the matrix A by scalar (–1) or changing the sign

of each element of A.

A = 1 2 3 1 2 3

, then A5 4 1 5 4 1

(5) Transpose of a matrix

Let A = (aij)m × n be any matrix, then its transpose denoted as A’ or AT, is

obtained by interchanging the row and columns of A i.e. first row of A becomes

just of A’, second row of A becomes second column of A’ and so on.

Also if 0(A) = m × n, then 0(A’) = n × m

Also aij = i, jth element of A = Aij

aij now becomes j, ith element of A’ = A’ji

A’ji = Aij or A’ij = Aji

(6) Difference of two matrices

If A and B are two matrices of the same or order, then we define A – B = A + (–B)

Properties of Matrix addition

Following properties can be easily deduced from the definitions note, wherever

we write addition between two matrices, it is assumed, they are compatible for

addition i.e. they satisfy the conditions of addition i.e. their orders are same.

(1) A + O = A

(2) A + (– A) = O or A – A = O, (O matrix)

(3) A + B = B + A

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(4) (A + B) + C = A + (B + C)

(5) K (A + B) = kA + kB

(6) (A’)’ = A

(7) (A + B)’ = A’ + B’

(8) (kA)’ = k.A’

(9) (K1 + K2) A = K1A + K2A

(10) (K1K2) A = K1(K2A) = K2(K1A)

Note : All the aboves properties can be easily verified.

Matrix multiplication

We introduce the concept of matrix multiplication in two steps.

Step-I (Compatibility)

Let A = (aij)m × n and B = (bij)n × p be two matrices then the product A × B is

defined.

i.e. A × b is defined and ca be obtained if.

number of columns in A = number of rows in B.

i.e. 0(A) = m × n, 0(B) = n × p.

then (A × B) is a matrix of order m × p.

A B A B

m n n p m p

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Step-II : If 0(A) = m × n, and 0(B) = n × p,

Let C = A × B, the product matrix C = A × B.

C =

11 12 1j 1p

11 12 1n21 22 2j 2p

i1 i2 in31 32 3j

n1 n2 nj np

b b b ba a a

b b b ba a a

b b b

b b b b

Cij = is jth entry of C

= sum of the pointwise products of ith row of A and jth column of B

= ai1bij + ai2b2j + ai3b3j + … + ainbnj = n

ir rjr 1

a b

Example : Consider A =

1 01 2 3

, B 2 12 3 1

1 4

0(A) = 2 × 3, 0(B) = 3 × 2

A × B is defined and will be of order 2 × 2

A × = 1 0

1 2 32 1

2 3 11 4

= 1 4 3 0 2 12 2 10

2 6 1 0 3 4 3 1

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Note : Here B × A also exists

0 B 3 2

0 A 2 3

0(B × A) = 3 × 3 0(A × B)

If should be noted that 0(B × A) may not be same 0(A × B).

Let us find B × A = 1 0

1 2 32 1

2 3 11 4

Example : A = 1 2 1 0

, B3 4 1 2

0(A) = 2 × 2 = 0(B)

Both A × B and B × A exist.

Consider A × B = 1 01 2

1 23 4

= 1 2 0 4 1 4

3 4 0 8 1 8

Let us find B × A = 1 21 0

3 41 2

= 1 0 2 0 1 2

1 6 2 8 7 10

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We clearly observe, even if 0(A × B) = 0(B × A)

but A × B B × A

Properties of Matrix multiplication

Wherever, we multiply two matrices, we assume, they are compatible for matrix

multiplication even if they are of the same order.

(1) A × B B × A

(2) A × O = O, O × A = O

(3) (A × B) × C = A × (B × C), associative property holds

(4) If O(A) = m × n, the A × In = A, Im × A = A

(5) (xA) × (yB) = (yA) × (xB) = (xy) A × B, where x and y are scalars.

(6) A × B = O without A or B being zero matrices.

A = 1 0 0 0

, B2 0 3 4

A × B = 0 0

0,0 0

Neither A nor B is a zero matrix, but A × B is a zero matrix.

(7) A × (B + C) = A × B + A × C

A B C A B A C1

A B C A C B C

These laws are called Distributive laws.

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(8) If A, B, C are three matrices and A = b, then AC = BC

But its converse is not true, i.e. AC = BC does not necessarily mean A = B.

Consider A = 1 0 4 0

, B2 0 5 0

and let C = 0 0

m n

we find A × C = 1 0 0 0

2 0 m n

= 0 0

0 0

B × C = 4 0 0 0 0 0

5 0 m n 0 0

Here we find AC = BC, but A B

(9) (A × B)’ = B’.A’

(Reversal rule of the transpose of the product of two matrices).

The show this, Let A =

1 41 0 2

, B 0 53 4 5

1 0

A × B = 1 4

1 0 20 5

3 4 51 0

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= 1 0 2 4 0 0 1 4

3 0 5 12 20 0 2 8

(A × B)’ =

'1 4 1 2

2 8 4 8

Now, B’ × A’ =

1 31 0 1

0 44 5 0

2 5

= 1 0 2 3 0 5 1 2

4 0 0 12 20 0 4 8

Thus (A × B)’ = B’ × A’.

Note : This rule can be extended product of three matrices as well, if

(A × B) × C exists, then

(A × B) × C = A × (B × C) = A × B × C

(denoted as the brackets can be inserted at will).

Extension of reversal rule for the product of three matrices states

(A × B × C)’ = C’ × B’ × A’.

Q.1. If A =

0 tan1 02

, I0 1

tan 02

then (I – A) cos sin

sin cos

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(a) I + A (b) I + 2A (c) I + 3A (d) none of these.

Consider

I – A =

0 tan 1 tan1 0 2 20 1

tan 0 tan 12 2

(I – A) cos sin

I Asin cos

=

1 tancos sin2sin cos

tan 12

=

cos sin tan sin cos tan2 2

cos tan sin sin tan cos2 2

=

cos cos sin sin sin cos cos sin2 2 2 2

cos cos2 2

sin cos cos sin cos cos sin sin2 2 2 2

cos cos2 2

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=

cos sin2 2

1 tancos cos2 2 2

tan 1sin cos22 2

cos cos2 2

=

0 tan1 0 2

I A0 1

tan 02

Q.2. A = ncos sin

, then A , n N.sin cos

(a) cos n sin n

sin n cos n

(b) cos n sin n

sin n cos n

(c) I3 (d) – I3

Sol. A = cos sin

sin cos

A2 = cos sincos sin

sin cossin cos

= 2 2

2 2

cos sin sin cos sin cos

sin cos sin cos sin cos

= cos 2 sin 2

sin 2 cos 2

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assume Am = cos m sin m

sin m cos m

AM + 1 = Am A = cos sincos m sin m

sin cossin m cos m

AM + 1 = cos m cos sin m sin cos m sin sin m cos

sin m cos cos m sin sin m sin cos m cos

=

cos m sin m

sin m cos m

=

cos m 1 sin m 1

sin m 1 cos m 1

The statement is true for n = m + 1.

Thus An = cos n sin n

n N.sin n cos n

Q.1. If x = 24 is a root of the equation,

log (x2 + 24a2)10 – log (a – 3)10 = 10ax

loga 3

then other root is

(a) 12, 16 (b) 16, 36 (c) 20, 16 (d) none of these

Sol. The given equation is

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log (x2 + 24a2) – log (a – 3) = 10ax

loga 3

2 2x 24a 10ax

log loga 3 1 3

2 2x 24a 10ax

, a 3a 3 1 3

(x – 6a) (x – 4a) = 0 x = 6a or 4a

If 6a = 24

a = 4,

other root is

x = 4a = 4 × 4 = 16

Q.2. The solution of the equation

22x 7x 5 x 1

x 1 2x 5log 6 log 6

is

(a) 2 and 0.9 (app) (b) 0.9 and 3 (c) 3 or 2 (d) 2 or – 2

Sol. The given equation is

22x 7x 5 x 1

x 1 2x 5log 6 log 6

The domain is x + 1 > 0, x 0 x > – 1, x 0 …(1)

and 2x + 5 . 0, x – 2 x > 5

,2

x – 2 …(2)

The given equation can be simplified to

22x 7x 5 x 1

x 1 2x 5log 6 log 6

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or x 1 2x 5 x 1x 1 x 1 2x 5log log 6 log 6

61 t 6

t

6

t 5t 0t

when 2x 5x 1log t

x 12x 5log t

t2 – 5t + 6 = 0

t = 2, 3

when t = 2

2x 5x 1log 2

(x + 2)2 = 2x + 5

x2 + 2x + 1 = 2x + 5

x2 = 4

x = 2

x –2

x 2

t = 3

2x 5x 1log 3

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(x + 1) = 2x + 5

x3 + 3x2 + 3x + 1 = 2x + 5

x3 + 3x2 + x – 4 = 0

which has a real root

x 0.9 (approx.)

f(1) = 5

f(0) = – 4

which is possible

Q.3. If

2x x3 3

3 5log log

4 4x 3

then x has

(a) one positive integral solution (b) one irrational value

(c) two positive rational values (d) none of these

Sol.

2x x3 3

3 5log log

1 24 4x 3 3

Apply log to base 3 to both sides.

2x x x 3

3 3 3 3

3 5 1log log log log

4 4 2

let x3log t, we get

3 23 5 1t t t

4 4 2

3t3 + 4t2 – 5t – 2 = 0

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(t – 1) [3t2 + 7t + 2] = 0

(t – 1) (t + 2) (3t + 1) = 0

t = 1, – 2, 1

3

x3

1log 1, 2,

3

x = 31, 3–2, 3–1/3

x = 3

1 13, ,

9 3

Hence (a), (b), (c) are all correct.

1 + y = 2

4y

where y = 4x 1x 1log

2

y 3 0y

y2 – 3y + 2 = 0 y = 1 or 2

when y = 1

4x 1

x 1log 1

4x + 1 = (x + 1) x = 0

which is not possible as it gives.

when y = 2

4x 1

x 1log 2

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(x + 1)2 = 4x + 1

x2 + 2x + 1 = 4x + 1

x2 – 2x = 0

x = 2 or 0

But x 0

x = 2 is the only solution.

Q.4. The solution of

2 24x 5x 1 x 2x 1

x 1 4x 1log 4 log

is

(a) 1

x2

(b) 2 (c) 3 (d) none of these

Sol. 2 24x 5x 1 x 2x 1

x 1 4x 1log 4 log

is solvable only if x + 1 > 0, x 0

x > – 1 and x 0 …(i)

also 4x + 1 > 0 and x 0

x > 1

4

Now

2 24x 5x 1 x 1

x 1 4x 1log 4 log

x 1 4x 1 x 1

4x 1x 1log 4 2 log

x 1 4x 1 x 1

x 1 x 1 4x 1log log 4 2 log

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Q.5. The solution set of the equation

2 3log x log x 4

x 2 x 2

is

(a) 1, 3

10 (b) 1

, 1000010

(c) 3, 10000

(d) 1, 3, 10000

10

Sol.

2 3log x log x 4x 2 x 2

2 3log x log x 4x 2 x 2 x 2

Both sides are equal x – 2 = 1 x 3

Now if x – 2 1, then

(log x)2 – 3 log x = 4 x = 1

10 or 10000

(log x)2 – 3 log x – 4 = 0

log x = – 1 or 4

log x = – 1 or 4

x = 10–1 or x = 104

Q.6. The solution of the equation

2

2

x 3x 5log 0 is

x x 7

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(a) {1, – 1} (b) {1, – 6} (c) {– 1, – 6} (d) none of these

Sol. 2

2

x 3x 5log 0

x x 7

x2 – x + 7 0 discreament = 1 – 49 < 0

log = 0 iff M = 1

2

2

x 3x 51

x x 7

iff x2 – 3x – 5 = ± (x2 – x + 7)

when x2 – 3x – 5 = x2 – x + 7

– 2x = 12

x 6

when x2 – 3x – 5 = – x2 + x – 7

2x2 – 4x + 2 = 0

x2 – 2x + 1 = 0

x 1

Q.7. The solution of the inequality

2x 13x 1

sin sin8 8

log log

is

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(a) 1

,3

(b) 1

, 0 3, 43

(c) 1

, 0 3,3

(d) none of these.

3x + 1 > 0 and x2 + 1 > 0, which is true x R.

x > 1

3 …(i)

We know m na alog log n > m when 0 < a < 1

since 0 sin 18

2x 13x 1 2

sin sin8 8

log log x 1 3x 1

x2 > 3x

x(x – 3) > 0 x < 0 or x > 3

But x > 1

3

Hence solution 1

3 < x < 0 and 3 < x <

1

, 0 3,3

Q.8. Solution of the me quality

3

3 10

3x 1log 3x 1 2 log 3

is

(a) 3 310 1 100 1

,3 3

(b) 3 10 1

, 33

(c) 3 100 1

, 33

(d) none of these

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Sol. For the solution of

3

3

3x 1 1010 3x 1

log 2 log 3

(3x + 1)3 1 and also |3x + 1| > 0

3x + 1 1 3x + 1 0

x 0 x 1

3

Now let 3

3x 1

10log a,

we get

1

a 2 3a

2

a 3 01

a2 – 3a + 2 < 0

1 < a < 2

1 < 33x 110log 2

101 < (3x + 1)3 < 102

3 310 3x 1 100

3 310 1 3x 100 1

3 310 1 100 1x

3 3

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Thus E = 3 310 1 100 1

,3 3

is the solution.

Clearly O E and 1

E3

Q.9. The real solution of

2x 2x2 2 1 2 22 log log log log 1 is

(a) 1 (b) 2 (c) 3 (d) 8.

Sol. Let x2log a and the given equation becomes

3 2x 2 x

2 2 1 2 2 22 log log log log log 1

a 22 1 2 2

32 log log log a 1

2

a2 2

32 log log a 1

2

2

2

alog 1

3a

2

2a

2.3

a2

a2 = 2a + 3

a2 – 2a – 3 = 0

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a = – 1, 3

x x2 2log 1 or log 3

But x2log 1 is rejected

x 32log 3 x 2 8

Q.10. The sum of the series to n terms

2 3a a a

b b b2 4 8log log log is

(a) na

b2log (b)

a

nb2log (c)

a

b2n log (d)

a

b2log

Sol. We know mx x xm a a

mlog log log

m

m

m

x maa

mlog log

m

2 3

2 3

a a a

b b b2 2 2

log log log

= a a a

b b b2 2 alog log log n terms

= a

ban log

Q.12. Solution of the set of equations

x yy x

3log log

2

xy = 27 is given by

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(a) x = 9, y = 3 (b) 1

x , 729 y27

(c) Both (a) and (b) (d) None of these.

Sol. Let x yy x

1log t, log

t

1 3

tt 2

2t2 – 2 = 3t

2t2 – 3t – 2 = 0

t = 2 or – 1

2

when t = 2

x 2ylog 2 x y

But xy = 27

y3 = 27 y = 3

x = 9

When t = 1

2

1 2xy

1log x y

2

x = 1

y

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x.y = 27

1y 27

y y = 272

y = 729

When t = 1

2

xy

1log

2

x = y–1/2

But xy = 27

y–1/2.y = 27

y1/2 = 27

y = 272 = 729

x = 1 1 1

27y 729

1

, 72927

Q.13. If x x xax bx cxlog , log , log are in H.P. and a, b, c, x (1, ) then a, b, c are in

(a) A.P. (b) G.P. (c) H.P. (d) none of these

Sol. xax ax a x a

x x x x

1 1 1log

log log log log 1

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Now x x xax bx cxlog , log , log are in H.P.

a b cx x xlog 1, log 1, log 1 are in A.P.

a b cx x xlog , log , log are in A.P.

b a cx x x2 log log log

2 acbx xlog log

b2 = ca a, b, c are in G.P.

option (b).

Q.14. The number of values of x [0, n], n Z that satisfy 1 cos x

sin xlog 2

is

(a) 0 (b) n (c) 2n (d) none of these

Sol. 1 cos x

sin xlog 2, sin x 1

as Base cannot be 1.

and 1 + cos x 0

2

sin x = 1 + cos x

sin2 x = 1 + cos x

1 – sin2 x + cos x = 0

cos x (cos x + 1) = 0

Either cos x = 0, 1 + cos x = 0

when cos x = 0, sin x = ± 1 or |sin x| = 1, which has been ruled out.

Now 1 + cos x = 0 is also not possible as logo is not deputed.

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Hence it has no solution option (a) is correct.

Q.15. If tan x cot xcos x sin xlog log 0, then the general solution for x is

(a) n , n Z4

(b) 2n , n Z

4

(c) 3

2n , n Z4

(b) none of these

Sol. tan x cot xcos x sin xlog log 0 sin x 1, cos x 1,

tan x > 0, cot x > 0

sin x > 0, cos x > 0

sin x cos x

cos x sin x

cos x sin xlog log 0

sin x cos x cos x sin xcos x cos x sin x sin xlog log log log 0

sin x cos xcos x sin xlog 1 log 1 0

1

t 2 0t

t2 – 2t + 1 = 0

t =1 sin x = cos x

tan x = 1,

let sin xcos xlog t

cos xsin x

1log

t

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clearly we will include the solution only in First quadrant.

where sin x > 0, cos x > 0 and tan x = 1

x = 2n , n Z4

option ‘b’ is correct.

Q.16. x 2 x 20.09 0.3log log ,

then x belongs to interval

(a) (2, 3] (b) (– , 3] (c) [3, ) (d) none of these.

Sol. x 2 x 20.09 0.3log log

x 2 x 210 100.09 0.310 10

log log

log log

2x 2

10 0.30.31010

1 1log 0

loglog

x 210 0.3

10

1log 0

2 log

But 0.310log 0

x 210log 0

x – 2 > Base as 10 > 1

x > 3

Symmetric Matrix

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A square matrix A = (aij)

n × n is said to be symmetric if A’ = A. In this case a

ij = a

ji i

and j.

For example A =

1 5 7

5 0 9

7 9 6

is a symmetric, as A’ = A.

Skew-symmetric Matrix

A square matrix A = (aij)

n × n is said to be a skew-symmetric matrix if A’ = – A. In this

aij = – a

ji i and j.

For example A =

0 4 7

4 0 6

7 6 0

is skew-symmetric

as we find A’ =

0 4 7

4 0 6 A.

7 6 0

Properties of Symmetric and Skew-symmetric matrices

1. If A and B are two symmetric skew-symmetric matrices of the see order then

so are

(i) A + B

(ii) A – B

(iii) kA

Explanation : If A is and B are both skew-symmetric

A’ = – A, B’ = – B

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X = A + B

X’ = (A + B)’ = A’ + B’ = (– A) + (– B)

= – [A + B] = – X

X = a + B is also skew-symmetric.

The other result follow similarly.

(2) If A is a square matrix, than

(i) A + A’ is a symmetric matrix

(ii) A – A’ is a skew-symmetric matrix

Explanation :

Since a is a square matrix, say 0(A) = n × n

0(A’) = n × n = 0(A)

A + A’ or A – A’ are also square matrices

Now let X = A + A’

X’ = (A + A’)’ = A’ + (A’)’ = A’ + A

= A + A’ = X

X’ = X X = A + A’ is a symmetric matrix

Now let Y = A – A’

Y’ = (A – A’)’ = A’ – (A’)’ = A’ – A

= – [A – A’] = – Y

Y = A – A’ is a skew-symmetric matrix.

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(3) If A and B are symmetric matrices of the same order, than AB (product) is

symmetric if and only if AB = BA.

Explanation : Given A and B are symmetric and AB = BA

A’ = A, B’ = B, AB = BA

Now let X = AB

X’ = (AB)’

= B’A’ (reversal rule)

= BA (as B’ = B, A’ = A)

= AB (as AB = BA)

= X

X = AB is symmetric matrix

Conversely if AB is symmetric

(AB)’ = AB

B’A’ = AB

BA = AB, Hence the converse is also true.

(4) Every square matrix A can be uniquely expressed as the sum of a symmetric

matrix and a skew-symmetric matrix.

Explanation : Let A be any square matrix, then

define P = 1

A A'2

and Q = 1

A A'2

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Now P’ = 1 1

A A' ' A' A P2 2

P is symmetric matrix

Q’ = 1 1 1

A A' ' A' A' ' A' A2 2 2

= 1

A A' Q2

Q is a skew-symmetric matrix

P + Q = 1 1

A A' A A'2 2

= 1 1 1

A A A' A' A O A2 2 2

A has been expressed as the sum of P and Q where P is a symmetric matrix and

Q is a skew-symmetric matrix.

To show uniqueness

Let A = B + C where B is symmetric and C is skew-symmetric

B’ = B and C’ = – C

But A’ = (B + C)’ = B’ + C’

= B – C

Now B + C = A and B – C = A’

B + C + B – C = A + A’

2B + O = A + A’

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2B = A + A’ B = 1

A A' P2

also (B + C) – (B – C) = A – A’

B + C – B + C = A – A’

2C = A – A’ C = 1

A A' Q2

This show P and Q are the only two matrices, which has the required property.

Hence it proves uniqueness.

(5) If P and Q are square matrices of the same order the P’QP is symmetric or

skew-symmetric according as Q is symmetric or skew-symmetric.

Explain : (i) Let Q be symmetric Q’ = Q

then X = P’QP

X’ = (P’QP)’ = P’Q’(P’)’

= P’QP = X

X = P’QP is symmetric.

(ii) Let Q be skew-symmetric then Q’ = – Q

Y = P’QP

Y’ = (P’QP)’ = P’Q’(P’)’

= P’(– Q)P = – P’QP = – Y

Y = P’QP is skew-symmetric.

(6) If A and B are symmetric matrices of the same order than

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(i) AB + BA is a symmetric matrix

(ii) AB – BA is a skew-symmetric matrix

Explanation : A’ = A, B’ = B.

(i) Now Let X = AB + BA

X’ = (AB + BA)’

= (AB)’ + (BA)’

= B’A’ + A’B’

= BA + AB = AB + BA = X

X = AB + BA is symmetric

(ii) Let Y = AB – BA

Y’ = (AB – BA)’ = (AB)’ – (BA)’

= B’A’ – A’B’ = BA – AB

= – (AB – BA) = – Y

Y = AB – BA is skew-symmetric.

(1) If A is any square matrix, then AA’ and A’A are both symmetric matrices.

Explanation : Let 0(A) = n × n 0(A’) = n × n

0(AA’) = 0(A’A) = n × n

Let X = A’A

X’ = (A’A)’ = A’(A’)’

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= A’A

= X

X = A’A is a symmetric matrix

Similarly Y = AA’ is also a symmetric matrix.

(2) Orthogonal Matrix

A square matrix A = (aij)n × n is said to an orthogonal matrix of

AA’= A’A = In.

For example let A = 1 2 1 2 1 2 1 2

, A'1 2 1 2 1 2 1 2

AA’ = 1 2 1 2 1 2 1 2

1 2 1 2 1 2 1 2

2

1 1 1 11 02 2 2 2

I1 1 1 1 0 1

2 2 2 2

Similarly A’A = I2

Hence A = 1 2 1 2

1 2 1 2

is an orthogonal matrix.

(3) Idempotent Matrix

A square matrix A is said to be idempotent if A2 = A.

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Consider A =

2 2 4

1 3 4

1 2 3

we find A2 =

2 2 4 2 2 4

1 3 4 1 3 4

1 2 3 1 2 3

=

2 2 4

1 3 4 A

1 2 3

A2 = A

A is an Idempotent matrix.

(4) Involuntary Matrix

A square matrix ‘A’ is said to be involuntary matrix if A2 = I

Consider A = 5 8

3 5

A2 = 5 8 5 8 25 24 40 40 1 0

3 5 3 5 15 15 24 25 0 1

= I2

A is involuntary matrix.

Nilpotent Matrix

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A square matrix A = (aij)n × n is said to be Nilpotent matrix of there exists a

positive integer m so that

Am = O.

In this case ‘m’ is called the index of the nilpotent matrix A.

For example A = 16 9

4 6

A2 = A – A = 6 9 6 9

4 6 4 6

= 36 36 54 54 0 0

24 24 36 36 0 0

A is a nilpotent matrix of index 2.

Conjugate of a matrix

If the elements of a matrix A are complex numbers, then the conjugate matrix of

A denoted as A , is obtained by replacing its elements by their corresponding

conjugate numbers.

i.e. A = (aij)n × n

then ijn n

A aij or A aij Aij

Triangulate of Matrix

The transpose of a conjugate A of square matrix is called triangulate of matrix A

and is denoted by AQ or A*

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A = A ,

If A = 2 i 5 i 2 i 3

, A3 1 i 5 i 1 i

Properties of Triangulate matrix

(1) (A) = A

(2) (A + B) = A + B

(3) (kA) = KA , where K is a complex number (scalar)

(4) (AB) = BA; Reversal rule.

Hermitian Matrix

A square matrix A = (aij)n × n is said to be Hermitian matrix, A = A i.e. aij = aji

For example A = 2 5 i

5 i 3

A = 2 5 i 2 5 i

A ' A.5 i 3 5 i 3

A is Hermitian matrix.

Skew-hermitian Matrix

A square matrix A = (aij)n × n is said to be skew-hermitian if A = – A i.e. if aij = –

aji

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A = 0 2 3i

3i 2 0

A = '

0 2 3iA

3i 2 0

= 0 3i 2 2 3i

A2 3i 0 3i 2 0

A = – A

and hence A is skew-hermitian matrix.

Elementary Transformation

The set of transformation listed below are called elementary row operations :

(1) Interchange of any two rows. If the elements of ith row the jth row are

interchanged , it is denoted as Ri Rj

(2) Multiplying the elements of any row by a nonzero constant. If we multiply ith

row by as constant ‘k’ it is denoted by Ri Ri kRi

(3) Adding the equimultiples of the elements of ith row to the elements of jth

row. If we add to the elements of jth row, K times the ith row, we denoted it by

Rj Rj + kRi

We have similar three elementary column operations.

Equivalent Matrices :

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Two matrices are said to be equivalent of one of them is obtained from the other

by a set of elementary operations. If A and B are equivalent matrices we denoted

by A ~ B.

e.g.

A =

2 3 1 1 0 4

1 0 4 ~ 2 3 1

5 7 6 5 7 6

Here R1 R2

~

1 0 4

2 2 3 0 1 8

5 7 6

Here we have operated as R2 R2 – 2R1

~

1 0 4

0 3 9

5 7 6

etc. etc.

Inverse of a square matrix :

Let A = (aij)n × n be any square matrix, if we can find ‘a’ matrix B, such that

A.B = B.A = I, then ‘B’ is called the inverse of A and is denoted by A–1.

Following properties can be easily established.

(1) 0(A) = (B) = n × n, if B = A–1.

(2) A–1 = B, then B–1 = A (A–1)–1 = A.

since A–1 = B AB = BA = I

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BA = BA = I

B–1 = A.

(3) (A.B)–1 = B–1.A–1

Explained : Let X = AB, and Y = B–1A–1

(AB) (B–1A–1) = A(BB–1) A–1

= (A.I) A–1

= AA–1 = I

XY = I similarly YX = I

X–1 = Y

(AB)–1 = B–1A–1

(ABC)–1 = C–1B–1A–1.

(4) If A–1 exists, then it is unique.

Let B and C be two inverses of A (if possible)

AB BA I

AC CA I

…*

But (BA) = B(AC) by associative law

(IC) = BI

C = B

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B and C are not two different matrices and hence A has a unique inverse.

Finding of an inverse with the help of Row/column operations.

Working Rule : We let A = IA

Then perform the row operations simultaneously an A (on the left side of

equality) and I (identity matrix) an the right so as to change A I (if possible).

We arrive as I = B.A,

Then by definition B becomes the inverse of A. Following examples will justify he

method.

Let A = 4 3

3 2

4 3 1 0

A.3 2 0 1

Let R1 R1 R2, we obtain

1 1 1 1

A,3 2 0 1

First we wanted to get 1 as the leading element on L.. Matrix.

Now, R2 R2 – 3R1, we get

1 1 1 1A,

0 1 3 4

We wanted to get 0 at a21.

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Multiply R2 by (– 1), R2 (– 1) R2

1 1 1 1A.

0 1 3 4

[We wanted to get 1 at a22]

Operate R1 R1 – R2

1 0 2 3A.

0 1 3 4

[Here we wanted to get 0 at a21]

Thus A–1 = 2 3

3 4

since I = BA

B = A–1.

To find the inverse of

A =

1 1 2

2 0 1

3 2 1

Let A = IA

1 1 2 1 0 0

2 0 1 0 1 0 A.

3 2 1 0 0 1

Operate R2 R2 – 2R1, R3 R3 – 3R1

Leading element is 1 we don’t disturb it.

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Thus we, want to make a21, a31 as zeros that is why we have made this

operation.

1 1 2 1 0 0

0 2 3 2 1 0 A.

0 5 5 3 0 1

Operate R2 2

1R ,

2

This is to make a22 = 1

1 1 2 1 0 0

0 1 3 2 1 1 2 0 A

0 5 5 3 0 1

Operate R3 R3 – 5R2

Here we make a32 equal to zero.

1 1 2 1 0 0

0 1 3 2 1 1 2 0 A

0 0 5 2 2 5 2 1

Operate R3 3

2R

5

This is being done be make a33 = 1.

1 1 2 1 0 0

0 1 3 2 1 1 2 0 A

0 0 1 4 5 1 2 5

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Operate R2 2 3 1 1 3

3R R , R R 2R

2

These two operation will make a31 and a32 as zeros.

1 1 0 3 5 2 4 5

0 1 0 1 5 1 3 5 A

0 0 1 4 5 1 2 5

Operate R1 R1 + R2

To make a12 = 0.

1 0 0 2 5 1 1 5

0 1 0 1 5 1 3 5 A

0 0 1 4 5 1 2 5

Q.1. If A = 23 1

, f x x 4x 41 1

and f(A) = 0 then A4 is equal to

(a) 32 32

32 32

(b) 48 32

32 16

(c) 48 32

32 32

(d) none of these

A2 = 3 1 3 1 9 1 3 1 8 4

1 1 1 1 3 1 1 1 4 0

f(A) = 28 4 3 1 1 0

A 4A 4I 4 44 0 1 1 0 1

= 8 12 4 4 4 0 0

04 4 0 4 4 0 0

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A2 = 4A – 4I

A3 = 4A2 – 4AI = 4A2 – 4A

= 4(4A – 4I) – 4A

= 16A – 16I – 4A = 12A – 16I

A4 = 12A2 – 16A

= 12[4A – 4I] – 16A

= 48A – 48I – 16A

= 32A – 48I

= 3 1 1 0 96 48 32

32 481 1 0 1 32 32 48

A4 = 48 32

32 16

Q.2. If

1 2 5 1 7

0 1 A 2 0 3 ,

1 3 5 1 8

then A is equal

(a) 1 1 1

2 0 3

(b)

1 2

1 0

1 3

(c) 1 0 3

2 1 1

(d) 1 0 3

2 1 1

Sol. On L.H.S. O (Ist matrix) = 3 × 2,

Right hand side the order 3 × 3

O(A) = 2 × 3

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let A = a b c

d e f

Thus

1 2 5 1 7a b c

0 1 2 0 3d e f

1 3 5 1 8

1 2 5 1 7a b c

0 1 2 0 3d e f

1 3 5 1 8

a 2d 5, b 2e 1, c 2f 7

d 2, e 0 f 3

a 3d 5, b 3e 1, c 3f 8

a = 1, b = – 1, c = 1

Thus, a b c 1 1 1

Ad e f 2 0 3

option (a) is correct.

Q.3. If A =

1 1 1

2 1 0 ,

1 0 0

then A2

(a) O (b) A (c) A’ (d) A–1.

A2 =

1 1 1 1 1 1 1 2 1 1 1 0 1 0 0

2 1 0 2 1 0 2 2 0 2 1 0 2 0 0

1 0 0 1 0 0 1 0 0 1 0 0 1 0 0

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=

0 0 1

0 1 2

1 1 1

A = IA

1 1 1 1 0 0

2 1 0 0 1 0 A

1 0 0 0 0 1

1 1 1 0 0 1

0 1 2 A 0 1 0

0 0 1 1 0 0

C2 C2 + C1, C3 C3 – C1

1 0 0 0 0 1

0 1 2 A 0 1 0

0 0 1 1 1 1

C3 C3 + 2C2

1 0 0 0 0 1

0 1 0 A 0 1 2

0 0 1 1 1 1

C2 (–1) C3

1 0 0 0 0 1

0 1 0 A 0 1 2

0 0 1 1 1 1

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A2 = 1

0 0 1

0 1 2 A .

1 1 1

Q.4.

2

2

2

0 c b a ab ac

c 0 a ab b bc

b a 0 ac bc c

(a) I3 (b)

abc 0 0

0 abc 0

0 0 abc

(c)

2

2

2

a 0 0

0 b 0

0 0 c

(d) none of these

=

2 2 2 2

2 2 2 2

2 2 2 2

0 abc ab 0 b c b c 0 bc bc 0 0 0

a c 0 a c abc 0 abc ac 0 ac 0 0 0 0

a b a b 0 ab ab 0 abc abc 0 0 0 0

Q.5. If A = 1 tan 2

tan 2 1

and B = A–1, then AB–10 =

(a) O (b) 1 0

0 1

(c) cos sin

sin cos

(d) cos sin

sin cos

We know, if = ad – bc 0,

1a b d b1

c d c a

0

0

0

2

2

2

c b

c a

b a

a ab ac

ab b bc

ac bc c

L

NMMM

O

QPPP

L

NMMM

O

QPPP

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B = A’ = 21 tan 2

, 1 tantan 2 1 2

B–1 =

1 tan1 12

A.

tan 12

AB–1 = 2

2

1 1A

1 tan2

1 tan 2 1 tan 2

tan 2 1 tan 2 1

= cos sin

sin cos

cos =

2

2

1 tan2

1 tan2

sin = 2

2 tan2

1 tan2

= 2

22

1 tan 2 2 tan 21

2 tan 2 1 tan 21 tan2

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=

2

2 2

2

2 2

1 tan 2 tan2 2

1 tan 1 tan2 2

2 tan 1 tan2 2

1 tan 1 tan2 2

Q.6. If W, W2 are the complex cube roots of unity, then

2 2

2 2

2 2 2

1 W W W W 1 1

W W 1 W 1 W W

W 1 W W W 1 W

(a)

0

0

0

(b)

1

1

1

(c) 2

1

W

W

(d)

2W

W

1

2 2

2 2

2 2 2

1 W W W W 1 1

W W 1 W 1 W W

W 1 W W W 1 W

=

2 2 2

2 2 2

2 2 2 2 2

1 W W W W 1 1 W 1 W 1

W W 1 W 1 W W 1 W W W

W W 1 W 1 W W 1 W W W

=

22 3

2 4 2

2 4 2

W W 1W W W

1 W W 1 W W

1 W W 1 W W

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=

0

0

0

Q.7. If A = 3 4

,1 1

the An =

(a) 1 2n 4

n 2 n

(b) 1 2n 4

2n 1 1 2n

(c) 1 2n 4n

n 1 2n

(d) none of these.

A2 = 3 4 3 4

1 1 1 1

= 9 4 12 4 5 8 1 2.2 4.2

3 1 4 1 5 3 2.1 1 22

A3 = 5 8 3 4 15 8 20 8

2 3 1 1 6 3 8 3

= 5 8 3 4 15 8 20 8

2 3 1 1 6 3 8 3

An = 1 2n 4n

n 1 2n

Q.8. If ln, mr, nr are the directions of three mutually perpendicular lines, then A =

1 1 1

2 2 2

3 3 3

l m n

l m n

l m n

is

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(a) a unit matrix (b) an orthogonal matrix (c) a skew-symmetric matrix (d) a

Hermitian matrix

We know

l1l2 + m1m2 + n1n2 = l1l3 + m1m3 + n1n3 = l2l3 + m2m3 + n2n3 = 0

Consider AAT = 1 2 31 1 1

1 2 32 2 2

1 2 33 3 3

l l ll m n

m m ml m n

n n nl m n

=

2 2 21 1 1 1 2 1 2 1 2 1 3 1 3 1 3

2 2 22 1 2 1 2 1 2 2 2 2 3 2 3 2 3

2 2 23 1 3 1 3 1 3 2 3 2 3 2 3 3 3

l m n l l m m n n l l m m n n

l l m m n n l m n l l m m n n

l l m m n n l l m m n n l m n

=

1 0 0

0 1 0 I

0 0 1

A is an orthogonal matrix.

Q.9. If A = 2

2

cos cos sin,

cos sin sin

2

2

cos cos sinB

cos sin sin

and A × B = O, then –

(a) 0 (b) odd multiple of 2

(c) multiple of (d) none of these.

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Sol. AB = 2 2

2 2

cos cos sin cos cos sin

cos sin sin cos sin sin

= 2 2 2 2

2 2 2 2

cos cos cos sin cos sin cos cos sin cos sin sin

cos sin cos sin cos sin cos sin cos sin sin sin

=

cos cos cos cos sin sin cos sin cos cos sin sin

sin cos cos cos sin sin singqsin cos cos sin sin

= cos cos cos sin

cossin cos sin sin

If A.B = 0

cos ( – ) = 0

– = odd multiple of .2

Q.10. The matrix A for which

3 4 1 1is

2 3 1 0

(a) 1 3

1 2

(b) 1 3

1 2

(c) 1 3

1 2

(d) 1 3

1 2

Sol. If B = a b

c d

then B–1 = d ba

,c aad bc

ad – bc 0

for B = 13 4 3 41

, B , ad bc 9 8 02 3 2 39 8

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B–1 = 3 4

,2 3

thus

given BA = 1 1

1 0

B–1BA = 11 1

B1 0

IA = 3 4 1 1 3 4 3 0

2 3 1 0 2 3 2 0

A = 1 3

1 2

Q.11. The value of x, for which

1 2 3 1

1 1 x 1 0 1 x 4

0 1 4 1

(a) 2 or – 2 (b) – 2 or 0 (c) 0, 2 (d) none of these

Sol.

1 2 3 1

1 1 x 1 0 1 x 4

0 1 4 1

1

1 1 2 x 3 1 4x x 4

1

22x x 4 4x 4

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2x 2x 4 4

2x 2x 4 4

x x 2 0

x 0 or 2

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