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Matrices and Determinants

1. DEFINITION OF MATRIXA system of m × n numbers arranged in the form of an ordered set of m rows and n columns is calledan m × n matrix. It can be read as m by n matrix. It is represents as A = [a

ij]

m × n and can be written in

expanded form as11 12 1n

21 22 2n

m1 m2 mn

a a . . . aa a . . . a

A

a a . . . a

e.g.,

2 1 0

1 1 3

6 5 1

is a 3by 3 matrix.

2. DIFFERENT TYPES OF MATRICES(i) Horizontal Matrix : Any matrix in which the number of columns is more than the number

of rows is called a horizontal matrix.

e.g.

4322

2798

5432

is a horizontal matrix.

Since here number of columns > number of rows.

(ii) Row Matrix: A matrix which has only one row and n columns is called a rowmatrix of length ne.g., [2 –1 3 0]

1×4 is a row matrix of length 4.

(iii) Vertical Matrix : Any matrix in which the number of rows is more than the numberof columns is called column matrix.

e.g.

98

76

54

32

is a column matrix.

Since here number of rows > number of columns.(iv) Column Matrix: A matrix which has only one column and m rows is called a column

matrix of length m.


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e.g.

1

3x

2

1

is a column matrix of length 4.

(v) Null or Zero Matrix : The matrix whose all elements are zero is called null matrix or zeromatrix. It is usually denoted by O.

e.g.

000

000

000

O 33

(vi) Square Matrix: A matrix for which the number of rows is equal to the number ofcolumns (each equal to n) is called a square matrix of order n.

e.g. A =

1 0 2 3

2 1 4 5

3 2 4 1

1 0 0 2

is a square matrix of order 4.

(vii) Diagonal Matrix: A square matrix of any order with zero elements every where,except on the main diagonal, is called a diagonal matrix.

e.g.

3 0 0

0 5 0

0 0 1

is a diagonal matrix of order 3.

(viii) Scalar matrix : A matrix whose diagonal elements are all equal and other entries

are zero, is called a scalar matrix e.g.k 0 0

A 0 k 00 0 k

(ix) Identity or Unit Matrix : A square matrix in which all the elements along themain diagonal (elements of the from a

ii) are unity is called

an identity matrix or a unit matrix. An identity matrix oforder n is denoted by I

n.

e.g. I3 =

1 0 0

0 1 0

0 0 1

(x) Triangular Matrix: A square matrix whose elements above the main diagonal or belowthe main diagonal are all zero is called a triangular matrix.

Note: (i) [aij]

n × n is said to be upper triangular matrix if i > j a

ij = 0,

(ii) [aij]

n × n is said to be lower triangular matrix if i < j a

ij = 0.


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e.g.

1 2 1 3

0 2 1 0

0 0 3 1

0 0 0 3

is an upper triangular matrix.

(xi) Sub Matrix : A matrix obtained by omitting some rows or some columns orboth of a given matrix A is called a sub matrix of A.

e.g., If

2 0 4

A 5 6 8

3 2 29

, then2 0

5 6

is a submatrix of A which

is obtained by omitting third row and third column of A.

3. EQUALITY OFTWO MATRICESTwo matrices A = [a

ij] and B = [b

ij] are said to be equal if they are of the same order and their

corresponding elements are equal. If two matrices A and B are equal, we write A = B.

4. ADDITION OF MATRICESIf A and B are two matrices of the same order m × n, then their sum is defined to be the matrix oforder m × n obtained by adding the corresponding elements of A and B.

e.g. If11 12 13 11 12 13

21 22 23 21 22 23

31 32 33 31 32 33

a a a b b b

A a a a and B b b b

a a a b b b

,

then11 11 12 12 13 13

21 21 22 22 23 23

31 31 32 32 33 33

a b a b a b

A B a b a b a b

a b a b a b

4.1 PROPERTIES OF MATRIXADDITION(i) Matrix Addition is Commutative

If A and B are two m × n matrices, then A + B = B + A.

(ii) Matrix addition is associativeIf A, B, C are three matrices, each of the order m × n, then (A + B) + C = A + (B + C).

(iii) Existence of additive identityIf O is the m × n null matrix, then A + O = A = O + A for every m × n matrix A. O is calledadditive identity.


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Illustration 1:If is an imaginary cube root of unity, show that

2 2 2

2 2 2

2 2 2

1 1 1

1 1 1

1 1 1

is a null matrix.

Solution:2 2 2

2 2 2

2 2 2

1 1 1

1 1 1

1 1 1

2 2 2

2 2 2

2 2 2

1 1 1 0 0 0

1 1 1 0 0 0

1 1 1 0 0 0

, which is a null matrix.

5. MULTIPLICATION OFA MATRIX BYA SCALARIf A = [a

ij]

m × n and is a scalar, then A = [a

ij]

m × n

e.g.

2 1 3

A 2 5 1

0 1 2

– 3A =6 3 96 15 3

0 3 6

.

6. MULTIPLICATION OFTWO MATRICESLet A = [a

ij]

m× n and B = [b

jk]

n×p be two matrices such that the number of columns in A is equal to

number of rows in B. Then the m × p matrix C = [cik]

m× p ,

where cik =

n

ij jkj 1

a .b (where i = 1, 2, 3 ... m, k = 1, 2, 3 ... p), is called the product of the matrices

A and B. We have

11 12 1p 11 12 1p11 12 1n

21 22 2p 12 22 2p21 22 2n

n1 n 2 np m1 m2 mpm1 m2 mn m n n p m p

b b ... b c c ... ca a ... a

b b ... b c c ... ca a ... a

... ... ... ... ... ... ... ...... ... ... ...

b b ... b c c ... ca a ... a

,

where Cij = a

i1 b

1j + a

i2 b

2j + ... + a

in b

nj =

n

ik kjk 1

a b

Properties of Matrix Multiplication


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(i) Matrix multiplication is associative

i.e., (AB)C = A(BC), A, B and C are m × n, n ×p and p × q matrices respectively.

(ii) Multiplication of matrices is distributive over addition of matricesi.e., A(B + C) = AB + AC

(iii) Existence of multiplicative identity of square matrices.If A is a square matrix of order n and I

n is the identity matrix of order n, then

A In = I

n A = A.

(iv) Whenever AB and BA both exist, it is not necessary that AB = BA.

(v) The product of two matrices can be a zero matrix while neither of them is a zero matrix.

e.g., If A =0 1 1 0 0 0

and B then AB0 0 0 0 0 0

, while neither A nor B is a

null matrix.

(vi) In the case of matrix multiplication of AB = 0, then it doesn’t necessarily imply that A = 0or B = 0 or BA = 0.

Illustration 2:Show that product of two upper (lower) triangular matrices is an upper (lower) triangular matrix.

Solution:Let A = [a

ij]

m× n and B = [b

jk]

n×p be two upper triangular matrices.

we know that aij = 0, when i > j and b

jk = 0 when j > k.

Now C = AB is of the type m × p, where C = [cik]

m×p, c

ik =

n

ij jkj 1

a b .

For i > k, cik = a

i1 b

1k + a

i2 b

2k + . . . + a

ik b

kk + a

ik + 1 b

k + 1k + . . . + b

nk = 0.

Hence AB is an upper triangular matrix.

DRILL EXERCISE - 1

1. Construct a 2 × 3 matrix whose elements are given by aij = i + 2j.

2. For A =

32

11 , show that A B is a diagonal matrix for any value of if =

52

1.

3. Show that the product of the matrices2

2cos cos sec

cos sin sin

and2

2cos cos sec

cos sin sin

is

the zero matrix, when and differ by an odd multiple of2

.


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4. For A =

32

11, show that AA2 4A + 5I

2 = 0.

5. If ‘A’ is a diagonal matrix then prove that AB = BA is possible only when B is a diagonal matrix.

7. TRACE OFA MATRIXLet A be a square matrix of order n. The sum of the diagonal elements of A is called the trace of A.

Trace (A) =n

ii 11 22 nni 1

a a a ... a .

8. TRANSPOSE OFA MATRIXThe matrix obtained from any given matrix A, by interchanging rows and columns, is called the transposeof A and is denoted by A or AAT .

e.g., If

3 2

1 2

A 4 5

7 8

, then2 3

1 4 7A

2 5 8

Properties of Transpose of a matrix

(i) (A ) A (ii) (A B) A B

(iii) ( A) A, being any scalar (iv) (AB) B A

DRILL EXERCISE - 2

1. If a, b, and c are the roots of an equation of the form 3 2x x p 0 , prove that the matrix

A =

a b cc a bb c a

satisfy AAT = I .

2. Let A =

cos

cos

cos

cos

x

x

x

x

1 0 0

0 2 1 0

0 0 3 1

1 0 0 4

L

N

MMMM

O

Q

PPPP. If trace A = -

1

2, find x .

3. Show that : (i) tr. (A + B) = tr. A + tr. B, A and B being of order n.

(ii) If A is of order m × n and B of order n × m, then tr. AB = tr. BA


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4. If A is 3 × 4 matrix and B is a matrix such that AB and BA are both defined, then find thetype of B.

5. If A =

481

744

418

9

1, then prove that AAT = I.

9. SPECIAL MATRICESSymmetric Matrix

A matrix which is unchanged by transposition is called a symmetric matrix. Such a matrix is necessarilysquare

e.g.,

2 1 3

1 4 1

3 1 5

Thus if A = [aij]

m × nis a symmetric matrix then m = n, a

ij = a

ji i.e., A A .

Skew Symmetric MatrixA square matrix A = [a

ij] is said to be skew symmetric, if a

ij = – a

ji for all i and j

e.g.

0 2 3 1

2 0 4 3

3 4 0 1

1 3 1 0

Thus if A = [aij]

m × n is a skew symmetric matrix, then m = n, a

ij = – a

ji i.e., A A .

Obviously diagonal elements of a square matrix are zero.

Note: Every square matrix can be uniquely expressed as the sum of symmetric and skew symmetric matrix.

i.e., A = )AA(2

1)AA(

2

1 , where )AA(2

1 and )AA(2

1 are symmetric and skew

symmetric parts of A.

Illustration 3:

Express A as the sum of a symmetric and a skew symmetric matrix, where A =

705

631

324

.

Solution: We have

A =

705

631

324


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4 1 5

A 2 3 0

3 6 7

Then AA

705

631

324

+

763

032

514

=

068

663

838

................(i)

and A A

705

631

324

–

763

032

514

=

0 1 2

1 0 6

2 6 14

................(ii)

Adding (1) and (ii) we get

2A =

1462

601

210

068

663

838

A=

731

302/1

12/10

034

332/3

42/34

Symmetric matrix Skew symmetric matrixOrthogonal Matrix

A square matrix A is said to be orthogonal, if AA = A A I , where I is a unit matrix.

Note: (i) If A is orthogonal, then A is also orthogonal.

(ii) If A and B are orthogonal matrices then AB and BA are also orthogonal matrices.

Illustration 4:


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If A =1 1 1

2 2 2

3 3 3

m n

m n

m n

l

l

l

, where < l1, m

1, n

1 > , < l

2 , m

2, n

2 > and < l

3, m

3, n

3 > are the direction of

cosines of three mutually perpendicular straight lines, then prove that AA= I.

Solution:Since < l

1, m

1, n

1 > , < l

2 , m

2, n

2 > and < l

3, m

3, n

3 > are the direction cosines of three mutually

perpendicular straight lines

2 2 21 1 1m n 1, l 2 2 2 2 2 2

2 2 2 3 3 3m n 1and n m 1 l l

and l1l2 + m

1m

2 + n

1n

2 = l

2l3 + m

2m

3 + n

2n

3 = l

3l1 + m

3m

1 + n

3n

1 = 0

We have A =1 1 1

2 2 2

3 3 3

m n

m n

m n

l

l

l

1 2 3

1 2 3

1 2 3

A m m m

n n n

l l l

.

21 2 3 1 1 1 1 1 1 1 1

21 2 3 2 2 2 1 1 1 1 1

21 2 3 3 3 3 1 1 1 1 1

m n m n

AA m m m m n n m m n

n n n m n n m n n

l l l l l l l

l l

l l

1 0 00 1 0 I0 0 1

Hence proved.

Idempotent Matrix: A square matrix A is called idempotent provided it satisfies the relation A2 = A.

e.g. The matrix2 2 4

A 1 3 41 2 3

is idempotent as

22 2 4 2 2 4 2 2 4

A 1 3 4 1 3 4 1 3 41 2 3 1 2 3 1 2 3

Involuntary Matrix

A matrix A such that A2 = I, is called involuntary matrix.

Illustration 5:


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Show that the matrix A =0 11 0

satisfies AA2 = –I. Hence or otherwise find the 16th power of the

matrix1 11 1

.

Solution:

2 0 1 0 1 1 0A I

1 0 1 0 0 1

Let B =1 1 0 1 1 01 1 1 0 0 1

= A + I

B2 = (A + I) (A + I) = A2 + 2A + I

Since A2 = –I, B2 = 2A

B16 = (B2)8 = (2A)8 = 28 (A2)4 = 28 (–I)4 = 281 0 256 00 1 0 256

.

Periodic MatrixA square matrix A is called periodic, if Ak + 1 = A, where k is a positive integer. If k is the least positiveinteger for which Ak + 1 = A, then k is said to be period of A. For k = 1, we get A2 = A and we calledit to be idempotent matrix.

Nilopotent MatrixA square matrix A is called a nilpotent matrix, if there exists a positive integer m such that Am =O. If mis the least positive integer such that Am = O, then m is called the index of the nilpotentmatrix A.

Illustration 6:

Show that the matrix1 1 3

5 2 6

2 1 3

is a nilpotent matrix of index 3.

Solution: Let2

1 1 3 0 0 0

A 5 2 6 A 3 3 9

2 1 3 1 1 3

3 2

0 0 0 1 1 3 0 0 0

A A .A 3 3 9 5 2 6 0 0 0

1 1 3 2 1 3 0 0 0

A3 = 0 i.e., A is a nilopotent matrix of index 3.

DRILL EXERCISE - 3

1. If A and B are two given square matrices, then show that

(i) BT AB is symmetric, if A is symmetric .


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(ii) BT AB is skew-symmetric, if A is skew symmetric.

2. Prove that the matrix

21 1 2 1 3

21 2 2 2 3

21 3 2 3 3

is idempotent, where 1 2 3, , are the direction

cosines of a line .

3.(i) Determine x, y, z such that0 2y zx y zx y z

LNMM

OQPP is orthogonal.

(ii) If A, B, A + I and A + B are idempotent matrices, show that AB = BA.

4. Find symmetric & skew symmetric parts of A =1 2 46 8 13 5 7

LNMM

OQPP .

5. The matrix

431

451

431

is a nilpotent matrix, then find the index of given matrix.

10. DETERMINANTEquations a

1x + b

1y = 0 and a

2x + b

2y = 0 in x and y have a unique solution if and only if

a1b

2 –a

2b

1 0. We write a1b

2 – a

2b

1 as

1 1

2 2

a b

a b and call it a determinant of order 2.

Similarly the equations a1x + b

1y + c

1z = 0, a

2x + b

2y + c

2z = 0 and a

3x + b

3y + c

3z = 0 have a unique

solution if a1(b

2 c

3 – b

3c

2) + b

1 (a

3c

2 – a

2c

3) + c

1 (a

2 b

3 – a

3 b

2) 0

i.e., 1 1 1

2 2 2

3 3 3

a b c

a b c 0

a b c

The number ai, b

i, c

i(i = 1, 2, 3) are called the elements of the determinant.

The determinant obtained by deleting the ith row and jth column is called the minor of the element atthe ith row and jth column. We shall denote it by M

ij. The cofactor of this element is (–1)i+j M

ij,

denoted by Cij.

Let A = [aij]

3×3 be a matrix, then the corresponding determinant (denoted by det A or | A |) is

11 12 13

21 22 23

31 32 33

a a a

a a a

a a a.

It is easy to see that | A | = a11

C11

+ a12

C12

+ a13

C13

(we say that we have expanded the determinant| A | along first row). Infect value of | A | can be obtained by expanding it along any row or along anycolumn. Further note that if elements of a row (column) are multiplied to the cofactors of other row

(column) and then added, then the result is zero : 11 21 12 22 13 23a C a C a C 0 .

10.1 PROPERTIES OF DETERMINANTS(i) The value of a determinant remains unaltered, if its rows are changed into columns and the

columns into rows.


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e.g.,1 1 1 1 2 3

2 2 2 1 2 3

3 3 3 1 2 3

a b c a a a

a b c b b b

a b c c c c

. Thus any property true for rows will also be true for

columns.

(ii) If all the elements of a row (or column) of a determinant are zero, then the value of thedeterminant is zero.

e.g.,

1 1 1 1 1

2 2

3 3 3 3 3

0 b c a b c

0 b c 0, 0 0 0 0

0 b c a b c

(iii) If any two rows (columns) of a determinant are identical, then the value of the determinantis zero.

e.g.,

1 1 1

2 2 2

3 3 3

a a c

a a c 0

a a c

(iv) The interchange of any two rows (columns) of a determinant results in change of it’s sign

i.e,

1 1 1 1 1 1

2 2 2 2 2 2

3 3 3 3 3 3

a b c b a c

a b c b a c

a b c b a c

(v) If all the elements of a row (column) of a determinant are multiplies by a non-zero constant,then the determinant gets multiplied by that constant.

e.g.,1 1 1 1 1 1

2 2 2 2 2 2

3 3 3 3 3 3

a kb c a b c

a kb c k a b c

a kb c a b c

and1 1 1 1 1 1

2 2 2 2 2 2

3 3 3 3 3 3

a b c ka kb kc

k a b c a b c

a b c a b c

(vi) If each element of a row (column) of a determinant is a sum of two terms, then determinantcan be written as sum of two determinant in the following way:

1 1 1 1 1 1 1 1 1 1

2 2 2 2 2 2 2 2 2 2

3 3 3 3 3 3 3 3 3 3

a b c d a b c a b d

a b c d a b c a b d

a b c d a b c a b d

In general

n n n

r 1 r 1 r 1n

2 2 2 2 2 2r 1

3 3 3 3 3 3

f (r) g(r) h(r)f (r) g(r) h(r)

a b c a b c

a b c a b c


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(vii) The value of a determinant remains unaltered under a column operation of the form

i i j kC C C C ( j, k i) or a row operation of the form

i i j kR R R R ( j, k i).

e.g.,1 1 1 1 1 1 1 1

2 2 2 2 2 2 2 2

3 3 3 3 3 3 3 3

a b c a b 2a 3c c

a b c a b 2a 3c c

a b c a b 2a 3c c

obtained after C

2 C2 + 2C

1 + 3C

3.

(viii) Product of two determinants

1 1 1 1 2 3

2 2 2 1 2 3

3 3 3 1 2 3

a b c

a b c m m m

a b c n n n

l l l

1 1 1 1 1 1 1 2 1 2 1 2 1 3 1 3 1 3

2 1 2 1 2 1 2 2 2 2 2 2 2 3 2 3 2 3

3 1 3 1 3 1 3 2 3 2 3 2 3 3 3 3 3 3

a b m c n a b m c n a b m c n



l l l

l l l

l l l

(row by column multiplication)

1 1 1 2 1 3 1 1 1 2 1 3 1 1 1 2 1 3

2 1 2 2 2 3 2 1 2 2 2 3 2 1 2 2 2 3

3 1 3 2 3 3 3 1 3 2 3 3 3 1 3 2 3 3

a b c a m b m c m a n b n c n



l l l

l l l

l l l

(row by row multiplication)

We can also multiply determinants column by row or column by column.

DRILL EXERCISE - 4

1. If a, b and c are positive and are the pth, qth and rth terms respectively of a G.P., then show without

expanding that

log a p 1

log b q 1 0

log c r 1

.

2. Prove that

cos( ) cos( ) cos( )

sin( ) sin( ) sin( )

sin( ) sin( ) sin( )

is independent of .


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3. Prove that

2 2

2 2

2 2

a ab c ca b c b

ab c b bc a c a

ac b bc a c b a

= 3 2 2 2 2 3( a b c ) .

4. Find the value of determinant

515653

1052615

552133

.

5. A mixture is to be made of three foods A, B, C. The three foods A, B, C contain nutrients P, Q, R asshown below :

Ounces per pound of NutrientFood

P Q RA 1 2 5B 3 2 1C 4 2 1How to form a mixture which will have ounces of P, 5 ounces of Q and 7 ounces of R ?

(ix) Limit of a determinant

Letx a x a x a

x a x a x a x a

x a x a x a

lim f (x) lim g(x) lim h(x)f (x) g(x) h(x)

(x) (x) m(x) n(x) , then lim (x) lim l(x) lim m(x) lim n(x) ,

u(x) v(x) w(x) lim u(x) lim v(x) lim w(x)

l

provided each of nine limiting values exist finitely.

(x) Differentiation of a determinant

Let

f (x) g(x) h(x)

(x) l(x) m(x) n(x) ,

u(x) v(x) w(x)

then

f (x) g (x) h (x) f (x) g(x) h(x) f (x) g(x) h(x)

(x) (x) m(x) n(x) (x) m (x) n (x) (x) m(x) n(x)

u(x) v(x) w(x) u(x) v(x) w(x) u (x) v (x) w (x)

l l l


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(xi) Integration of a Determinant

Let

f (x) g(x) h(x)

(x) a b c

l m n

, where a, b, c, l, m and n are constants,

then

b b b

a a ab

a

f (x)dx g(x)dx h(x)dx

(x)dx a b c

m n

l

Note that if more than one row (column) of (x) are variable, then in order to find

b

a

(x)dx first

we evaluate the determinant (x) by using the properties of determinants and then we integrate it.

DRILL EXERCISE - 5

1. If f(x) =

cos sin( ) cos( )

sin( ) cos( ) sin( )

sin( ) sin( )

x x x x x x

x x x x x x

x x

2 2 2

2 2 2

22 0 2

c h then find f ( )0 .

2. Let f(x) =1xxtan

x2xxsin2

1xxcos2

, then findx

)0('flim

0x.

3. If a, b, c are in A.P. and f(x) =

12x3cx

11x2bx

11xax

2

2

2

then f(x) is

4. If r r rf (x),g (x), h (x) where r = 1, 2, 3 are polynomials in x, such that

fr(a) = g

r(a) = h

r (a) ; r = 1, 2, 3 and

1 2 3

1 2 3

1 2 3

f (x) f (x) f (x)

F(x) g (x) g (x) g (x)

h (x) h (x) h (x)

, then find F (a) .

5. Ifu

y ,v where u and v are functions of x, show that

23

2

u v 0d y

v u v vdx

u v 2v

.


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11. SPECIAL DETERMINANTS

(i) Skew symmetric Determinant

A determinant of a skew symmetric matrix of odd order is zero. e.g.,

0 b c

b 0 a 0

c a 0

(iii) Circulant DeterminantA determinant is called circulant if its rows (columns) are cyclic shifts of the first row (columns).

e.g.,

a b c

b c a

c a b. It can be show that its value is – (a3 + b3 + c3 – 3abc) .

(iv)2 2 2

1 1 1

a b c (a b) (b c) (c a)

a b c

(v)3 3 3

1 1 1

a b c (a b) (b c) (c a) (a b c)

a b c

(iv)2 2 2

3 3 3

1 1 1

a b c

a b c = (a – b) (b – c) (c – a) (ab + bc + ca)

Illustration 7:If A, B and C are the angles of a triangle and

2 2 2

1 1 1

1 sin A 1 sin B 1 sin C 0

sin A sin A sin B sin B sin C sin C

, prove that the ABC is isosceles.

Solution :

Let2 2 2

1 1 1

1 sin A 1 sin B 1 sin C

sin A sin A sin B sin B sin C sin C

Applying 2 2 1C C C and 3 3 1C C C , we get


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2

1 0 0

1 sin A sin B sin A sin C sin A

sin A sin A (sin B sin A)(sin B sin A 1) (sin C sin A)(sin C sin A 1)

Expanding along R1

sin B sin A sin C sin A

(sin B sin A) (sin B sin A 1) (sin C sin A) (sin C sin A 1)

= (sinB – sinA) (sinC – sinA)1 1

sin B sin A 1 sin C sin A 1

Now 0 (sinB – sinA) (sinC – sinA) (sinC – sinB) = 0 sinB = sinA or sinC = sinA or sinC = sinB

B = A or C = A or C = BIn all the three cases, we will have an isosceles triangle.

12. ADJOINT OFA SQUARE MATRIXLet A = [a

ij]

n×n be an n × n matrix. The transpose Bof the matrix B = [A

ij]

n ×n, where AA

ij denotes the

cofactor of the element aij in the determinant |A|, is called the adjoint of the matrix A and is denoted by

the symbol adj A.Thus the adjoint of a matrix A is the transpose of the matrix formed by the cofactors of A i.e., if

11 12 1n 11 21 1n

21 22 2n 12 22 n2

n1 n2 nn 1n 2n nn

a a a A A A

a a a A A AA , then AdjA

a a a A A A

It is easy to see that A(adjA) = (adj A)A = |A|. In.

13. INVERSE OFA SQUARE MATRIXLet A be any n–rowed square matrix. Then a matrix B, if exists, such that AB = BA = I

n, is called

the inverse of A. Inverse of A is usually denoted by A–1(if exists).We have |A| I

n = A(adjA)

|A| AA–1 = (adjA). Thus the necessary and sufficient condition for a square matrix A to possess

the inverse is that |A| 0 and then AA–1 =Adj(A)

| A |. A square matrix A is called non-singular if

|A| 0. Hence a square matrix A is invertible if and only if A is non-singular..

Properties of Inverse of a Matrix(i) Every invertible matrix possesses a unique inverse.(ii) If A and B are invertible matrices of the same order, then AB is invertible and

(AB)–1 = B–1 A–1.(iii) If A is an invertible square matrix, then AT is also invertible and (AT)–1 = (A–1)T.(iv) If A is a non-singular square matrix of order n, then |adjA| = |A|n–1

(v) If A and B are non-singular square matrices of the same order, thenadj (AB) = (adj B) (adj A)


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Illustration 8:

If A =

1 2 2

2 1 2

2 2 1

, then show that AA2 – 4A – 5I = 0, where I and 0 are the unit matrix and the null

matrix of order 3 respectively. Use this result to find A–1 .

Solution :

Given A =

1 2 2

2 1 2

2 2 1

A2 =

1 2 2 1 2 2

2 1 2 2 1 2

2 2 1 2 2 1

=

9 8 8

8 9 8

8 8 9

2

9 8 8 1 2 2 1 0 0

A 4A 5I 8 9 8 4 2 1 2 5 0 1 0

8 8 9 2 2 1 0 0 1

9 8 8 4 8 8 5 0 0

8 9 8 8 4 8 0 5 0

8 8 9 8 8 4 0 0 5

=

0 0 00 0 00 0 0

A2 – 4A – 5I = 0or 5I = A2 – 4Aon multiplying by A–1 , we get

5A–1 = A – 4I1 2 2 4 0 0

2 1 2 0 4 0

2 2 1 0 0 4

3 2 2

2 3 2

2 2 3

1

3 2 21

A 2 3 25

2 2 3

3/ 5 2 / 5 2 / 5

2 / 5 3/ 5 2 / 5

2 / 5 2 / 5 3/ 5

.

DRILL EXERCISE - 6

1. Prove that Adj (Adj (A)) = |A| n–2 A. Deduce that |Adj (Adj (A))| =2)1n(|A| .

2. If A =1 23 4

, show that A satisfies the equation AA2 – 5A – 2I = 0. Using this result determine A5.


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3. If A and B are non-singular matrices of order n × n, prove that(a) (AB)–1 = B–1A–1

(b) If A is non-singular then (A–1)–1 = A.(c) If A is non-singular, then (A–1)´ = (A´)–1.

(d) If the matrix product AB of two square matrices is zero, then either A = 0 or B = 0 or both A and B are singular matrices.

4. Find the condition for the matrix

012

)cb(cb

)ba(ba

to be non invertible.

5. If a matrices A satisfies a relation A2 + A – I = 0, prove that A–1 exists and A–1 = I + A, I beingidentity matrix.

14. SYSTEM OF LINEAR SIMULTANEOUS EQUATIONSConsider the system of linear non-homogeneous simultaneous equations in three unknowns x, yand z, given by a

1x + b

1y + c

1z = d

1, a

2x + b

2y + c

2z = d

2 and a

3x + b

3y + c

3z = d

3 ,

Let1 1 1 1

2 2 2 2

3 3 3 3

a b c x d

A a b c ,X y ,B d ,

a b c z d

Let | A | =1 1 1 1 1 1

2 2 2 x 2 2 2

3 3 3 3 3 3

a b c d b c

a b c , d b c

a b c d b c

, obtained on replacing first column of by B.

Similarly let1 1 1 1 1 1

y 2 2 2 z 2 2 2

3 3 3 3 3 3

a d c a b d

a d c and a b d

a d c a b d

.

It can be shown that AX = B, x y zx , y. , z .

14.1 Determinant Method of SolutionWe have the following two cases :

Case I

If 0, then the given system of equations has unique solution, given by

x y zx / , y / and z / .

Case II

If 0, then two sub cases arise:


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(a) at least one of x , y zand is non-zero, say x 0. Now in x. x , L.H.S. is zero

and R.H.S. is not equal to zero. Thus we have no value of x satisfying x. x . Hence

given system of equations has no solution.

(b) x y z 0. In the case the given equations are dependent. Delete one or two

equation from the given system (as the case may be) to obtain independent equation(s).The remaining equation(s) may have no solution or infinitely many solution(s).

For example in x + y + z = 3, 2x + 2y + 2z = 9, 3x + 3y + 3z = 12, x y z 0

and hence equations are dependent (infact third equation is the sum of first two equations).Now after deleting the third equation we obtain independent equations x + y + z = 3,2x + 2y + 2z = 9, which obviously have no solution (infact these are parallel planes) where

as in x + y + z = 3, 2x – y + 3z = 4, 3x + 4z = 7, x = y z 0 and hence

equations are dependent (infact third equation is the sum of first two equations). Now afterdeleting any equation (say third) we obtain independent equations x + y + z = 3,2x – y + 3z = 4, which have infinitely many solutions (infact these are non parallel planes)

For let z = R, then7 4

x3

and

2y

3

. Hence we get infinitely many solutions.

14.2 Matrix Method of Solution

(a) 0, then AA–1 exists and hence AX = B AA–1 (AX) = A–1 B x = AA–1 B and thereforeunique values of x, y and z are obtained.

(b) : We have AX = B ((adj A)A)X = (adj A)B X = (adj A)B.

If = 0, then X = 03 × 1

, zero matrix of order 3 × 1. Now if (adj A)B = 0, then the systemAX = B has infinitely many solution, else no solution.

Note : A system of equation is called consistent if it has a least one solution. If the system has no solution,then it is called inconsistent.

Illustration 9:Solve the system of equations, x + 2y + 3z = 1 ; 2x + 3y + 2z = 2 ; 3x + 3y + 4z = 1with the help of matrix inversion.

Solution :The given system of equations in the matrix form can be written as

1 2 3 x 1

2 3 2 y 2

3 3 4 z 1

AX = B

where

1 2 3 x 1

A 2 3 2 ,X y and B 2

3 3 4 z 1

.

Now |A| = 1(12 – 6) – 2 (8 – 6) + 3(6 – 9) = 6 – 4 – 9 = –7 0.Hence the given system has unique solution.


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Let C be the matrix of cofactors of elements in |A|. Then11 12 13

21 22 23

31 32 33

C C C

C C C C

C C C

Here

6 2 3

C 1 5 3

5 4 1

Adj A = =

6 1 5

C 2 5 4

3 3 1

1

6 1 5AdjA 1

A 2 5 4| A | 7

3 3 1

6 / 7 1/ 7 5 / 7

2 / 7 5 / 7 4 / 7

3/ 7 3/ 7 1/ 7

1

6 / 7 1/ 7 5 / 7 1

A B 2 / 7 5 / 7 4 / 7 2

3/ 7 3/ 7 1/ 7 1

x 3/ 7

y 8 / 7

z 2 / 7

( AA–1 B = X )

x = –3/7, y = 8/7, z = –2/7

DRILL EXERCISE - 7

1. Solve the following system of equations by Determinant Method.

(i) x + y + z = 6, x + 2y + 3z = 14, x + 4y + 7z = 30

(ii) x + y + z = 6, x – y + z = 2, 2x + y – z = 1

(iii) x + y + z = 3, x + 2y + 3z = 4, 2x + 3y + 4z = 8.


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2. If x = cy + bz, y = az + cx, z = bx + ay where x, y, z are not all zeros, prove by using consistencyof equations that a2 + b2 + c2 + 2 abc = 1.

3. Solve the following system of equation by Matrix Method

(i) 5x + 2y = 4; 7x + 3y = 5

(ii) 2x + y + z = 1 ;x – 2y – z =2

3; 3y – 5z = 9

4. Solve the following system of equation by Matrix Method(i) 2x + 3y + 3z = 5; x – 2y + z = –4; 3x – y – 2z = 3(ii) x – y + 2z = 7; 3x + 4y – 5z = –5; 2x – y + 3z = 12

5. Investigate for what values of , the equationsx + y + z = 6x + 2y + 3z = 10

x + 2y + z have(i) no solution(ii) unique solution(iii) infinitely many solutions

15. SYSTEM OF LINEAR HOMOGENEOUS SIMULTANEOUS EQUATIONSConsider the system of linear homogeneous simultaneous equations in three unknowns x, y and z,given by a

1x + b

1y + c

1z = 0, a

2x + b

2y + c

2z = 0 and a

3x + b

3y + c

3z = 0.

In this case, system of equations is always consistent as x = y = z = 0 is always a solution. If thesystem has unique solution (the case when coefficient determinant 0), then x = y = z = 0 is theonly solution (called trivial solution). However if the system has coefficient determinant = 0, thenthe system has infinitely many solutions. Hence in this case we get solutions other than trivialsolution also and we say that we have non-trivial solutions.

Illustration 10:Solve : 2x + 3ky + (3k + 4) z = 0

x + (k + 4) y + (4k + 2) z = 0

x + 2(k + 1)y + (3k + 4) z = 0.

Solution : The given system of equations is AX = 0, where A =

2 3k 3k 41 k 4 4k 21 2k 2 3k 4

and X =

xyz

.

Now | A | = 0 k = 2

Hence if k 2, then | A | 0 the given system has only trivial solutioni.e., x = y = z = 0.

However if k = – 2 or 2, then | A | = 0 and we consider the cases separately


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Case I k = 2,

the given equations become

2x + 6y + 10z = 0 x + 3y + 5z = 0 . . . (i)

x + 6y + 10z = 0 . . . (ii)

x + 6y + 10z = 0 . . . (iii)

Hence independent equation are (i) and (ii). Solving we get,

x = 0, 3y + 5z = 0 x = 0, y = , z =35 , R .

Case II k = – 2The given equations becomes

2x – 6y – 2z = 0 x – 3y – z = 0 . . . (iv)

x + 2y – 6z = 0 . . . (v)

x – 2y – 2z = 0 . . . (vi)

We observe that45

(iv) +15

(v) = (vi). Thus equations are dependent. Consider independent

equations x – 3y – z = 0 and x – 2y – 2z = 0. Let z = R, then x – 3y = and x – 2y = 2

y = , x = 4 . Hence x = 4 , y = and z = , R.

DRILL EXERCISE - 8

1. If A =

111

312

111

, Find AA–1. Using A–1, solve the following system of linear equation.

x + 2y + z = 4 ; –x + y + z = 0 ; x – 3y + z = 2.

2. Show that the equations 3x + 4y + 5z = a ; 4x + 5y + 6z = b ; 5x + 6y + 7z = c, do not have a solutionunless a + c = 2b.

3. Solve the following system of homogeneous equations2x + 3y – z = 0 ; x – y – 2z = 0 ; 3x + y + 3z = 0

4. Find the product of the following matrices

A =

111

517

315

and B =

312

123

211

and use it to solve the following system of linear equations.

x + y + 2z = 1 ; 3x + 2y + z = 7; 2x + y + 3z = 2

5. Show that the following system of equations is inconsistent,


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3x – y + 2z = 3 ; 2x + y + 3z = 5 ; x – 2y – z = 1.

16. APPLICATION OF MATRICESMatrix instead of being a quantity is an operator to be used for transformation of vectors or variables.The addition and multiplication of matrices are compositions of operators instead of scalar quantities.

16.1 Geometrical Applications(i) Reflection : If the point (x, y) is reflected in y-axis the new coordinates are

xx , yy ...........(i)These equations can be put in the form

y0x1x y1x0y In terms of matrices, they can be written as

y

x

10

01

y

x

Hence the reflection in y-axis is obtained by pre-multiplying the column vector

y

x with

10

01. Similarly the reflection in x-axis is obtained by pre-multiplying with

10

01.

(ii) Reflection about the line y = x : If we reflect the point (x, y) in the line y = x, y ischanged into x and x into y, so that the new coordinates are :

y x , x y , ...........(i)

and sox 0 1 x

y 0 1 y

Similarly the reflection of (x, y) in the line x + y = 0 is obtained by

yx , xy

orx 0 1 x

y 1 0 y

(iii) Rotation about the origin : If the coordinates of a point P are (x, y), O being theorigin and the line OP is rotated about O as centre through an angle in the anti-clockwisedirection, the new coordinates of P are

sinsinrcoscosr)cos(rx

= sinycosx

and sincosrcossinr)sin(ry

= sinxcosy

so that

y

x

cossin

sincos

y

x

where ),r( are the polar coordinates of (x – y). If the rotation is in clockwise direction,we put for .

DRILL EXERCISE - 9

1. If (x1 – x

2)2 + (y

1 – y

2)2 =a2

(x2 – x

3)2 + (y

2 – y

3)2 = b2

(x1 – x

3)2 + (y

1 – y

3)2 = c2, then prove thatwww.StudySteps.in of 43

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answer key

Drill Exercise -1

1.

864

753

Drill Exercise - 2

2. x =2

9

k, k I , k n 9 , n I 4. 3 × 4, 3 × 3

Drill Exercise -3

3. (i) x y z LNM

OQP

1

2

1

6

1

3, ,

4.1 4 7 24 8 37 2 3 7

0 2 1 22 0 2

1 2 2 0

/

/&

/

/

LNMM

OQPP

LNMM

OQPP

5. 2

Drill Exercise -4

4. 325215 5. x = 1, y = 1, z = 1

Drill Exercise -5

1. 2 2. –2 3. 0

Drill Exercise -6

2.

30462337

155810694. =

2

1


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Drill Exercise -7

1. (i) x = k – 2, y = 8 – k, z = k, k R. (ii) x =1, y = 2, z = 3(iii) no solution

3. (i) x = 2, y = –3 (ii) x = 1, y =2

1, z = –

2

3

4. (i) x = 1, y = 2, z = –1 (ii) x = 2, y = 1, z = 3

5. (i) 3, 10 (ii) 3, R (iii) 3, 10

Drill Exercise -8

1. x =5

9, y =

5

2, z =

5

73. x = y = z = 0 4. x = 2, y = 1, z = –1

Drill Exercise -9

2. x = 3, y = 1, z = 2

SOLVED SUBJECTIVE EXAMPLES

Example 1 :

If A =0 tan /2

tan /2 0

and I is a 2 × 2 unit matrix, then prove that

cos sinI A (I A)

sin sin

Solution :

1 0I

0 1

and given0 tan / 2

Atan / 2 0

1 tan / 2

I Atan / 2 1

... (1)

R.H.S. = (I – A)cos sin

sin cos

1 tan / 2 cos sin

tan / 2 1 sin cos


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2

2 2

2

2 2

1 tan / 2 2 tan / 21 tan / 2 1 tan / 2 1 tan / 2

tan / 2 1 2 tan / 2 1 tan / 2

1 tan / 2 1 tan / 2

2 2 2

2 2 2 2

2 2 2

2 2 2 2

1 tan / 2 2 tan / 2 2 tan / 2 tan / 2(1 tan / 2)

1 tan / 2 1 tan / 2 1 tan / 2 1 tan / 2

tan / 2(1 tan / 2 2 tan / 2 2 tan / 2 1 tan / 2

1 tan / 2 1 tan / 2 1 tan / 2 1 tan / 2

2 2

2 2

2 2

2 2

(1 tan / 2) tan / 2(1 tan / 2)

(1 tan / 2 (1 tan / 2

tan / 2(1 tan / 2) (1 tan / 2)

(1 tan / 2) (1 tan / 2)

1 tan / 2

tan / 2 1

= I + A = L.H.S. {from (1)}

Example 2 :

If

0 1 0

A 0 0 1

p q r

, show that AA3 = pI + qA + rA2

Solution :We have A2 = AA

0 1 0 0 1 0

0 0 1 0 0 1

p q r p q r

2

0 0 1

p q r

pr p qr q r

3 2

2

0 0 1 0 1 0

A A .A p q r 0 0 1

pr p qr q r p q r

2

2 2 2 3

p q r

pr p qr q r

pq r p pr q qr p 2qr r


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and pI + qA + rA2 =2

1 0 0 0 1 0 0 0 1

p 0 1 0 q 0 0 1 r p q r

0 0 1 p q r pr p qr q r

2

2 2 2 3

p 0 0 0 q 0 0 0 r

0 p 0 0 0 q pr qr r

0 0 p pq q qr pr pr qr qr r

2

2 2 2 3

p 0 0 0 q 0 0 0 r

0 0 pr p 0 qr 0 q r

0 pq pr 0 q pr qr p qr qr r

2

2 2 2 3

p q r

pr p qr q r

pq pr q pr qr p 2qr r

Example 3 :

Verify that A =

1 2 21

2 1 23

2 2 1

is an orthogonal matrix.

Solution :

1 2 21

A 2 1 23

2 2 1

1 2 21

A 2 1 23

2 2 1

1 2 2 1 2 21 1

AA 2 1 2 2 1 23 3

2 2 1 2 2 1

1 4 4 2 2 4 2 4 21

2 2 4 4 1 4 4 2 29

2 4 2 4 2 2 4 4 1

9 0 01

0 9 09

0 0 9

1 0 0

0 1 0

0 0 1

= I

Hence A is an orthogonal matrix.

Example 4 :If A and B are two non-zero square matrices of the same order n such that AB = O, then


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| A | = | B | = 0. Further show that | C | = 0 | adj C| = 0, for any square matrix C.

Solution :If | A | 0, then AA–1 exists. Hence AB = 0 AA–1 (AB) = A–1O B = O, which is not the case.Hence | A | = 0. Similarly | B | = 0.If |C| = O, then C (adj C) = |C| I = 0 |adj C| = 0, as shown earlier conversely if |adj C| = 0 thenC(adj C) = |C| I |C|. |adj C| = |C|n |C| = 0 .

Example 5 :In the equations x = cy + bz, y = az + cx, z = bx + ay, where x, y, z are not all zero, prove that(i) a2 + b2 + c2 + 2abc = 1

(ii)2 2 2

2 2 2

x y z

1 a 1 b 1 c

.

Solution :(i) Since x, y and z are not all zero,

the given equationsx – cy – bz = 0 . . . (1)cx – y + az = 0 . . . (2)

and bx + ay – z = 0 . . . (3)have a nontrivial solution.

1 c b0 i.e. c 1 a

b a 1

= 0

or 1[1 – a2] + c[– c – ab] – b[ca + b] = 0 [expressing by R1]

or 1 – a2 – c2 – abc – abc – b2 = 0 . . . (4)or a2 + b2 + c2 + 2abc = 1(ii) By cross-multiplication (1) and (2), we have

2

x y zca b bc a 1 c

squaring

2 2 2

2 2 2 2 2 2 2 2 22

x y z

c a 1 a c b c 1 b c 1 c

[Using (4)]

or 2 2 2

22 2 2 2 2

x y z

1 a 1 c 1 b 1 c 1 c

Hence2 2 2

2 2 2

x y z

1 a 1 b 1 c

.

Example 6 :


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Leta b

A0 1

, where a 0. Show that for

nn

n

b(a 1)a

n N,A (a 1)

0 1

Solution :We have to show by mathematical induction

Step I : For n = 1,

1

(a 1)a b

(a 1)A

0 1

a b

0 1

Hence the result is true for n = 1

Step II: Assume the result to be true for some k 0.

kk

k

(a 1)a b

A (a 1)

0 1

Step III: For n = k + 1

Ak+1 = Ak. A

kk b(a 1)

a a b(a 1)

0 10 1

kk k b(a 1)

a .a 0 a .b .1(a 1)

0 0 0 1.1

k 1k 1 a 1

a b(a 1)

0 1

Hence the result is true for k + 1. Therefore by the principle of induction, the result is true for all

n N .

Example 7 :By the method of matrix inversion, solve the system.

1 1 1 x u 9 12

2 5 7 y v 52 15

2 1 1 z w 0 1

Solution :

We have

1 1 1 x y 9 2

2 5 7 y v 52 15

2 1 1 z w 0 1

or AX = B or X = A–1 B . . . (i)


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Where

1 1 1 x u 9 2

A 2 5 7 ,X y v and B 52 15

2 1 1 z w 0 9

|A| = 1 (–5 – 7) –1 (–2 – 14) + 1 (2 – 10) = –12 + 16 – 8 = 4 0 Let C be the matrix of cofactors of elements of |A|.

11 12 13

21 22 23

31 32 33

C C C

C C C C

C C C

5 7 2 7 2 5

1 1 2 1 2 1

1 1 1 1 1 1

1 1 2 1 2 1

1 1 1 1 1 1

5 7 2 7 2 5

12 16 8

2 3 1

2 5 3

Adj

12 2 2

A C 16 3 5

8 1 3

1

12 2 2Adj.A 1

A 16 3 5| A | 4

8 1 3

Now, 1

12 2 2 9 21

A B 16 3 5 52 154

8 1 3 0 1

4 41

12 84

20 4

1 1

3 2

5 1

from (1) X = A–1B

x u 1 1

y v 3 2

z w 5 1

On equating the corresponding elements, we havex = 1, u = –1y = 3, v = 2z = 5, w = 1

Example 8 :

If a, b and c are all different and if

2 3

2 3

2 3

a a 1 a

b b 1 b 0,

c c 1 c

prove that abc = –1.Solution :

2 3 2 2 3

2 3 2 2 3

2 3 2 2 3

a a 1 a a a 1 a a a

D b b 1 b b b 1 b b b

c c 1 c c c 1 c c c


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2 2

2 2

2 2

a a 1 1 a a

b b 1 abc 1 b b

c c 1 1 c c

2 2

1 2 21 3

2 2

1 a a 1 a a

( 1) 1 b b abc 1 b b [C C in 1st det .]

1 c c 1 c c

2 2

2 2 22 3

2 2

1 a a 1 a a

( 1) 1 b b abc 1 b b [C C in 1st det .]

1 c c 1 c c

Thus (abc + 1). = 0 abc = – 1, as

2

2

2

1 a a

1 b b

1 c c

0 (a, b and c are all different).

Example 9 :For what value of k the following system of equations :x + ky + 3z = 03x + ky - 2z = 02x + 3y – 4z = 0 possess a non-trivial solution. For that value of k, find all the solutions of the system.

Solution :For the nontrivial solution

1 k 3

3 k 2 0

2 3 4

k = 33/2

Putting the value of k in the given equations, the equations become

33x y 3z 0

2 ... (i)

333x y 2z 0

2 ... (ii)

2x + 3y - 4z = 0 ... (iii)Multiply (i) by 3 and subtract from (ii) to get

–33y – 11z = 0or z = –3y ... (iv)

Now let y = , z = –3from (iii), 2x + 3 + 12 = 0

15

x2

, R


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Example 10 :Let x

1 = 3y

1 + 2y

2– y

3 , y

1 = z

1 – z

2 + z

3

x2 = –y

1 + 4y

2 + 5y

3, y

2 = z

2 + 3z

3

x3 = y

1 – y

2 + 3y

3, y

3 = 2z

1 + z

2.

Express x1, x

2, x

3 in terms of z

1, z

2 , z

3.

Solution :

We have1 1 1

2 2 2

3 3 3

x y z3 2 1 3 2 1 1 1 1x 1 4 5 y 1 4 5 0 1 3 z

1 1 3 1 1 3 2 1 0x y z

1 1

2 2

3 3

x z1 2 9x 9 10 11 z

7 1 2x z

x1 = z

1 – 2z

2 + 9z

3, x

2 = 9z

1 + 10z

2 + 11z

3, x

3 = 7z

1 + z

2 – 2z

3

Example 11 :

If (x)

x x x

x x x

x x v x

, show that (x) 0 and (x) (0) Sx, where S denote the

sum of all the cofactors of all elements in (0) and dash denotes the derivative with respect to x.

Solution :

1 x x x 1 x x x 1

(x) 1 x x x 1 x x x 1

1 x v x x 1 v x v x x 1

Applying 2 2 1C C xC and 3 3 1 1 1 2 3 3 2C C xC in first and C C xC ,C C xC in sec-

ond and 1 1 3 2 2 3C C xC and C C xC in third to get

1 1 1

(x) 1 1 1

1 v 1 v 1

(x) 0 .

If S is the sum of all the cofactors of all elements in (0) , then it can be seen that (x) = S.

on integrating (x) Sx c

(0) 0 c

Hence (x) Sx (0).

Example 12 :If , are the roots of the equation ax 2 + bx + c = 0 and s

n = n n , evaluate


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1 2

1 2 3

2 3 4

3 1 s 1 s1 s 1 s 1 s1 s 1 s 1 s

in terms of a, b, and c only..

Solution :2 2

2 2 3 3

2 2 3 3 4 4

1 1 1 1 1

1 1 1

1 1 1

2

2 2 2

1 1 11 1 11 1

1 1

= 21 1

= 2 21

=

2 2

2

b c b 4c1

a a aa

= 2 2

4

a b c b 4ac

a

.

Example 13 :Prove that

2a a b a cb a 2b b cc a c b 2c

= 4(b + c)(c + a) (a + b)

Solution :

Let2a a b a c

b a 2b b cc a c b 2c

Putting a + b = 0

b = – a

then2a 0 a c0 2a c a

c a c a 2c

Expanding along R1

= – 2a{– 4ac – (c – a)2} – 0 + (a + c){0 – 2a (c + a)}= 2a (c + a)2 – 2a (c + a)2


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= 0

Hence a + b is a factor of Similarly b + c and c + a are the factors of .

On expansion of determinant we can see that each term of the determinant is a homogeneous expres-sion in a, b, c of degree 3 and also R.H.S is a homogeneous expression of degree 3.

Let = k(a + b) (b + c) (c + a)

or


= k (a + b) (b + c) (c + a)

If we choose a = 0, b = 1, c = 2, we get

0 – 1 (– 4 – 6) + 2( 3 + 4) = 6k k = 4

Hence


= 4(a + b) (b + c) (c + a)

Example 14 :If f(x) is a polynomial of degree < 3, prove that

2

2

2

1 a f (a) /(x a) 1 a af (x)

1 b f (b) /(x b) 1 b b(x a) (x b) (x c)

1 c f (c) /(x c) 1 c c

Solution :

f (x) A B C

(x a) (x b) (x c) (x a) (x b) (x c)

... (1) (say)

On comparing the various powers of x, we get

f (a)A

(a b) (c a)

f (b)B

(a b) (b c)

f (c)C

(b c) (c a)

Now from (1) we have


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f (a) f (b) f (c)(c b) (a c) (b a)

f (x) (x a) x b (x c)(x a) (x b) (x c) (a b) (b c) (c a)

2

2

2

1 a f (a) /(x a)

1 b f (b) /(x b)

1 c f (c) /(x c)

1 a a

1 b b

1 c c

Example 15 :If A, B and C are the angles of a triangle, show that

(i)sin 2A sin C sin B

sin C sin 2B sin A 0

sin B sin A sin 2C

(ii)

1 cos B cosC cos B cos B

cosC cos A 1 cos A cos A 0

1 cos B 1 cos A 1

Solution :

(i)

sin 2A sin C sin B

L.H.S sin C sin 2B sin A

sin B sin A sin 2C

2ka cos A kc kb

kb 2kbcos B ka

kb ka 2kccosC

(from sine rule)

3

2a cos A c b

k c 2bcos B a

b a 2ccosc

3

a cos A a cos A a cos B bcos A a cosC ccos A

k a cos B bcos A bcos B bcos B bcosC ccos B

a cosC ccos A bcosC ccos B ccosC ccosC

3

cos A a 0 a cos A 0

k cos B b 0 b cos B 0

cosC c 0 c cosC 0

= 0 × 0 = 0 = R.H.S.

(ii) L.H.S. =

1 cos B cosC cos B cos B

cosC cos A 1 cos A cos A

1 cos B 1 cos A 1

Applying 1 1 3 2 2 3C C C :C C C


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1 cosC cos B

cosC 1 cos A

cos B cos A 1

a cosC cos B1

a cosC 1 cos Aa

a cos B cos A 1

Applying 1 1 2 3C C bC cC

0 cosC cos B1

0 1 cos A 0 R.H.S.a

0 cos A 1


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SOLVED OBJECTIVE EXAMPLES

Example 1 :

If2 2

2 3

3 1x 4x x

x 2 1x x

, then x =

(a) 1 (b) – 1(c) – 2 (d) 3

Solution :x2 – 4x = –3 x2 – 4x + 3 = 0 x = 1, 3x2 = 1 x = 1x2 = –x + 2 x2 + x – 2 = 0 x = –2, 1x3 = 1 x = 1, , 2

Common value of x is 1.Hence (a) is correct.

Example 2 :

If the trace of the matrix A =2

2

x 1 0 2 5

3 x 2 4 1

1 2 x 3 1

2 0 4 x 6

is 0, then x is equal to

(a) –2, 3 (b) 2, –3(c) –3, 2 (d) 3, –2

Solution :

Trace of matrix is defined asn

2ii

i 1

a 2x 2x 12 0

x = –3, 2Hence (c) is correct

Example 3 :If A and B are square matrices of order 3, then(a) adj (AB) = –adjA + adj B (b) (A + B)–1 = A–1 + B–1

(c) AB = 0 |A| = 0 or |B| = 0 (d) AB = 0 |A| = 0 and |B| = 0Solution :

If AB = 0, then at least one of A and B is necessarily singular.Hence (c) is correct.

Example 4 :If A and B are any two square matrices of the same order, then

(a) (AB) A B (b) adj(AB) = adj(A) adj(B)

(c) (AB) B A (d) AB = 0 A = O or B = OSolution :

It is a known fact that AB B A .


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Hence (c) is the correct answer.Example 5 :

Let A be a square matrix of order n & k be a scalar. Then |kA| equals(a) k |A| (b) |k| |A|(c) kn |A| (d) none of these

Solution :KA is the matrix, in which all the entries of A are multiplied by K.Hence |KA| = Kn |A|, taking K common from all the columns.Hence (c) is correct.

Example 6 :

If A =1 tan x

,tan x 1

then the value of1A A is

(a) cos4x (b) sec2x(c) –cos4x (d) 1

Solution :1 tan x

Atan x 1

12

1 tan x1A

tan x 11 tan x

,1 cos 2x sin 2x

A Asin 2x cos 2x

| AA–1| = 1Hence (d) is correct.

Example 7 :The digits A, B and C are such that the three digit numbers A88, 6B8, 86C are divisible by 72 then the

determinant

A 6 88 B 68 8 C

is divisible by

(a) 72 (b) 144(c) 288 (d) 216

Solution :

R3 1 2 3

A 6 8 A 6 8100R 10R R 8 B 6 8 B 6

8 8 C A88 6BC 86C

which is divisible by 72. Hence (a) is correct.

Example 8 :

Maximum value of

2 2

2 2

2 2

1 sin x cos x 4sin 2x

sin x 1 cos x sin 2x

sin x cos x 1 4sin 2x

is

(a) 4 (b) 6


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(c) 2 (d) none of theseSolution :

Applying 1 1 2C C C we get

2

2

2

2 cos x 4sin 2x

2 1 cos x 4sin 2x

1 cos x 1 4sin 2x

Applying 2 2 1 3 3 1R R R and R R R we get ,

22 cos x 4sin 2x

0 1 0

1 1 1

2 4sin 2x 6

Hence maximum value is 6 and (b) is the correct answer.

Example 9 :

If

mr

2 mr

2 2 2 2

2r 1 C 1

m 1 2 m 1

sin (m ) sin (m) sin (m 1)

then

m

rr 0 is equal to

(a) m2 – 1 (b) 2m

(c) zero (d) none of theseSolution :

mr

2 mr

2 2 2 2

2r 1 C 1

m 1 2 m 1


m m mm

rr 0 r 0 r 0

m2 m

rr 0 2 2 2 2

(2r 1) C 1

m 1 2 m 1


2 m

2 m

2 2 2 2

m 1 2 m 1

m 1 2 m 1 0


Hence (c) is correct.

Example 10 :The system of linear equations x + y + z = 2, 2x + y – z = 3, 3x + 2y + kz = 4 has a unique solutionif(a) k 0 (b) –1 < k < 1(c) – 2 < k < 2 (d) k = 0

Solution :The system of equations has a unique solution if

1 1 12 1 1 0 k 03 2 k


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Hence (a) is the correct answer.Example 11 :

If A, B and C are the angles of a non-right angled triangle ABC, then the value of

tan A 1 1

1 tan B 1

1 1 tan C is equal to

(a) 1 (b) 2(c) –1 (d) –2

Solution :Given determinant is equal to; tanA (tanB. tanC – 1) – 1 (tanC – 1) + 1 (1 – tanB)= tanA. tanB. tanC – tanA – tanB – tanC + 2 = 2(as tanA = tan A)Hence (b) is correct.

Example 12 :

If x, y, z are non zero real numbers, then the values of

2

2

2

1x yz

x1

y zxy

1z xy

z

depends upon

(a) x only (b) y only(c) z only (d) none of these

Solution :Multiplication of R

1 by x, R

2 by y and R

3 by z, reduces the given determinant to,

3 3

3 3

3 3

1 x xyz 1 x 11

1 y xyz 1 y 1 0xyz

1 z xyz 1 z 1

Hence (d) is the correct answer.

Example 13 :

Let f(x) = 2

cos x x 1

2sin x x 2x

tan x x 1

. The value ofx 0

f (x)lim

x is equal to

(a) 1 (b) –1(c) 0 (d) none of these

Solution :

x 0

f (x)lim

x

x 0

f (x) f (0)lim f (0)

x 0


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2

sin x 1 0

f (x) 2sin x x 2x

tan x x 1

2

2

cos x x 1 cos x x 1

2cos x 2x 2 2sin x x 2x

tan x x 1 sin x 1 0

Thusx 0

f (x)lim

x = 2

2

0 1 0 cos x x 1

f (0) 0 0 0 2sin x x 2x

0 0 1 sin x 1 0

1 0 1

0 0 0

1 1 0

= 0

Hence (c) is correct.

Example 14 :

If

p b c

a p, b q, c r and a q c 0

a b r

then the value ofp q r

p a q b r c

is equal to

(a) –1 (b) 1

(c) –2 (d) 2

Solution :

1 1 2 2 3R R R , R R reduces the determinant to,

p a b q 0

0 q b c r 0

a b r

(p – a) (q – b) r + (r – c)b) + (q – b) ((r – c)a) = 0 (p – a)((q – b)(r - c) – (r–c)a) = 0 (p – a)((q – b)(r–c + c) – – c)(q – b– q)) – (q – b) (r – c) (p – a – p) = 0 ( p – a) (q – b) (r – c) + c(p – a) (q – b) – (p – a) q – b) (r – c) + q(p – a) (r – c)

–(p – a) (q – b) (r – c) + p(q – b) (r – c) = 0Dividing through out by (p – a) (q – b) (r – c) we get,

c q p1

r c q b p a

c q p

1 2r c q b p a

r q p2

r c q b p a

Hence (d) is correct.

Example 15 :Let a, b and c be positive real numbers. The following system of equations in x, y and z

2 2 2 2 2 2 2 2 2

2 2 2 2 2 2 2 2 2

x y z x y z x y z1, 1, 1

a b c a b c a b c has


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(a) no solution (b) unique solution

(c) infinitely many solutions (d) finitely many solutions

Solution :

Let2 2

2 2

x yX, Y

a b and

2

2

zZ

c , then the given system of equations is

X + Y – Z = 1, X – Y + Z = 1, – X + Y + Z = 1

The coefficient of matrix A =

1 1 11 1 1 | A | 01 1 1

On solving, we get X = Y = Z = 1Hence x = a, y = b, z = cThus (d) is the correct answer.


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