MATRICES - · PDF filecrackiitjee.in MATRICES
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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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iitjee
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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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iitjee
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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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iitjee
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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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iitjee
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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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iitjee
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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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iitjee
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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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iitjee
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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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iitjee
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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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iitjee
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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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iitjee
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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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iitjee
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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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iitjee
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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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iitjee
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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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iitjee
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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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iitjee
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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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iitjee
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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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iitjee
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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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iitjee
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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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iitjee
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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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