狠狠撸

Addition of Matrices
Individual elements are added
Both matrices must be of the same order
If A =
?11 ?12 ?13
?21 ?22 ?23
is a 2 × 3 matrix
and B =
?11 ?12 ?13
?21 ?22 ?23
is another 2 × 3 matrix.
Then, A + B =
?11 + ?11 ?12 + ?12 ?13 + ?13
?21 + ?21 ?22 + ?22 ?23 + ?23
i.g. : If A =
1 2 ?4
2 3 7
and B =
4 ?1 1
3 1 0
Then, A + B =
1 + 4 ?1 + 2 ?4 + 1
2 + 3 3 + 1 7 + 0
A + B =
5 1 ?3
5 4 7

Example 6
Given A = 3 1 ?1
2 3 0
and B =
2 5 1
?2 3
1
2
, find A + B
A + B = 3 1 ?1
2 3 0
+
2 5 1
?2 3
1
2
=
3 + 2 1 + 5 ?1 + 1
2 ? 2 3 + 3 0 +
1
2
=
2 + 3 1 + 5 0
0 6
1
2

Multiplication of Matrice by a scalar
If A = [aij] m × n is a matrix and k is a scalar,
then kA = k [aij]m × n = [k (aij)] m × n, that is, (i, j)th element of
kA
is kaij for all possible values of i and j.
Every element is multiplied by the scalar.
Eg: If A =
3 1 1.5
5 7 ?3
2 0 5
, then
3A = 3
3 1 1.5
5 7 ?3
2 0 5
3A =
3 × 3 3 × 1 3 × 1.5
3 × 5 3 × 7 3 × ?3
3 × 2 3 × 0 3 × 5
=
9 3 4.5
3 5 21 ?9
6 0 15

Multiplication of Matrice by a scalar
Negative of a matrix
Denoted by –A.
We define –A = (-1) A
Eg: A =
3 1
?5 ?
,
then -A = (-1) A = (-1)
3 1
?5 ?
=
?3 ?1
5 ??

Example 7
If A =
1 2 3
2 3 1
and B =
3 ?1 3
?1 0 2
then find 2A – B.
Finding 2A
2A = 2
1 2 3
2 3 1
=
1 × 2 2 × 2 2 × 3
2 × 2 2 × 3 1 × 2
=
2 4 6
4 6 2
Now, finding 2A – B
2A – B =
2 4 6
4 6 2
?
3 ?1 3
?1 0 2
=
2 ? 3 4 ? (?1) 6 ? 3
4 ? (?1) 6 ? 0 2 ? 2
=
?1 4 + 1 3
4 + 1 6 0
=
?1 5 3
5 6 0

Now, finding 2A – B
2A – B =
2 4 6
4 6 2
?
3 ?1 3
?1 0 2
=
2 ? 3 4 ? (?1) 6 ? 3
4 ? (?1) 6 ? 0 2 ? 2
=
?1 4 + 1 3
4 + 1 6 0
=
?1 5 3
5 6 0

Ex 3.2,1
Let A =
2 4
3 2
, B =
1 3
?2 5
, C =
?2 5
3 4
Find each of the following
(i) A + B
A + B =
2 4
3 2
+
1 3
?2 5
=
2 + 1 4 + 3
3 ? 2 2 + 5
=
3 7
1 7

Ex 3.2,1
Let A =
2 4
3 2
, B =
1 3
?2 5
, C =
?2 5
3 4
(ii) A – B
A – B =
2 4
3 2
?
1 3
?2 5
=
2 ? 1 4 ? 3
3 ? (?2) 2 ? 5
=
1 1
3 + 2 ?3
=
1 1
5 ?3

Ex3.2,1
Let A =
2 4
3 2
, B =
1 3
?2 5
, C =
?2 5
3 4
(iii)3A – C
Finding 3A
3A = 3
2 4
3 2
=
3 × 2 3 × 4
3 × 3 3 × 2
=
6 12
9 6
Hence
3A – C =
6 12
9 6
?
?2 5
3 4
=
6 ? (?2) 12 ? 5
9 ? 3 6 ? 4

Hence
3A – C =
6 ? (?2) 12 ? 5
9 ? 3 6 ? 4
=
6 + 2 7
6 2
=
8 7
6 2

Hence
3A – C =
6 12
9 6
-
?2 5
3 4
=
6 ? (?2) 12 ? 5
9 ? 3 6 ? 4
=
6 + 2 7
6 2
=
8 7
6 2

Ex3.2, 2
Compute the following:
(iii)
?1 4 ?6
8 5 16
2 8 5
+
12 7 ?6
8 0 5
3 2 4
?1 4 ?6
8 5 16
2 8 5
+
12 7 ?6
8 0 5
3 2 4
=
?1 + 12 4 + 7 ?6 + 6
8 + 8 5 + 0 16 + 5
2 + 3 8 + 2 5 + 4
=
11 11 0
16 5 21
5 10 9

Ex3.2, 2
(i)
a b
?b a
+
a b
b a
a b
?b a
+
a b
b a
=
a + a b + b
b + b a + a
=
2a 2b
0 2a

Ex3.2,2
(ii)
a2 + b2 b2 + c2
a2 + c2 a2 + b2
+
2ab 2bc
?2ac ?2ab
a2 + b2 b2 + c2
a2 + c2 a2 + b2
+
2ab 2bc
?2ac ?2ab
=
a2 + b2 + 2ab b2 + c2 + 2bc
a2 + c2 ? 2ac a2 + b2 ? 2ab
=
a + b 2 b + c 2
a ? c 2 a ? b 2
Using ( a + b)2 = a2 + b2 + 2ab)
& (a – b)2 = a2 + b2 – 2ab

Ex3.2, 2
(iv)
cos2 x sin2 x
sin2 x cos2 x
+
sin2 x cos2 x
cos2 x sine2 x
cos2 x sin2 x
sin2 x cos2 x
+
sin2 x cos2 x
cos2 x sin2 x
=
cos2 x + sin2 x sin2 x + cos2 x
sin2 x + cos2 x cos2 x + sin2 x
=
1 1
1 1
( sin2 x + cos2 x = 1)

Ex3.2, 5
If A =
2
3
1
5
3
1
3
2
3
4
3
7
3
2
2
3
and B =
2
5
3
5
1
1
5
2
5
4
5
7
5
6
5
2
5
,then compute 3A – 5B
Given A =
2
3
1
5
3
1
3
2
3
4
3
7
3
2
2
3
, B =
2
5
3
5
1
1
5
2
5
4
5
7
5
6
5
2
5
3A – 5B
= 3
2
3
1
5
3
1
3
2
3
4
3
7
3
2
2
3
- 5
2
5
3
5
1
1
5
2
5
4
5
7
5
6
5
2
5

3A – 5B = 3
2
3
1
5
3
1
3
2
3
4
3
7
3
2
2
3
- 5
2
5
3
5
1
1
5
2
5
4
5
7
5
6
5
2
5
=
2
3
× 3 1 × 3
5
3
× 3
1
3
× 3
2
3
× 3
4
3
× 3
7
3
× 3 2 × 3
2
3
× 3
–
2
5
× 5
3
5
× 5 1 × 5
1
5
× 5
2
5
× 5
4
5
× 5
7
5
× 5
6
5
× 5
2
5
× 5
=
2 3 5
1 2 4
7 6 2
–
2 3 5
1 2 4
7 6 2
=
2 ? 2 3 ? 3 5 ? 5
1 ? 1 2 ? 2 4 ? 4
7 ? 7 6 ? 6 2 ? 2
=
0 0 0
0 0 0
0 0 0

Hence, 3A – 5B =
0 0 0
0 0 0
0 0 0

Ex3.2, 11
If x
2
3
+ y
?1
1
=
10
5
, find values of x and y.
x
2
3
+ y
?1
1
=
10
5
2?
3?
+
??
? =
10
5
2? ? ?
3? ? ?
=
10
5
Since the matrices are equal.
corresponding elements are equal
2x ? y = 10
3x + y = 5
…(1)
…(2)

Adding (1) & (2)
(2x – y) + (3x + y) = 10 + 5
2x – y + 3x + y = 15
2x + 3y – y + y = 15
5x + 0 = 15
x =
15
5
x = 3
Putting value of x in (1)
2x – y = 10
2(3) – y = 10
6 – y = 10
– y = 10 – 6
– y = 4
y = – 4
Hence, x = 3 & y = – 3

Ex3.2, 4
If A =
1 2 ?3
5 0 2
1 ?1 1
, B =
3 ?1 2
4 0 5
2 0 3
and , C =
4 1 2
0 3 2
1 ?2 3
then compute (A+B) and (B – C) . Also, verify that
A+(B - C) = (A+B) – C
Calculating A + B
A + B =
1 2 ?3
5 0 2
1 ?1 1
+
3 ?1 2
4 0 5
2 0 3
=
1 + 3 2 ? 1 ?3 + 2
5 + 4 0 + 2 2 + 5
1 + 2 1 + 0 1 + 3
=
4 1 ?1
9 2 7
3 ?1 4

Calculating B – C
B – C =
3 ?1 2
4 2 5
2 0 3
–
4 1 2
0 3 2
1 ?2 3
=
3 ? 4 2 ? 1 ?3 + 2
4 ? 0 0 + 2 2 + 5
2 ? 1 0 ? (?2) 3 ? 3
=
?1 ?2 0
4 ?1 3
1 2 0
We need to verify
A + (B – C) = (A + B) – C
Taking L.H.S
A + (B – C) =
1 2 ?3
5 0 2
1 ?1 1
+
?1 ?2 0
4 ?1 3
1 2 0

A + (B – C ) =
1 2 ?3
5 0 2
1 ?1 1
+
?1 ?2 0
4 ?1 3
1 2 0
=
1 ? 1 2 ? 2 ?3 + 0
5 + 4 0 ? 1 2 + 3
1 + 1 ?1 + 2 1 + 0
=
0 0 ?3
9 ?1 5
2 1 1
Taking R.H.S
(A + B) – C =
4 1 ?1
9 2 7
3 ?1 4
?
4 1 2
0 3 2
1 ?2 3
=
4 ? 4 1 ? 1 ?1 ? 2
9 ? 0 2 ? 3 7 ? 2
3 ? 1 ?1 + 2 4 ? 3

(A + B) – C =
4 ? 4 1 ? 1 ?1 ? 2
9 ? 0 2 ? 3 7 ? 2
3 ? 1 ?1 + 2 4 ? 3
=
0 0 ?3
9 ?1 5
2 1 1
= L.H.S
Hence L.H.S = R.H.S
Hence proved

Ex3.2, 6
Simplify cos θ
cos θ sin θ
?sin θ cos θ
+ sin θ
sin θ ?cos θ
cos θ sin θ
cos θ
cos θ sin θ
?sin θ cos θ
+ sin θ
sin θ ?cos θ
cos θ sin θ
=
cos θ(cos θ) cos θ(sin θ)
cos θ (????θ) cos θ(cos θ)
+
sin θ(sin θ) sin θ(?cos θ)
sin θ(cos θ) sin θ(sin θ)
=
cos2 θ cos θ sin θ
?cos θ sin θ cos2 θ
+
sin2 θ ?cos θ sin θ
sin θ cos θ sin2 θ
=
cos2 θ + sin2 θ sin θ cos θ ? cos θ sin θ
?cos θ sin θ + sin θ cos θ cos2 θ sin2 θ
=
1 0
0 1
(∵ cos2θ + sin2θ = 1)

Ex3.2, 22
Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 ×
p, n × 3 , and p × k respectively.
If n = p, then the order of the matrix 7X – 5Z is
(A)p × 2 (B) 2 × n (C) n × 3 (D) p × n
7X – 5Z
= 7 X 2 × ? - 5 Z 2 ×?
This is possible only when
Order of X = Order of Z
2 × n = 2 × p
Therefore, n = p
So, the matrix 7X – 5Z = 7 X 2 × ? - 5 Z 2 × ?
Hence, the order of matrix 7X ? 5Z is 2 × n.
Hence, correct answer is B
Given order of X is 2 × n
and order of Z is 2 × p

Example 9,
Find X and Y , if X + Y =
5 2
0 9
and X – Y =
3 6
0 ?1
It is given that
X + Y =
5 2
0 9
X – Y =
3 6
0 ?1
`Adding (1) and (2)
(X + Y) + (X – Y) =
5 2
0 9
+
3 6
0 ?1
X + X + Y – Y =
5 + 3 2 + 6
0 + 0 9 ? ( ?1)
… (1)
… (2)

X + X + Y – Y =
5 + 3 2 + 6
0 + 0 9 ? ( ?1)
2X =
8 8
0 8
X =
1
2
8 8
0 8
X =
8
2
8
2
0
8
2
X =
4 4
0 4

Putting X =
4 4
0 4
in (1)
X + Y =
5 2
0 9
Y =
5 2
0 9
– X
Y =
5 2
0 9
-
4 4
0 4
Y =
5 ? 4 2 ? 4
0 ? 0 9 ? 4
Y =
1 ?2
0 5
Hence, X =
4 4
0 4
, Y =
1 ?2
0 5

Ex3.2, 7
Find X and Y, if
(i) X + Y =
7 0
2 5
and X – Y =
3 0
0 3
Let X + Y =
7 0
2 5
X – Y =
3 0
0 3
Adding (1) and (2)
X + Y + X – Y =
7 0
2 5
+
3 0
0 3
X + Y + X – Y =
7 + 3 0 + 0
2 + 0 5 + 3
2x + 0 =
10 0
2 8
…(1)
…(2)

2x + 0 =
10 0
2 8
2X =
10 0
2 8
X =
1
2
10 0
2 8
=
10
2
0
2
2
2
8
2
=
5 0
1 4

Putting value of X in (1)
X + Y =
7 0
2 5
Y =
7 0
2 5
– X
Y =
7 0
2 5
–
5 0
1 4
Y =
7 ? 5 0 ? 0
2 ? 1 5 ? 4
Y =
2 0
1 1
Hence X =
5 0
1 4
& Y =
2 0
1 1

Ex3.2, 7
Find X and Y, if
(ii) 2X + 3Y =
2 3
4 0
and 3X + 2Y =
2 ?2
?1 5
Let 2X + 3Y =
2 3
4 0
3X + 2Y =
2 ?2
?1 5
Multiplying (1) by 3
3 × (2X+ 3Y) = 3
2 3
4 0
6X + 9Y =
2 × 3 3 × 3
4 × 3 0 × 3
6X + 9Y =
6 9
12 0
…(1)
…(2)
…(3)

Multiplying (2) by 2
2 × (3X + 2Y) = 2 ×
2 ?2
?1 5
6X + 4Y =
2 × 2 ?2 × 2
?1 × 2 5 × 2
6X + 4Y =
4 ?4
?2 10
Subtracting (3) from (4),
(6X + 9Y ) – (6X + 4Y) =
6 9
12 0
-
4 ?4
?2 10
6X + 9Y – 6X – 4Y =
4 ? 6 9 ? (?4)
12 ? (?2) 0 ? 10
9Y – 4Y + 6X – 6X =
2 9 + 4
12 + 2 ?10
5Y + 0 =
2 13
14 ?10
… (4)

5Y + 0 =
2 13
14 ?10
Y =
1
5
2 13
14 ?10
Y =
2
5
13
5
14
5
?
10
5
Putting value of Y in (1)
2X + 3
2
5
13
5
14
5
?2
=
2 3
4 0
2X +
2
5
× 3
13
5
× 3
14
5
× 3 ?2 × 3
=
2 3
4 0

2X +
2
5
× 3
13
5
× 3
14
5
× 3 ?2 × 3
=
2 3
4 0
2X +
6
5
39
5
42
5
?6
=
2 3
4 0
2X =
2 3
4 0
?
6
5
?
39
5
42
5
?6
2X =
2 ?
6
5
3 ?
39
5
4 ?
42
5
0 ? (?6)
2X =
2 ×5?6
5
3 ×5?39
5
4 ×5?42
5
6

2X =
2 ×5?6
5
3 ×5?39
5
4 ×5?42
5
6
2X =
10?6
5
15?39
5
20?42
5
6
X =
1
2
4
5
?
?24
5
?22
5
6
X =
4
5
×
1
2
?24
5
×
1
2
?22
5
×
1
2
6 ×
1
2
X =
2
5
?12
5
?11
5
3

Hence, X =
2
5
?12
5
?11
5
3
, Y =
2
5
13
5
14
5
?
10
5

Ex3.2, 8
Find X, if Y =
3 2
1 4
and 2X + Y =
1 0
?3 2
Given 2X + Y =
1 0
?3 2
2X =
1 0
?3 2
– Y
Putting value of Y
2X =
1 0
?3 2
–
3 2
1 4
2X =
1 ? 3 0 ? 2
?3 ? 1 2 ? 4
2X =
?2 ?2
?4 ?2
X =
1
2
?2 ?2
?4 ?2

X =
1
2
?2 ?2
?4 ?2
X =
?2
2
?2
2
?4
2
?2
2
X =
?1 ?1
?2 ?1
Hence, X =
?1 ?1
?2 ?1

Example 8
If A =
8 0
4 ?2
3 6
and B =
2 ?2
4 2
?5 1
then find the matrix X,
such that 2A + 3X = 5B.
Given that 2A + 3X = 5B
Putting values
2
8 0
4 ?2
3 6
+ 3X = 5
2 ?2
4 2
?5 1
8 × 2 0 × 2
4 × 2 ?2 × 2
3 × 2 6 × 2
+ 3X =
2 × 5 ?2 × 5
4 × 5 2 × 5
?5 × 5 1 × 5
16 0
8 ?2
6 12
+ 3X =
10 ?10
20 10
?25 5

16 0
8 ?2
6 12
+ 3X =
10 ?10
20 10
?25 5
3X =
10 ?10
20 10
?25 5
–
16 0
8 ?4
6 12
3X =
10 ? 16 ?10 ? 0
20 ? 8 10 ? (?4)
?25 ? 6 5 ? 12
3X =
?6 ?10
12 14
?31 ?7
X =
1
3
?6 ?10
12 14
?31 ?7
X =
?6
3
?10
3
?12
3
14
3
?31
3
?7
3

X =
?6
3
?10
3
?12
3
14
3
?31
3
?7
3
X =
?2
?10
3
4
14
3
?31
3
?7
3
Hence, X =
?2
?10
3
4
14
3
?31
3
?7
3

Ex3.2,9
Find x and y, if 2
1 3
0 ?
+
? 0
1 2
=
5 6
1 8
Given that
2
1 3
0 ?
+
? 0
1 2
=
5 6
1 8
1 × 2 3 × 2
0 × 2 ? × 2
+
? 0
1 2
=
5 6
1 8
2 6
0 2?
+
? 0
1 2
=
5 6
1 8
2 + ? 6 + 0
0 + 1 2? + 2
=
5 6
1 8
Since matrices are equal.
Corresponding elements are equal

Since matrices are equal, corresponding elements are equal
Therefore,
2 + y = 5
2x + 2 = 8
Solving (1)
2 + y = 5
y = 5 – 2
y = 3
Solving (2)
2x + 2 = 8
2x = 8 – 2
2x = 6
x =
6
2
x = 3
…(1)
…(2)

Ex 3.2,10
Solve the equation for x, y, z and t, if
2
? ?
? ? + 3
? 0
1 2
= 3
3 5
4 6
2
? ?
? ? + 3
1 ?1
0 2
= 3
3 5
4 6
? × 2 ? × 2
? × 2 ? × 2
+
1 × 3 ?1 × 3
0 × 3 2 × 3
=
3 × 3 5 × 3
4 × 3 6 × 3
2? 2?
2? 2?
+
3 ?3
0 6
=
9 15
12 18
2? + 3 2? ? 3
2? + 0 2? + 6
=
9 15
12 18

2x + 3 = 9
2y = 12
2z – 3 = 15
2t + 6 = 18
Solving equation (1)
2x + 3 = 9
2x = 9 – 3
x =
6
2
x = 3
…(1)
…(2)
…(3)
…(4)

2y = 12
y =
12
2
y = 6
2z – 3 = 15
2z = 15 + 3
2z = 18
z =
18
2
x = 9

2t + 6 = 18
2t = 18 – 6
2t = 12
t =
12
2
t = 6
Hence x = 3 , y = 6 , z = 9 & t = 6

Ex3.2,12
Given 3
x z
z w
=
x 6
?1 2w
+
4 x + y
z + w 3
find the values
of x, y, z and w.
3
x z
z w
=
x 6
?1 2w
+
4 x + y
z + w 3
x × 3 z × 3
z × 3 w × 3
=
x + 4 6 + x + y
?1 + z + w 2w + 3
3x 3z
3z 3w
=
x + 4 6 + x + y
1 ? z + w 2w + 3

3x = x + 4
3y = 6 + x + y
3z = 1 – z + w
3w = 2w + 3
3x = x + 4
3x – x = 4
2x = 4
x =
4
2
x = 2
…(1)
…(2)
…(3)
…(4)

3y = 6 + x + y
3y – y = 6 + x
2y = 6 + x
Putting x = 2
2y = 6 + 2
2y = 8
2y =
8
2
y = 4

3w = 2w + 3
3w – 2w = 3
w = 3
3z = – 1 + z + w
3z – z = – 1 + w
2z = – 1 + w
Putting w = 3
2z = – 1 + 3
2z = 2
z =
2
2
z = 1
Hence, x = 2, y = 4 , w = 3 & z = 1

Ex3.2,13
If F (x) =
cos ? ?sin ? 0
sin ? cos ? 0
0 0 1
, Show that F(x) F(y) = F(x + y)
F (x) =
cos ? ?sin ? 0
sin ? cos ? 0
0 0 1
We need to show
F(x) F(y) = F(x + y)
Taking L.H.S.
Finding F(x)
F (x) =
cos ? ?sin ? 0
sin ? cos ? 0
0 0 1

Finding F(y)
Replacing x by y in F(x)
F (y) =
cos ? ?sin ? 0
sin ? cos ? 0
0 0 1
Now,
F(x) F(y)
=
cos ? ?sin ? 0
sin ? cos ? 0
0 0 1
cos ? ?sin ? 0
sin ? cos ? 0
0 0 1
=
cos ? cos ? + ?sin ? sin ? + 0 cos ?(? sin ?) + (? sin ?) cos + 0 0 + 0 + 0 × 1
?sin ? cos ? + cos ? sin ? + 0 sin ? (? sin ?) + cos ? cos ? + 0 0 + 0 + 0 × 1
0 × cos ? + 0 + sin ? + 0 × 1 0 × (? sin ?) + 0 × cos ? + 0 0 + 0 + 1 × 1
=
cos ? cos ? ?sin ? . sin ? ?cos ? ? sin ? ? sin ? cos ? 0
sin ? cos ? + cos ? sin ? ? sin ? ? sin ? + cos ? cos ? 0
0 0 1

=
cos ? cos ? ?sin ? . sin ? ?cos ? ? sin ? ? sin ? cos ? 0
sin ? cos ? + cos ? sin ? ? sin ? ? sin ? + cos ? cos ? 0
0 0 1
We know that cos x cos y – sin x sin y = cos (x + y)
& sin x cos y + cos x sin y = sin (x + y)
=
cos(? + ?) ?[cos ? sin ? + sin ? cos ?] 0
sin(? + ?) cos ? cos ? ? sin ? sin ? 0
0 0 1
=
cos(? + ?) ? sin(? + ?) 0
sin(? + ?) cos(? + ?) 0
0 0 1

Taking R.H.S
F(x + y)
Replacing x by (x + y) in F(x)
=
cos(? + ?) ? sin(? + ?) 0
sin(? + ?) cos(? + ?) 0
0 0 1
= L.H.S.
Hence proved

Example 10
Find the values of x and y from the following equation:
2
x 5
7 y ? 3
+
3 ?4
1 2
=
7 6
15 14
2
x 5
7 y ? 3
+
3 ?4
1 2
=
7 6
15 14
x × 2 5 × 2
7 × 2 (y ? 3) × 2
+
3 ?4
1 2
=
7 6
15 14
2x 10
14 2? ? 6
+
3 ?4
1 2
=
7 6
15 14
2x + 3 10 ? 4
14 + 1 2? ? 6 + 2
=
7 6
15 14
2x + 3 6
15 2? ? 4
=
7 6
15 14

2x + 3 6
15 2? ? 4
=
7 6
15 14
2x + 3 = 7
& 2y – 4 = 14
Solving (1)
2x = 7 – 3
2x = 4
x =
4
2
x = 2
…(1)
…(2)

Solving (2)
2y – 4 = 14
2y = 14 + 4
2y = 418
y =
18
2
y = 9
Hence x = 2 & y = 9

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