class12 ch

1. Addition of Matrices
Individual elements are added
Both matrices must be of the same order
If A =
𝑎11 𝑎12 𝑎13
𝑎21 𝑎22 𝑎23
is a 2 × 3 matrix
and B =
𝑏11 𝑏12 𝑏13
𝑏21 𝑏22 𝑏23
is another 2 × 3 matrix.
Then, A + B =
𝑎11 + 𝑏11 𝑎12 + 𝑏12 𝑎13 + 𝑏13
𝑎21 + 𝑏21 𝑎22 + 𝑏22 𝑎23 + 𝑏23
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

2. 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

3. 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

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

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

6. 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

7. 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

8. 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
(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

9. Ex3.2,1
Let A =
2 4
3 2
, B =
1 3
−2 5
, C =
−2 5
3 4
Find each of the following
(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

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

11. 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

12. 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

13. Ex3.2, 2
Compute the following:
(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

14. Ex3.2,2
Compute the following:
(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

15. Ex3.2, 2
Compute the following:
(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)

16. 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

17. 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

18. Hence, 3A – 5B =
0 0 0
0 0 0
0 0 0

19. 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
2𝑥
3𝑥
+
−𝑦
𝑦 =
10
5
2𝑥 − 𝑦
3𝑥 − 𝑦
=
10
5
Since the matrices are equal.
corresponding elements are equal
2x − y = 10
3x + y = 5
…(1)
…(2)

20. 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

21. 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

22. 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

23. 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

24. (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

25. 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 (−𝑠𝑖𝑛θ) 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)

26. 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 × 𝑛 - 5 Z 2 ×𝑛
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 × 𝑛 - 5 Z 2 × 𝑛
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

27. 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)

28. 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

29. 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

30. 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)

31. 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

32. 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

33. 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)

34. 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)

35. 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

36. 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

37. 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

38. Hence, X =
2
5
−12
5
−11
5
3
, Y =
2
5
13
5
14
5
−
10
5

39. 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

40. 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

41. 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

42. 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

43. 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

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

45. 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)

46. Hence x = 3
& y = 3

47. Ex 3.2,10
Solve the equation for x, y, z and t, if
2
𝑥 𝑧
𝑦 𝑡 + 3
𝑦 0
1 2
= 3
3 5
4 6
2
𝑥 𝑧
𝑦 𝑡 + 3
1 −1
0 2
= 3
3 5
4 6
𝑥 × 2 𝑧 × 2
𝑦 × 2 𝑡 × 2
+
1 × 3 −1 × 3
0 × 3 2 × 3
=
3 × 3 5 × 3
4 × 3 6 × 3
2𝑥 2𝑧
2𝑦 2𝑡
+
3 −3
0 6
=
9 15
12 18
2𝑥 + 3 2𝑧 − 3
2𝑦 + 0 2𝑡 + 6
=
9 15
12 18
Since matrices are equal.
Corresponding elements are equal

48. Since matrices are equal, corresponding elements are equal
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)

49. Solving equation (2)
2y = 12
y =
12
2
y = 6
Solving equation (3)
2z – 3 = 15
2z = 15 + 3
2z = 18
z =
18
2
x = 9

50. Solving equation (4)
2t + 6 = 18
2t = 18 – 6
2t = 12
t =
12
2
t = 6
Hence x = 3 , y = 6 , z = 9 & t = 6

51. 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
Since matrices are equal.
Corresponding elements are equal

52. Since matrices are equal, corresponding elements are equal
3x = x + 4
3y = 6 + x + y
3z = 1 – z + w
3w = 2w + 3
Solving equation (1)
3x = x + 4
3x – x = 4
2x = 4
x =
4
2
x = 2
…(1)
…(2)
…(3)
…(4)

53. Solving equation (2)
3y = 6 + x + y
3y – y = 6 + x
2y = 6 + x
Putting x = 2
2y = 6 + 2
2y = 8
2y =
8
2
y = 4

54. Solving equation (4)
3w = 2w + 3
3w – 2w = 3
w = 3
Solving equation (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

55. Ex3.2,13
If F (x) =
cos 𝑥 −sin 𝑥 0
sin 𝑥 cos 𝑥 0
0 0 1
, Show that F(x) F(y) = F(x + y)
F (x) =
cos 𝑥 −sin 𝑥 0
sin 𝑥 cos 𝑥 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 𝑥 −sin 𝑥 0
sin 𝑥 cos 𝑥 0
0 0 1

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

57. =
cos 𝑥 cos 𝑦 −sin 𝑥 . sin 𝑦 −cos 𝑥 − sin 𝑦 − sin 𝑥 cos 𝑦 0
sin 𝑥 cos 𝑦 + cos 𝑥 sin 𝑦 − sin 𝑥 − sin 𝑦 + cos 𝑥 cos 𝑦 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(𝑥 + 𝑦) −[cos 𝑥 sin 𝑦 + sin 𝑥 cos 𝑦] 0
sin(𝑥 + 𝑦) cos 𝑥 cos 𝑦 − sin 𝑥 sin 𝑦 0
0 0 1
=
cos(𝑥 + 𝑦) − sin(𝑥 + 𝑦) 0
sin(𝑥 + 𝑦) cos(𝑥 + 𝑦) 0
0 0 1

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

59. 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 2𝑦 − 6
+
3 −4
1 2
=
7 6
15 14
2x + 3 10 − 4
14 + 1 2𝑦 − 6 + 2
=
7 6
15 14
2x + 3 6
15 2𝑦 − 4
=
7 6
15 14

60. 2x + 3 6
15 2𝑦 − 4
=
7 6
15 14
Since matrices are equal.
Corresponding elements are equal
2x + 3 = 7
& 2y – 4 = 14
Solving (1)
2x = 7 – 3
2x = 4
x =
4
2
x = 2
…(1)
…(2)

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