Post on 05-Jul-2020
Class-XII-Maths Matrices
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EXERCISE 3.1
1. In the matrix π΄ = [
2 5 19 β7
35 β25
212
β3 1 β5 17
], write:
(i) The order of the matrix,
(ii) The number of elements,
(iii) Write the elements π13, π21, π33, π24, π23.
Solution:
Given:
π΄ = [
2 5 19 β7
35 β25
212
β3 1 β5 17
]
Number of rows = 3
Number of columns = 4
Step 1:
(i) The order of the matrix=number of rows Γ number of columns
Hence the order of the given matrix = 3 Γ 4
Step 2:
(ππ) Since, the order of matrix is 3 Γ 4
So, the number of elements = 3 Γ 4 = 12
(πππ) From the given matrix, we can observe the elements:
π13 = 19
π21 = 35
π33 = β5
π24 = 12
π23 =5
2
Back of Chapter Questions
CBSE NCERT Solutions for Class 12 Maths Chapter 03
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2. If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?
Solution:
Given:
The number of elements = 24
Step 1:
Therefore, the possible orders are as follows:
1 Γ 24, 2 Γ 12, 3 Γ 8, 4 Γ 6, 6 Γ 4,8 Γ 3, 12 Γ 2 and 24 Γ 1
Step 2:
Now, if it has 13 elements,
then the possible orders are 13 Γ 1 and 1 Γ 13
3. If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?
Solution:
Given:
Total number of elements in matrix = 18
Step 1:
Therefore, the possible orders are as follows:
1 Γ 18, 2 Γ 9, 3 Γ 6, 6 Γ 3, 9 Γ 2 and 18 Γ 1
Step 2:
So, if it has 5 elements,
then the possible orders are 5 Γ 1 and 1 Γ 5
4. Construct a 2 Γ 2 matrix, π΄ = [πππ], whose elements are given by:
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(i) πππ =(π+π)2
2 [2 marks]
(ii) πππ =π
π [2 marks]
(iii) πππ =(π+2π)2
2 [2 marks]
Solution:
Given:
π΄ = [πππ]
Since, it is a 2 Γ 2 matrix
π΄ = [π11 π12
π21 π22]
Step 1:
(i) Here, πππ =(π+π)2
2,
So, the elements of matrix are:
π11 =(1+1)2
2= 2
π12 =(1+2)2
2=
9
2
π21 =(2+1)2
2=
9
2
π22 =(2+2)2
2= 8
Therefore, the required matrix = [2
9
29
28]
Step 2:
(ii) Here ,πππ =π
π
So, the elements of matrix are:
π11 =1
1= 1
π12 =1
2
π21 =2
1= 2
π22 =2
2= 1
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Therefore, the required matrix=[1
1
2
2 1] [2 marks]
Step 3:
(πππ) Here, πππ =(π+2π)2
2,
The elements of matrix are:
π11 =(1+2(1))2
2=
(1+2)2
2=
32
2=
9
2
π12 =(1+2(2))2
2=
(1+4)2
2=
52
2=
25
2
π21 =(2+2(1))
2
2=
(2+2)2
2=
(4)2
2=
16
2= 8
π22 =(2+2(2))2
2=
(2+4)2
2=
(6)2
2=
36
2= 18
Therefore, the required matrix = [9
2
25
2
8 18] [2 marks]
5. Construct a 3 Γ 4 matrix, whose elements are given by:
(i) πππ =1
2| β 3π + π| [3 marks]
(ii) πππ = 2π β π [3 marks]
Solution:
Given:
Since, it is a 3 Γ 4 matrix
π΄ = [
π11 π12 π13 π14
π21 π22 π23 π24
π31 π32 π33 π34
]
Step 1:
(i) Here, πππ =1
2| β 3π + π|
So, the elements of matrix are:
π11 =1
2|β3(1) + 1| =
1
2|β3 + 1| =
1
2|β2| =
1
2(2) = 1 [
π
π Mark]
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π12 =1
2| β 3(1) + 2| =
1
2| β 3 + 2| =
1
2| β 1| =
1
2(1) =
1
2 [
π
π Mark]
π13 =1
2| β 3(1) + 3| =
1
2|0| =
1
2(0) = 0 [
π
π Mark]
π14 =1
2| β 3(1) + 4| =
1
2| β 3 + 4| =
1
2|1| =
1
2(1) =
1
2 [
π
π Mark]
π21 =1
2| β 3(2) + 1| =
1
2| β 6 + 1| =
1
2| β 5| =
1
2(5) =
5
2
[π
π Mark]
π22 =1
2| β 3 Γ 2 + 2| =
1
2| β 6 + 2| =
1
2| β 4| =
1
2(4) = 2 [
π
π Mark]
π23 =1
2| β 3(2) + 1| =
1
2| β 6 + 3| =
1
2| β 3| =
1
2(3) =
3
2 [
π
π Mark]
π24 =1
2| β 3 Γ 2 + 2| =
1
2| β 6 + 4| =
1
2| β 2| =
1
2(2) = 1 [
π
π Mark]
π31 =1
2| β 3(3) + 1| =
1
2| β 9 + 1| =
1
2| β 8| =
1
2(8) = 4 [
π
π Mark]
π32 =1
2| β 3(3) + 2| =
1
2| β 9 + 2| =
1
2| β 7| =
1
2(7) =
7
2 [
π
π Mark]
π33 =1
2| β 3 Γ 3 + 3| =
1
2| β 9 + 3| =
1
2| β 6| =
1
2(6) = 3 [
π
π Mark]
π34 ==1
2| β 3 Γ 3 + 4| =
1
2| β 9 + 4| =
1
2| β 5| =
1
2(5) =
5
2 [
π
π Mark]
Therefore, the required matrix A =
[ 1
1
20
1
25
22
3
21
47
23
5
2]
Step 2:
(ii) Here, πππ = 2π β π,
So, the elements of matrix are:
π11 = 2(1) β 1 = 2 β 1 = 1 [π
π Mark]
π12 = 2(1) β 2 = 2 β 2 = 0 [π
π Mark]
π13 = 2(1) β 3 = 2 β 3 = β1 [π
π Mark]
π14 = 2(1) β 4 = 2 β 4 = β2 [π
π Mark]
π21 = 2(2) β 1 = 4 β 1 = 3 [π
π Mark]
π22 = 2(2) β 2 = 4 β 2 = 2 [π
π Mark]
π23 = 2(2) β 3 = 4 β 3 = 1 [π
π Mark]
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π24 = 2(2) β 4 = 4 β 4 = 0 [π
π Mark]
π31 = 2(3) β 1 = 6 β 1 = 5 [π
π Mark]
π32 = 2(3) β 2 = 6 β 2 = 4 [π
π Mark]
π33 = 2(3) β 3 = 6 β 3 = 3 [π
π Mark]
π34 = 2(3) β 4 = 6 β 4 = 2 [π
π Mark]
Therefore, the required matrix A = [1 0 β1 β23 2 1 05 4 3 2
]
6. Find the values of π₯, π¦ and π§ from the following equations:
(i) [4 3π₯ 5
] = [π¦ π§1 5
] [2 Marks]
(ii) [π₯ + π¦ 25 + π§ π₯π¦
] = [6 25 8
] [2 Marks]
(iii) [
π₯ + π¦ + π§π₯ + π§π¦ + π§
] = [957] [2 Marks]
Solution:
(i) Step 1:
Given: [4 3π₯ 5
] = [π¦ π§1 5
].
If two matrices are equal, then their corresponding elements are also equal. [π
π Mark]
Step 2:
Comparing the corresponding elements, we get
4 = y
3 = π§
π₯ = 1
Therefore, 4 = y, 3 = π§ and π₯ = 1 [ππ
π Mark]
(ii)Step 1:
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Given: [π₯ + π¦ 25 + π§ π₯π¦
] = [6 25 8
]
If two matrices are equal, then their corresponding elements are also equal. [π
π Mark]
Step 2:
π₯ + π¦ = 6
5 + π§ = 5
π₯π¦ = 8 [π
π Mark]
Step 3:
By solving these above equations, we get
π₯ = 2, π¦ = 4 and π§ = 0 OR
π₯ = 4, π¦ = 2 and π§ = 0
Therefore, π₯ = 2, π¦ = 4 and π§ = 0 or π₯ = 4, π¦ = 2 and π§ = 0 [π Mark]
(iii) Step 1:
Given: [
π₯ + π¦ + π§π₯ + π§π¦ + π§
] = [957]
If two matrices are equal, then their corresponding elements are also equal. [π
π Mark]
Step 2:
Comparing the corresponding elements, we get
π₯ + π¦ + π§ = 9
π₯ + π§ = 5
π¦ + π§ = 7 [π
π Mark]
Step 3:
By solving these above equations, we get
π₯ = 2,
π¦ = 4
π§ = 3
Therefore, π₯ = 2, π¦ = 4 and π§ = 3 [π Mark]
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7. Find the value of π, π, π and π from the equation:
[π β π 2π + π2π β π 3π + π
] = [β1 50 13
] [3 marks]
Solution:
Step 1:
If two matrices are equal, then their corresponding elements are also equal.
Therefore, on comparing the corresponding elements, we get
π β π = β1β¦ (1)
2π β π = 0β¦ (2)
2π + π = 5β¦ (3)
3π + π = 13β¦ (4) [1π
π Marks]
Step 2:
Now,
By solving equation (1) and (2), we get
π = 1, π = 2
Putting the value of π in equation (3), we get
π = 3
Putting the value of π in the equation (4), we get
π = 4
Hence, the required values are π = 1, π = 2, π = 3 and π = 4 [1π
π Marks]
8. π΄ = [πππ]πΓπ is a square matrix, if
(A) π < π
(B) π > π
(C) π = π
(D) None of these
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Solution:
(C)
Step 1:
π΄ = [πππ]πΓπ means matrix A is of order m Γ n
In a square matrix the number of column is same as the number of rows.
Number of rows = Number of columns
π = π
Hence, the option (πΆ) is correct.
9. Which of the given values of π₯ and π¦ make the following pair of matrices equal
[3π₯ + 7 5π¦ + 1 2 β 3π₯
] , [0 π¦ β 28 4
] [2 marks]
(A) π₯ =β1
3, π¦ = 7
(B) Not possible to find
(C) π¦ = 7, π₯ =β2
3
(D) π₯ =β1
3, π¦ =
β2
3
Solution:
(B)
Step 1:
Here [3π₯ + 7 5π¦ + 1 2 β 3π₯
] = [0 π¦ β 28 4
]
If two matrices are equal, then their corresponding elements are also equal. [π
π Mark]
Step 2:
Comparing the corresponding elements, we get
3π₯ + 7 = 0
5 = π¦ β 2
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π¦ β 1 = 8
2 β 3π₯ = 4 [π
π Mark]
Step 3:
By solving these above equations, we get
π¦ = 7, π₯ = β7
3 and π₯ = β
2
3,
Here, the value of π₯ is not unique.
Therefore, the option (π΅) is correct.
10. The number of all possible matrices of order 3 Γ 3 with each entry 0 or 1 is:
(A) 27
(B) 18
(C) 81
(D) 512 [2 Marks]
Solution:
(D)
Given:
It is a 3 Γ 3 matrix.
Step 1:
A = [
π11 π12 π13
π21 π22 π23
π31 π32 π33
]
The total number of elements in a matrix of order 3 Γ 3 = 9
Step 2:
If each entry is 0 or 1, then total number of permutations for each element = 2
Therefore, the total permutation for 9 elements = 29 = 512
Hence, the option (D) is correct.
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EXERCISE 3.2
1. Let π΄ = [2 43 2
], π΅ = [1 3β2 5
], πΆ = [β2 5 3 4
]
Find each of the following:
(i) π΄ + π΅ [2 marks]
(ii) π΄ β π΅ [2 marks]
(iii) 3π΄ β πΆ [3 marks]
(iv) AB [3 marks]
(v) BA [3 marks]
Solution:
Given:
π΄ = [2 43 2
], π΅ = [1 3β2 5
], πΆ = [β2 5 3 4
]
(i) π΄ + π΅
Step 1:
π΄ + π΅ = [2 43 2
] + [1 3β2 5
]
Step 2:
= [2 + 1 4 + 33 β 2 2 + 5
] = [3 71 7
]
Hence, π΄ + π΅ = [3 71 7
]
(ii) A β B
Step 1:
A β B = [2 43 2
] β [1 3
β2 5]
Step 2:
= [2 β 1 4 β 33 + 2 2 β 5
] = [1 15 β3
]
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Hence, A β B = [1 15 β3
]
(iii) 3π΄ β πΆ
Step 1:
3π΄ β πΆ = 3 [2 43 2
] β [β2 53 4
]
Step 2:
= [6 129 6
] β [β2 53 4
]
Step 3:
= [6 + 2 12 β 59 β 3 6 β 4
] = [8 76 2
]
Hence, 3π΄ β πΆ = [8 76 2
]
(iv) π΄π΅
Step 1:
π΄π΅ = [2 43 2
] [1 3β2 5
]
= [2 Γ 1 + 4 Γ β2 2 Γ 3 + 4 Γ 53 Γ 1 + 2 Γ β2 3 Γ 3 + 2 Γ 5
]
Step 2:
= [2 β 8 6 + 203 β 4 9 + 10
]
Step 3:
= [β6 26β1 19
]
Hence, π΄π΅ = [β6 26β1 19
]
(v) π΅π΄
Step 1:
π΅π΄ = [1 3β2 5
] [2 43 2
]
= [1 Γ 2 + 3 Γ 3 1 Γ 4 + 3 Γ 2β2 Γ 2 + 5 Γ 3 β2 Γ 4 + 5 Γ 2
]
Step 2:
= [2 + 9 4 + 6
β4 + 15 β8 + 10]
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Step 3:
= [11 1011 2
]
Hence, π΅π΄ = [11 1011 2
]
2. Compute the following:
(i) [π πβπ π
] + [π ππ π
] [2 marks]
(ii) [π2 + π2 π2 + π2
π2 + π2 π2 + π2] + [2ππ 2ππβ2ππ β2ππ
] [2 marks]
(iii) [β1 4 β68 5 162 8 5
] + [12 7 68 0 53 2 4
] [2 marks]
(iv) [cos2π₯ sin2π₯sin2π₯ cos2π₯
] + [sin2π₯ cos2π₯
cos2π₯ sin2π₯] [2 marks]
Solution:
(i) Given:
[π πβπ π
] + [π ππ π
]
Step 1:
= [π + π π + πβπ + π π + π
]
Step 2:
= [2π 2π0 2π
]
Therefore, [π πβπ π
] + [π ππ π
] = [2π 2π0 2π
]
(ii) Given:
[π2 + π2 π2 + π2
π2 + π2 π2 + π2] + [2ππ 2ππβ2ππ β2ππ
]
Step 1:
= [π2 + π2 + 2ππ π2 + π2 + 2ππ
π2 + π2 β 2ππ π2 + π2 β 2ππ]
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Step 2:
= [(π + π)2 (π + π)2
(π β π)2 (π β π)2]
Therefore, [π2 + π2 π2 + π2
π2 + π2 π2 + π2] + [2ππ 2ππβ2ππ β2ππ
] = [(π + π)2 (π + π)2
(π β π)2 (π β π)2]
(iii) Given:
[β1 4 β68 5 162 8 5
] + [12 7 68 0 53 2 4
]
Step 1:
= [β1 + 12 4 + 7 β6 + 68 + 8 5 + 0 16 + 52 + 3 8 + 2 5 + 4
]
Step 2:
= [11 11 016 5 215 10 9
]
Therefore, [β1 4 β68 5 162 8 5
] + [12 7 68 0 53 2 4
] = [11 11 016 5 215 10 9
]
(iv) Given:
[πππ 2π₯ π ππ2π₯π ππ2π₯ πππ 2π
] + [π ππ2π πππ 2π
πππ 2π₯ π ππ2π₯]
Step 1:
= [πππ 2π₯ + π ππ2π π ππ2π₯ + πππ 2π₯π ππ2π₯ + πππ 2π πππ 2π₯ + π ππ2π₯
]
Step 2:
= [1 11 1
]
Therefore, [πππ 2π₯ π ππ2π₯π ππ2π₯ πππ 2π
] + [π ππ2π πππ 2π
πππ 2π₯ π ππ2π₯] = [
1 11 1
]
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3. Compute the indicated products.
(i) [π πβπ π
] [π βππ π
] [2 marks]
(ii) [123] [2 3 4] [2 marks]
(iii) [1 β22 3
] [1 2 32 3 1
] [2 marks]
(iv) [2 3 43 4 54 5 6
] [1 β3 50 2 43 0 5
] [2 marks]
(v) [2 13 2
β1 1] [
1 0 1β1 2 1
] [2 marks]
(vi) [3 β1 3β1 0 2
] [2 β31 03 1
] [2 marks]
Solution:
(i) Given:
[π πβπ π
] [π βππ π
]
Step 1:
= [π Γ π + π Γ π π Γ (βπ) + π Γ π
βπ Γ π + π Γ π (βπ) Γ (βπ) + π Γ π]
Step 2:
= [ π2 + π2 βππ + ππβππ + π2 π2 + π2 ]
= [π2 + π2 0
0 π2 + π2]
Hence, [π πβπ π
] [π βππ π
] = [π2 + π2 0
0 π2 + π2]
(ii) Given:
[123] [2 3 4]
Step 1:
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= [1 Γ 2 1 Γ 3 1 Γ 42 Γ 2 2 Γ 3 2 Γ 43 Γ 2 3 Γ 3 3 Γ 4
]
Step 2:
= [2 3 44 6 86 9 12
]
Hence, [123] [2 3 4] = [
2 3 44 6 86 9 12
]
(iii) Given:
[1 β22 3
] [1 2 32 3 1
]
Step 1:
= [1 Γ 1 + (β2) Γ 2 1 Γ 2 + (β2) Γ 3 1 Γ 3 + (β2) Γ 1
2 Γ 1 + 3 Γ 2 2 Γ 2 + 3 Γ 3 2 Γ 3 + 3 Γ 1]
= [1 β 4 2 β 6 3 β 22 + 6 4 + 9 6 + 3
]
Step 2:
= [β3 β4 18 13 9
]
Hence, [1 β22 3
] [1 2 32 3 1
] = [β3 β4 18 13 9
]
(iv) Given:
[2 3 43 4 54 5 6
] [1 β3 50 2 43 0 5
]
Step 1:
= [2 Γ 1 + 3 Γ 0 + 4 Γ 3 2 Γ (β3) + 3 Γ 2 + 4 Γ 0 2 Γ 5 + 3 Γ 4 + 4 Γ 53 Γ 1 + 4 Γ 0 + 5 Γ 3 3 Γ (β3) + 4 Γ 2 + 5 Γ 0 3 Γ 5 + 4 Γ 4 + 5 Γ 54 Γ 1 + 5 Γ 0 + 6 Γ 3 4 Γ (β3) + 5 Γ 2 + 6 Γ 0 4 Γ 5 + 5 Γ 4 + 6 Γ 5
]
Step 2:
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= [2 + 0 + 12 β6 + 6 + 0 10 + 12 + 203 + 0 + 15 β9 + 0 + 0 15 + 16 + 254 + 0 + 18 β12 + 10 + 0 20 + 20 + 30
]
= [14 0 4218 β1 5622 β2 70
]
Hence, [2 3 43 4 54 5 6
] [1 β3 50 2 43 0 5
] = [14 0 4218 β1 5622 β2 70
]
(v) Given:
[2 13 2β1 1
] [1 0 1
β1 2 1]
Step 1:
= [
2 Γ 1 + 1 Γ (β1) 2 Γ 0 + 1 Γ 2 2 Γ 1 + 1 Γ 13 Γ 1 + 2 Γ (β1) 3 Γ 0 + 2 Γ 2 3 Γ 1 + 2 Γ 1β1 Γ 1 + 1 Γ (β1) β1 Γ 0 + 1 Γ 2 β1 Γ 1 + 1 Γ 1
]
Step 2:
= [1 2 31 4 5β2 2 0
]
Hence, [2 13 2β1 1
] [1 0 1
β1 2 1] = [
1 2 31 4 5β2 2 0
]
(vi) Given:
[3 β1 3β1 0 2
] [2 β31 03 1
]
Step 1:
= [3 Γ 2 + (β1) Γ 1 + 3 Γ 3 3 Γ (β3) + (β1) Γ 0 + 3 Γ 1β1 Γ 2 + 0 Γ 1 + 2 Γ 3 β1 Γ (β3) + 0 Γ 0 + 2 Γ 1
]
Step 2:
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= [6 β 1 + 9 β9 β 0 + 3
β2 + 0 + 6 3 + 0 + 2]
= [14 β64 5
]
Hence, [3 β1 3β1 0 2
] [2 β31 03 1
] = [14 β64 5
]
4. If π΄ = [1 2 β35 0 21 β1 1
], π΅ = [3 β1 24 2 52 0 3
] and πΆ = [4 1 20 3 21 β2 3
], then compute (π΄ + π΅)
and (π΅βπΆ). Also, verify that π΄ + (π΅ β πΆ) = (π΄ + π΅) β πΆ. [3 marks]
Solution:
Given:
π΄ = [1 2 β35 0 21 β1 1
], π΅ = [3 β1 24 2 52 0 3
] and πΆ = [4 1 20 3 21 β2 3
]
Step 1:
So, π΄ + π΅ = [1 2 β35 0 21 β1 1
] + [3 β1 24 2 52 0 3
]
= [1 + 3 2 β 1 β3 + 25 + 4 0 + 2 2 + 51 + 2 β1 + 0 1 + 3
]
β΄ π΄ + π΅ = [4 1 β19 2 73 β1 4
]
Step 2:
Now, π΅ β πΆ = [3 β1 24 2 52 0 3
] β [4 1 20 3 21 β2 3
]
= [3 β 4 β1 β 1 2 β 24 β 0 2 β 3 5 β 22 β 1 0 β (β2) 3 β 3
]
β΄ π΅ β πΆ = [β1 β2 04 β1 31 2 0
]
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Step 3:
πΏπ»π = π΄ + (π΅ β πΆ) = [1 2 β35 0 21 β1 1
] + [β1 β2 04 β1 31 2 0
] = [0 0 β39 β1 52 1 1
]
π π»π = (π΄ + π΅) β πΆ = [4 1 β19 2 73 β1 4
] β [4 1 20 3 21 β2 3
] = [0 0 β39 β1 52 1 1
]
β΄ πΏπ»π = π π»π
Hence, π΄ + (π΅ β πΆ) = (π΄ + π΅) β πΆ = [0 0 β39 β1 52 1 1
]
5. If π΄ =
[ 2
31
5
31
3
2
3
4
37
32
2
3]
and =
[ 2
5
3
51
1
5
2
5
4
57
5
6
5
2
5]
, then compute 3π΄ β 5π΅. [3 Marks]
Solution:
Given:
π΄ =
[ 2
31
5
31
3
2
3
4
37
32
2
3]
and π΅ =
[ 2
5
3
51
1
5
2
5
4
57
5
6
5
2
5]
Step 1:
3π΄ β 5π΅ = 3
[ 2
31
5
31
3
2
3
4
37
32
2
3]
β 5
[ 2
5
3
51
1
5
2
5
4
57
5
6
5
2
5]
Step 2:
=
[ 2
3Γ 3 1 Γ 3
5
3Γ 3
1
3Γ 3
2
3Γ 3
4
3Γ 3
7
3Γ 3 2 Γ 3
2
3Γ 3]
β
[ 2
5Γ 5
3
5Γ 5 1 Γ 5
1
5Γ 5
2
5Γ 5
4
5Γ 5
7
5Γ 5
6
5Γ 5
2
5Γ 5]
Step 3:
Class-XII-Maths Matrices
20 Practice more on Matrices www.embibe.com
= [2 3 51 2 47 6 2
] β [2 3 51 2 47 6 2
]
= [2 β 2 3 β 3 5 β 51 β 1 2 β 2 4 β 47 β 7 6 β 6 2 β 2
] = [0 0 00 0 00 0 0
] = 0
Therefore, 3π΄ β 5π΅ = [0 0 00 0 00 0 0
]
6. Simplify cosπ [cosπ sinπβsinπ cosπ
] + sinπ [sinπ βcosπcosπ sinπ
] [3 Marks]
Solution:
Given:
πππ π [πππ π π πππβπ πππ πππ π
] + π πππ [π πππ βπππ ππππ π π πππ
]
Step 1:
= [cosπ(cosπ) cosπ(sinπ)
cosπ(βsinπ) cosπ(cosπ)] + [
sinπ(sinπ) sinπ(βcosπ)
sinπ(cosπ) sinπ(sinπ)]
= [πππ 2π π ππππππ πβπ ππππππ π πππ 2π
] + [π ππ2π βπ ππππππ π
π ππππππ π π ππ2π]
Step 2:
= [πππ 2π + π ππ2π π ππππππ π β π ππππππ πβπ ππππππ π + π ππππππ π πππ 2π + π ππ2π
]
= [cos2π + sin2π 00 cos2π + sin2π
] (β΅ cos2π + sin2π = 1)
Step 3:
= [1 00 1
] = πΌ
Hence, cosπ [cosπ sinπβsinπ cosπ
] + sinπ [sinπ βcosπcosπ sinπ
] = [1 00 1
]
Class-XII-Maths Matrices
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7. Find π and π, if
(i) π + π = [7 02 5
] and π β π = [3 00 3
] [3 Marks]
(ii) 2π + 3π = [2 34 0
] and 3π + 2π = [2 β2β1 5
] [3 Marks]
Solution:
(i) Given:
π + π = [7 02 5
] β― (1)
and π β π = [3 00 3
] β¦. (2)
Step 1:
Adding equation (1) and (2), we get
π + π + π β π = [7 02 5
] + [3 00 3
]
2π = [10 02 8
]
β π = [5 01 4
] [1π
π Mark]
Step 2:
Putting the value of π in equation (1), we get
[5 01 4
] + π = [7 02 5
]
π = [7 02 5
] β [5 01 4
]
β π = [2 01 1
]
Therefore, π = [5 01 4
] and π = [2 01 1
] [1π
π Mark]
(ii) Given:
2π + 3π = [2 34 0
]β¦ (1)
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22 Practice more on Matrices www.embibe.com
3π + 2Y = [2 β2β1 5
] β―(2)
Step 1:
Multiply equation (1) by 3 and equation (2) by 2, on subtracting, we get
3 (2π + 3π) β 2(3π + 2π) = 3 [2 34 0
] β 2 [2 β2β1 5
]
β 6π + 9π β 6π β 4π = [6 912 0
] β [4 β4β2 10
]
β 5π = [2 1314 β10
]
β π = [
2
5
13
514
5β2
] [1π
π Mark]
Step 2:
Putting the value of π in equation (1), we get
2π + 3 [
2
5
13
514
5β2
] = [2 34 0
]
β 2π = [2 34 0
] β [
6
5
39
542
5β6
]
= [2 β
6
53 β
39
5
4 β42
50 + 6
] = [
4
5β
24
5
β22
56
]
β π = [
2
5β
12
5
β11
53
]
Hence, π = [
2
5β
12
5
β11
53
] and π = [
2
5
13
514
5β2
] [1π
π Mark]
8. Find X, if π = [3 21 4
] and 2π + π = [1 0β3 2
] [3 Marks]
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Solution:
Given:
π = [3 21 4
]
Step 1:
2π + π = [1 0β3 2
]
β 2π + [3 21 4
] = [1 0β3 2
] [β : π = [3 21 4
]]
[1π
π Mark]
Step 3:
β 2π = [1 0β3 2
] β [3 21 4
] = [β2 β 2β4 β 2
]
β π = [β1 β 1β2 β 1
]
Hence, the value of π = [β1 β 1β2 β 1
][1π
π Mark]
9. Find π₯ and π¦, if 2 [1 30 π₯
] + [π¦ 01 2
] = [5 61 8
] [3 Marks]
Solution:
Given:
2 [1 30 π₯
] + [π¦ 01 2
] = [5 61 8
]
Step 1:
β [2 60 2π₯
] + [π¦ 01 2
] = [5 61 8
]
β [2 + π¦ 6 + 00 + 1 2π₯ + 2
] = [5 61 8
]
Step 2:
If two matrices are equal, then their corresponding elements are also equal.
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Comparing the corresponding elements, we get
2 + π¦ = 5 and 2π₯ + 2 = 8
Step 3:
By solving these equations, we get
π¦ = 3 and π₯ = 3
Therefore, the values of π¦ = 3 and π₯ = 3.
10. Solve the equation for π₯, π¦, π§ and π‘, if 2 [π₯ π§π¦ π‘] + 3 [
1 β10 2
] = 3 [3 54 6
] [3 Marks]
Solution:
Given:
2 [π₯ π§π¦ π‘] + 3 [
1 β10 2
] = 3 [3 54 6
]
Step 1:
[π₯ Γ 2 π§ Γ 2π¦ Γ 2 π‘ Γ 2
] + [1 Γ 3 β1 Γ 30 Γ 3 2 Γ 3
] = [3 Γ 3 5 Γ 34 Γ 3 6 Γ 3
]
β [2π₯ 2π§2π¦ 2π‘
] + [3 β30 6
] = [9 1512 18
]
Step 2:
If two matrices are equal, then their corresponding elements are also equal.
Comparing the corresponding elements, we get
2π₯ + 3 = 9
2π§ β 3 = 15
2π¦ = 12
2π‘ + 6 = 18
Step 3:
By solving these equations, we get
π₯ = 3, π§ = 9, π¦ = 6 and π‘ = 6
Hence, the values of π₯ = 3, π§ = 9, π¦ = 6 and π‘ = 6
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11. If π₯ [23] + π¦ [
β11
] = [105
], find the values of π₯ and π¦. [3 Marks]
Solution:
Given:
π₯ [23] + π¦ [
β11
] = [105
]
Step 1:
β [2π₯3π₯
] + [βπ¦π¦ ] = [
105
]
β [2π₯ β π¦3π₯ + π¦
] = [105
]
Step 2:
If two matrices are equal, then their corresponding elements are also equal.
Comparing the corresponding elements, we get
2π₯ β π¦ = 10
3π₯ + π¦ = 5
Step 3:
Adding both the equations, we get
5π₯ = 15
β π₯ = 3
Putting the value of π₯ in the equation 3π₯ + π¦ = 5, we get
3(3) + π¦ = 5
β π¦ = β4
Therefore, the values x and y are 3 and -4 respectively.
12. Given 3 [π₯ π¦π§ π€
] = [π₯ 6β1 2π€
] + [4 π₯ + π¦π§ + π€ 3
], find the values of π₯, π¦, π§ and π€. [3
Marks]
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Solution:
Given:
3 [π₯ π¦π§ π€
] = [π₯ 6β1 2π€
] + [4 π₯ + π¦π§ + π€ 3
]
Step 1:
[3 Γ π₯ 3 Γ π¦3 Γ π§ 3 Γ π€
] = [π₯ + 4 6 + π₯ + π¦
β1 + π§ + π€ 2π€ + 3]
β [3π₯ 3π¦3π§ 3π€
] = [π₯ + 4 6 + π₯ + π¦1 + π§ + π€ 2π€ + 3
]
Step 2:
If two matrices are equal, then their corresponding elements are also equal.
Comparing the corresponding elements, we get
3π₯ = π₯ + 4
3π¦ = 6 + π₯ + π¦
3π§ = β1 + π§ + π€
3π€ = 2π€ + 3
Step 3:
By solving these equations, we get
β π₯ = 2, π¦ = 4, π§ = 1 and π€ = 3
Therefore, the values of π₯ = 2, π¦ = 4, π§ = 1 and π€ = 3
13. If πΉ(π₯) = [cos π₯ β sinπ₯ 0sinπ₯ cosπ₯ 00 0 1
], show that πΉ(π₯)πΉ(π¦) = πΉ(π₯ + π¦). [3 Marks]
Solution:
Given:
πΉ(π₯) = [cos π₯ β sinπ₯ 0sinπ₯ cos π₯ 00 0 1
]
Step 1:
For πΉ(π¦), replace x by y in πΉ(π₯)
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27 Practice more on Matrices www.embibe.com
πΉ(π¦) = [cosπ¦ β sin π¦ 0sin π¦ cosπ¦ 00 0 1
] [π
π Mark]
Step 2:
By taking LHS = πΉ(π₯)πΉ(π¦)
= [πππ π₯ βπ πππ₯ 0π πππ₯ πππ π₯ 00 0 1
] [cos π¦ β sinπ¦ 0 0sinπ¦ cos π¦ 0
0 0 1] [
π
π Mark]
Step 3:
= [
cosπ₯ cosπ¦ + (βsinπ₯)sinπ¦ + 0 cosπ₯(βsinπ¦) + (βsinπ₯)cosπ¦ + 0 0 + 0 + 0 Γ 1
sinπ₯ cosπ¦ + cosπ₯ sinπ¦ + 0 sinπ₯(βsinπ¦) + cosπ₯ cosπ¦ + 0 0 + 0 + 0 Γ 1
0 Γ cosπ¦ + 0 Γ sinπ¦ + 0 Γ 1 0 Γ (βsinπ¦) + 0 Γ cosπ¦ + 0 0 + 0 + 1 Γ 1
]
= [cos(π₯ + π¦) β[cosπ₯ sinπ¦ + sinπ₯ cosπ¦] 0
sin(π₯ + π¦) cosπ₯ cosπ¦ β sinπ₯ sinπ¦ 00 0 1
]
β΅ cosπ₯ cosπ¦ β sinπ₯ sinπ¦ = cos(π₯ + π¦)sinπ₯ cosπ¦ + cosπ₯ sinπ¦ = sin(π₯ + π¦)
[1Mark]
Step 4:
= [πππ (π₯ + π¦) βπ ππ(π₯ + π¦) 0π ππ(π₯ + π¦) πππ (π₯ + π¦) 00 0 1
] = πΉ(π₯ + π¦) = π π»π
πΏπ»π = π π»π
Hence it is proved πΉ(π₯)πΉ(π¦) = πΉ(π₯ + π¦).
14. Show that
(i) [5 β16 7
] [2 13 4
] β [2 13 4
] [5 β16 7
] [3 Marks]
(ii) [1 2 30 1 01 1 0
] [β1 1 00 β1 12 3 4
] β [β1 1 00 β1 12 3 4
] [1 2 30 1 01 1 0
] [4 Marks]
Solution:
(i)Step 1:
By taking πΏπ»π = [5 β16 7
] [2 13 4
]
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28 Practice more on Matrices www.embibe.com
= [5 Γ 2 + (β1) Γ 3 5 Γ 1 + (β1) Γ 4
6 Γ 2 + 7 Γ 3 6 Γ 1 + 7 Γ 4] [
π
π Mark]
Step 2:
= [10 β 3 5 β 412 + 21 6 + 28
]
= [7 133 34
]
Step 3:
Now,
By taking π π»π = [2 13 4
] [5 β16 7
]
= [2 Γ 5 + 1 Γ 6 2 Γ (β1) + 1 Γ 73 Γ 5 + 4 Γ 6 6 Γ 1 + 7 Γ 4
] [π
π Mark]
Step 4:
= [10 + 6 β2 + 715 + 24 β3 + 28
] = [16 539 25
]
Hence, [5 β16 7
] [2 13 4
] β [2 13 4
] [5 β16 7
]
(ii)Step 1:
By taking πΏπ»π = [1 2 30 1 01 1 0
] [β1 1 00 β1 12 3 4
]
= [1 Γ (β1) + 2 Γ 0 + 3 Γ 2 1 Γ 1 + 2 Γ (β1) + 3 Γ 3 1 Γ 0 + 2 Γ 1 + 3 Γ 40 Γ (β1) + 1 Γ 0 + 0 Γ 2 0 Γ 1 + 1 Γ (β1) + 0 Γ 3 0 Γ 0 + 1 Γ 1 + 0 Γ 41 Γ (β1) + 1 Γ 0 + 0 Γ 2 1 Γ 1 + 1 Γ (β1) + 0 Γ 3 1 Γ 0 + 1 Γ 1 + 0 Γ 4
]
Step 2:
= [β1 + 0 + 6 1 β 0 + 9 0 + 2 + 120 + 0 + 0 0 β 1 + 0 0 + 1 + 0
β1 + 0 + 0 1 β 1 + 0 0 + 1 + 0]
πΏπ»π = [5 8 140 β1 1β1 0 1
]
Step 3:
Now,
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By taking π π»π = [β1 1 00 β1 12 3 4
] [1 2 30 1 01 1 0
]
=
[β1 Γ 1 + 1 Γ 0 + 0 Γ 1 β1 Γ 2 + 1 Γ 1 + 0 Γ 1 β1 Γ 3 + 1 Γ 0 + 0 Γ 00 Γ 1 + (β1) Γ 0 + 1 Γ 1 0 Γ 2 + (β1) Γ 1 + 1 Γ 1 0 Γ 3 + (β1) Γ 0 + 1 Γ 02 Γ 1 + 3 Γ 0 + 4 Γ 1 2 Γ 2 + 3 Γ 1 + 4 Γ 1 2 Γ 3 + 3 Γ 0 + 4 Γ 0
]
Step 4:
= [β1 + 0 + 0 β2 + 1 + 0 β3 + 0 + 00 + 0 + 1 0 β 1 + 1 0 + 0 + 02 + 0 + 4 4 + 3 + 4 6 + 0 + 0
]
π π»π = [β1 β1 β31 0 06 11 6
]
β΄ πΏπ»π β π π»π
Hence, [1 2 30 1 01 1 0
] [β1 1 00 β1 12 3 4
] β [β1 1 00 β1 32 3 4
] [1 2 30 1 01 1 0
]
15. Find π΄2 β 5π΄ + 6πΌ, if π΄ = [2 0 12 1 31 β1 0
] [4 marks]
Solution:
Given:
π΄ = [2 0 12 1 31 β1 0
]
Step 1:
We know that, π΄2 = π΄. π΄
Class-XII-Maths Matrices
30 Practice more on Matrices www.embibe.com
β π΄2 = [2 0 12 1 31 β1 0
] [2 0 12 1 31 β1 0
]
=
[
2 Γ 2 + 0 Γ 2 + 1 Γ 1 2 Γ 0 + 0 Γ 1 + 1 Γ (β1) 2 Γ 1 + 0 Γ 3 + 1 Γ 02 Γ 2 + 1 Γ 2 + 3 Γ 1 2 Γ 0 + 1 Γ 1 + 3 Γ (β1) 2 Γ 1 + 1 Γ 3 + 3 Γ 01 Γ 2 + (β1) Γ 2 + 0 Γ 1 1 Γ 0 + (β1) Γ 1 + 0 Γ (β1) 1 Γ 1 + (β1) Γ 3 + 0 Γ 0
]
Step 2:
= [4 + 0 + 1 0 + 0 β 1 2 + 0 + 04 + 2 + 3 0 + 1 β 3 2 + 3 + 02 β 2 + 0 0 β 1 + 0 1 β 3 + 0
]
π΄2 = [5 β1 29 β2 50 β1 β2
]
Step 3:
Therefore, π΄2 β 5π΄ + 6πΌ
= [5 β1 29 β2 50 β1 β2
] β 5 [2 0 12 1 31 β1 0
] + 6 [1 0 00 1 00 0 1
]
Step 4:
= [5 β1 29 β2 50 β1 β2
] β [10 0 510 5 155 β5 0
] + [6 0 00 6 00 0 6
]
= [5 β 10 + 6 β1 β 0 + 0 2 β 5 + 09 β 10 + 0 β2 β 5 + 6 5 β 15 + 00 β 5 + 0 β1 + 5 + 0 β2 β 0 + 6
]
= [1 β1 β3
β1 β1 β10β5 4 4
]
Hence, π΄2 β 5π΄ + 6πΌ = [1 β1 β3
β1 β1 β10β5 4 4
]
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31 Practice more on Matrices www.embibe.com
16. If π΄ = [1 0 20 2 12 0 3
], prove that π΄3 β 6π΄2 + 7π΄ + 2πΌ = 0 [5 Marks]
Solution:
Given:
π΄ = [1 0 20 2 12 0 3
]
Step 1:
We know that, π΄2 = π΄. π΄
β π΄2 = [1 0 20 2 12 0 3
] [1 0 20 2 12 0 3
]
= [1 Γ 1 + 0 Γ 0 + 2 Γ 2 1 Γ 0 + 0 Γ 2 + 2 Γ 0 1 Γ 2 + 0 Γ 1 + 2 Γ 30 Γ 1 + 2 Γ 0 + 1 Γ 2 0 Γ 0 + 2 Γ 2 + 1 Γ 0 0 Γ 2 + 2 Γ 1 + 1 Γ 32 Γ 1 + 0 Γ 0 + 3 Γ 2 2 Γ 0 + 0 Γ 2 + 3 Γ 0 2 Γ 2 + 0 Γ 1 + 3 Γ 3
]
Step 2:
= [1 + 0 + 4 0 + 0 + 0 2 + 0 + 60 + 0 + 2 0 + 4 + 0 0 + 2 + 32 + 0 + 6 0 + 0 + 0 4 + 0 + 9
]
π΄2 = [5 0 82 4 58 0 13
]
Step 3:
Now,
π΄3 = π΄2π΄ = [5 0 82 4 58 0 13
] [1 0 20 2 12 0 3
]
= [5 Γ 1 + 0 Γ 0 + 8 Γ 2 5 Γ 0 + 0 Γ 2 + 8 Γ 0 5 Γ 2 + 0 Γ 1 + 8 Γ 3 2 Γ 1 + 4 Γ 0 + 5 Γ 2 2 Γ 0 + 4 Γ 2 + 5 Γ 0 2 Γ 2 + 4 Γ 1 + 5 + 3 8 Γ 1 + 0 Γ 0 + 13 Γ 2 8 Γ 0 + 0 Γ 2 + 13 Γ 0 8 Γ 2 + 0 Γ 1 + 13 Γ 3
]
= [5 + 0 + 16 0 + 0 + 0 10 + 0 + 242 + 0 + 10 0 + 8 + 0 4 + 4 + 158 + 0 + 26 0 + 0 + 0 16 + 0 + 39
]
π΄3 = [21 0 3412 8 2334 0 55
]
Step 4:
Class-XII-Maths Matrices
32 Practice more on Matrices www.embibe.com
Therefore, π΄3 β 6π΄2 + 7π΄ + 2πΌ
πΏπ»π = [21 0 3412 8 2334 0 55
] β 6 [5 0 82 4 58 0 13
] + 7 [1 0 20 2 12 0 3
] + 2 [1 0 00 1 00 0 1
]
= [21 0 3412 8 2334 0 55
] β [
6(5) 0(5) 8(6)
2(6) 4(6) 5(6)
8(6) 0(6) 13(6)] + [
1(7) 0(7) 2(7)
0(7) 2(7) 1(7)
2(7) 0(7) 3(7)] +
[
2(1) 2(0) 2(0)
2(0) 1(2) 0(2)
2(0) 0(2) 1(2)]
Step 5:
= [21 β 30 + 7 + 2 0 β 0 + 0 + 0 34 β 48 + 14 + 012 β 12 + 0 + 0 8 β 24 + 14 + 2 23 β 30 + 7 + 034 β 48 + 14 + 0 0 + 0 + 0 + 0 55 β 78 + 21 + 2
]
= [0 0 00 0 00 0 0
] = 0
β΅ πΏπ»π = π π»π
Hence proved.
17. If π΄ = [3 β24 β2
] and πΌ = [1 00 1
], find π so that π΄2 = ππ΄ β 2πΌ [3 marks]
Solution:
Given:
π΄ = [3 β24 β2
] and πΌ = [1 00 1
]
Step 1:
π΄2 = ππ΄ β 2πΌ
We know that, π΄2 = π΄. π΄
β [3 β24 β2
] [3 β24 β2
] = π [3 β24 β2
] β 2 [1 00 1
]
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33 Practice more on Matrices www.embibe.com
Step 2:
β [3 Γ 3 + (β2) Γ 4 3 Γ (β2) + (β2) Γ (β2)
4 Γ 3 + (β2) Γ 4 4 Γ (β2) + (β2) Γ (β2)] = [
3π β2π4π β2π
] β [2 00 2
]
β [1 β24 β4
] = [3π β 2 β2π
4π β2π β 2]
Step 3:
Here, matrices are equal.
Comparing the corresponding elements, we get
β 4π = 4
β΄ π = 1
Hence, the value of π is 1.
18. If π΄ = [0 β tan
πΌ
2
tanπΌ
20
] and πΌ is the identity matrix of order 2, show that
πΌ + π΄ = (πΌ β π΄) [cos πΌ β sinπΌsinπΌ cosπΌ
] [5 Marks]
Solution:
Given:
π΄ = [0 β tan
πΌ
2
tanπΌ
20
]
Step 1:
By considering LHS = πΌ + π΄
= [1 00 1
] + [0 β tan
πΌ
2
tanπΌ
20
] = [1 β tan
πΌ
2
tanπΌ
21
]
Step 2:
Now,
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34 Practice more on Matrices www.embibe.com
By taking RHS = (πΌ β π΄) [πππ πΌ βπ πππΌπ πππΌ πππ πΌ
]
= ([1 00 1
] β [0 β tan
πΌ
2
tanπΌ
20
]) [πππ πΌ βπ πππΌπ πππΌ πππ πΌ
]
= [1 β tan
πΌ
2
tanπΌ
21
] [πππ πΌ βπ πππΌπ πππΌ πππ πΌ
]
Step 3:
= [cosπΌ + tan
πΌ
2sinπΌ βsinπΌ + tan
πΌ
2cos πΌ
β tanπΌ
2cos πΌ + sinπΌ tan
πΌ
2sin πΌ + cosπΌ
]
Step 4:
=
[ πππ πΌ +
sinπΌ
2
πππ πΌ
2
π πππΌ βπ πππΌ +π ππ
πΌ
2
πππ πΌ
2
πππ πΌ
βπ ππ
πΌ
2
πππ πΌ
2
πππ πΌ + π πππΌπ ππ
πΌ
2
πππ πΌ
2
π πππΌ + πππ πΌ]
[π
π Mark]
Step 5:
=
[ πππ πΌ πππ
πΌ
2+π ππ
πΌ
2π πππΌ
πππ πΌ
2
βπ πππΌ πππ πΌ
2+π ππ
πΌ
2πππ πΌ
πππ πΌ
2
βπππ πΌ π πππΌ
2+πππ
πΌ
2π πππΌ
πππ πΌ
2
π πππΌ π πππΌ
2+πππ
πΌ
2πππ πΌ
πππ πΌ
2 ]
[π
π Mark]
Step 6:
=
[ cos(πΌβ
πΌ
2)
πππ πΌ
2
βsin(πΌβπΌ
2)
πππ πΌ
2
sin(πΌβπΌ
2)
πππ πΌ
2
cos(πΌβπΌ
2)
πππ πΌ
2 ]
[π
π Mark]
Step 7:
= [1 β tan
πΌ
2
tanπΌ
21
] = πΏπ»π [π
π Mark]
β΅ πΏπ»π = π π»π
Hence, it is proved.
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19. A trust fund has βΉ 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide βΉ 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of: [5 Marks]
(a) βΉ1800
(b) βΉ2000
Solution:
Given:
Trust fund= βΉ 30,000
First bond pays interest = 5% per year
Second bond pays interest = 7% per year
Step 1:
Let the amount invested in first bond = βΉ π₯
The amount invested in second bond = βΉ(30000 β π₯)
(π) If the total annual interest is βΉ1800, then
Investment in Bonds (in βΉ) Annual Interest Rate Interest (in β€]
[π₯
30,000 β π₯] [5%7%
] [1800]
Step 2:
By solving, we get
π₯ Γ 5% + (30000 β π₯) Γ 7%=1800
β5π₯
100+
7
100(30000 β π₯) = 1800
β 5π₯ + 210000 β 7π₯ = 180000
β β2π₯ = β30000
π₯ = 15000
Step 3:
So, amount investment at 5% = βΉ15000
The amount investment at 7% = βΉ(30000 β π₯)
= βΉ(30000 β 15000) = βΉ15000
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36 Practice more on Matrices www.embibe.com
Therefore, the amount invested in first bond is βΉ15000 and second bond is βΉ15000 [π
π
Mark]
Step 4:
(b) If the total annual interest is βΉ2000, then
Investment in Bonds (in βΉ] Annual Interest Rate Interest (inβΉ)
[π₯
30,000 β π₯] [5%7%
] [2000]
Step 5:
By solving, we get
π₯ Γ 5% + (30000 β π₯) Γ 7% = 2000
β5π₯
100+
7
100(30000 β π₯) = 2000
β 5π₯ + 210000 β 7π₯ = 200000
β β2π₯ = β10000
β π₯ = 5000
Step 6:
So, amount investment at 5% = βΉ5000
The amount investment at 7% = βΉ(30000 β π₯)
= βΉ(30000 β 5000) = βΉ25000
Therefore, the amount invested in first bond is βΉ 5000 and second bond is βΉ25000. [π
π
Mark]
20. The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are βΉ 80, βΉ 60 and βΉ 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.
[3 Marks]
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Solution:
Given:
Step 1:
Number of chemistry books = 10 dozen = 10 Γ 12 = 120
Number of physics books = 8 dozen = 8 Γ 12 = 96
Number of economics books = 10 dozen = 10 Γ 12 = 120 [π
π Mark]
Step 2:
Number of books
Chemistry Physics Economics
Selling Price per book Total amount (in βΉ)
[120 96 120] [806040
] [π₯]
[1Mark]
Step 3:
Now,
The shopkeeper receives the total amount of = Number of books ΓSelling price per book
= [120 96 120] [806040
]
= [180 Γ 80 + 96 Γ 60 + 120 Γ 40]
= 120 Γ 80 + 96 Γ 60 + 120 Γ 40
= 9600 + 5760 + 4800
= βΉ 20160 [1π
π Marks]
Hence, the bookshop will receive βΉ20160 from selling all the books.
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21. Assume X, Y, Z, W and P are matrices of order 2 Γ n, 3 Γ k, 2 Γ p, n Γ 3 and p Γ k,
respectively.
The restriction on π, π and π so that ππ + ππ will be defined are: [2 Marks]
(A) π = 3, π = π
(B) π is arbitrary, π = 2
(C) π is arbitrary, π = 3
(D) π = 2, π = 3
Solution:
Given:
The order of π = π Γ π
The order of π = 3 Γ π.
Step 1:
So, ππ will be defined if, π = 3.
Hence, the order of ππ is π Γ π.
Step 2:
Order of π = π Γ 3 and order of π = 3 Γ π
According to order, WY is defined and its order is π Γ π.
ππ + ππ is defined, if the order of PY and WY are equal.
β π Γ π = π Γ π
β π = π,
Therefore, the option (A) is correct.
22. Assume X, Y, Z, W and P are matrices of order 2 Γ n, 3 Γ k, 2 Γ p, n Γ 3 and p Γ k,
respectively.
If π = π, then the order of the matrix 7π β 5π is: [2 marks]
(A) π Γ 2
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(B) 2 Γ π
(C) π Γ 3
(D) π Γ π
Solution:
Given:
The order of π = 2 Γ π
The order of π = 2 Γ π
π = π
Step 1:
During the addition or subtraction, the order of matrix doesnβt change.
Therefore, the order of 7π β 5π = order of π = order of π
Step 2:
β 2 Γ π = 2 Γ π
= 2 Γ π
Hence, the option (π΅) is correct
EXERCISE 3.3
1. Find the transpose of each of the following matrices:
(i) [
51
2
β1
]
(ii) [1 β12 3
]
(iii) [β1 5 6
β3 5 62 3 β1
]
Solution:
(i) Let π΄ = [
51
2
β1
]
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π΄β² = [
51
2
β1
]
β²
= [51
2β1]
Hence the transpose of the given matrix is [5 1
2 β 1]
(ππ) Let π΅ = [1 β12 3
]
π΅β² = [1 β12 3
]β²
= [1 2
β1 3]
Hence the transpose of the given matrix is [1 2β1 3
]
(iii) Let πΆ = [β1 5 6
β3 5 62 3 β1
]
πΆβ² = [β1 5 6
β3 5 62 3 β1
]
β²
= [β1 β3 25 5 36 6 β1
]
Hence the transpose of the given matrix is [β1 β3 25 5 36 6 β1
]
2. If π΄ = [β1 2 35 7 9β2 1 1
] and π΅ = [β4 1 β51 2 01 3 1
], then verify that
(i) (π΄ + π΅)β² = π΄β² + π΅β², [3 Marks]
(ii) (π΄ β π΅)β² = π΄β² β π΅β² [3 Marks]
Solution:
Given:
π΄ = [β1 2 35 7 9β2 1 1
] and π΅ = [β4 1 β51 2 01 3 1
]
(i) (π΄ + π΅)β² = π΄β² + π΅β² Step 1:
(π΄ + π΅)
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= [β1 2 35 7 9
β2 1 1] + [
β4 1 β51 2 01 3 1
] = [β1 β 4 2 + 1 3 β 55 + 1 7 + 2 9 + 0
β2 + 1 1 + 3 1 + 1]
(π΄ + π΅) = [β5 3 β26 9 9
β1 4 2]
Step 2:
Hence (π΄ + π΅)β = [β5 β6 β13 9 4
β2 9 2] β¦.(1) [
π
π Mark]
Step 3:
Now,
π΄β² + π΅β² = [β1 2 35 7 9
β2 1 1]
β²
+ [β4 1 β51 2 01 3 1
]
β²
[π
π Mark]
Step 4:
= [β1 5 β22 7 13 9 1
] + [β4 1 11 2 3
β5 0 1] = [
β1 β 4 5 + 1 β2 + 12 + 1 7 + 2 1 + 33 β 5 9 + 0 1 + 1
] = [β5 6 β13 9 4
β2 9 2] β¦.(2)
From the equation (1) and (2], we get
(π΄ + π΅)β = π΄β² + π΅β² [1Mark]
Hence it is proved.
(ii) (π΄ β π΅)β² = π΄β² β π΅β²
Step 1:
(π΄ β π΅)
= [β1 2 35 7 9β2 1 1
] β [β4 1 β51 2 01 3 1
] = [β1 + 4 2 β 1 3 + 55 β 1 7 β 2 9 β 0
β2 β 1 1 β 3 1 β 1]
(π΄ β π΅) = [3 1 84 5 9
β3 β2 0] [1Mark]
Step 2:
Thus, (π΄ β π΅)β = [3 4 β31 5 β28 9 0
] β― (1) [π
π Mark]
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Step 3:
Aβ² β π΅β² = [β1 2 35 7 9
β2 1 1]
β²
β [β4 1 β51 2 01 3 1
]
β²
[π
π Mark]
Step 4:
= [β1 5 β22 7 13 9 1
] β [β4 1 11 2 3
β5 0 1] = [
β1 + 4 5 β 1 β2 β 12 β 1 7 β 2 1 β 33 + 5 9 β 0 1 β 1
] = [3 4 β31 5 β28 9 0
] β― (2)
From the equation (1) and (2), we get
(π΄ + π΅)β² = π΄β² + π΅β² [1Mark]
Hence it is proved.
3. If π΄β² = [3 4β1 20 1
] and π΅ = [β1 2 11 2 3
], then verify that
(i) (π΄ + π΅)β² = π΄β² + π΅β² [3 Marks]
(ii) (π΄ β π΅)β² = π΄β² β π΅β² [3 Marks]
Solution:
Given:
π΄β² = [3 4β1 20 1
] and π΅ = [β1 2 11 2 3
]
(i) (π΄ + π΅)β² = π΄β² + π΅β² Step 1:
π΄β² = [3 4
β1 20 1
]
β π΄ = [3 β1 04 2 1
] [β΅ (π΄β²)β² = π΄] [π
π Mark]
Step 2:
(π΄ + π΅) = [3 β1 04 2 1
] + [β1 2 11 2 3
]
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43 Practice more on Matrices www.embibe.com
= [3 β 1 β1 + 2 0 + 14 + 1 2 + 2 1 + 3
] = [2 1 15 4 4
]
Step 3:
Therefore (π΄ + π΅)β² = [2 51 41 4
] β¦. (1) [π
π Mark]
Step 4:
π΄β² + π΅β = [3 β1 04 2 1
]β²
+ [β1 2 11 2 3
]β²
= [3 4
β1 20 1
] + [β1 12 21 3
] = [3 β 1 4 + 1
β1 + 2 2 + 20 + 1 1 + 3
]
π΄β² + π΅β = [2 51 41 4
] β― (2)
From the equation (1) and (2), we get
(π΄ + π΅)β = π΄β² + π΅β²
Hence, it is proved.
(ii) (π΄ β π΅)β² = π΄β² β π΅β²
Step 1:
π΄β² = [3 4β1 20 1
]
β π΄ = [3 βπ 04 2 1
] [β΅ (π΄β²)β² = π΄] [π
π Mark]
Step 2:
(A β B) = [3 β1 04 2 1
]β[β1 2 11 2 3
]
= [3 + 1 β1 β 2 0 β 14 β 1 2 β 2 1 β 3
] = [4 β3 β13 0 β2
]
Step 3:
Therefore (A β B)β = [4 3β3 0β1 β2
] β― (1) [π
π Mark]
Step 4:
π΄β² β π΅β² = [3 β1 04 2 1
]β²
β [β1 2 11 2 3
]β²
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44 Practice more on Matrices www.embibe.com
= [3 4
β1 20 1
] β [β1 12 21 3
] = [3 + 1 4 β 1
β1 β 2 2 β 20 β 1 1 β 3
]
= [4 3
β3 0β1 β2
]β― (2)
From the equation (1) and (2], we get
(π΄ β π΅)β = π΄β² β π΅β²
Hence, it is proved.
4. If π΄ = [β2 31 2
] and π΅ = [β1 01 2
], then find (π΄ + 2π΅)β² [2 Marks]
Solution:
Given:
A = [β2 31 2
] and π΅ = [β1 01 2
]
Step 1:
(π΄ + 2π΅) = [β2 31 2
] + 2 [β1 01 2
]
= [β2 31 2
] + [β2 02 4
]
Step 2:
= [β2 β 2 3 + 01 + 2 2 + 4
] = [β4 33 6
] [π
π Mark]
Step 3:
(π΄ + 2π΅)β = [β4 33 6
]β²
= [β4 33 6
] [π
π Mark]
Therefore, (π΄ + 2π΅)β = [β4 33 6
]
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5. For the matrices π΄ and π΅, verify that (π΄π΅)β² = π΅β²π΄β², where
(i) π΄ = [1β43
], π΅ = [β1 2 1] [3 marks]
(ii) π΄ = [012], π΅ = [1 5 7] [3 marks]
Solution:
(i) Given:
π΄ = [1β43
] , π΅ = [β1 2 1]
Step 1:
So, π΄π΅ = [1β43
] [β1 2 1]
= [1 Γ (β1) 1 Γ 2 1 Γ 1β4 Γ (β1) β4 Γ 2 β4 Γ 13 Γ (β1) 3 Γ 2 3 Γ 1
] = [β1 2 14 β8 β4
β3 6 3]
Step 2:
π΄π΅β² = [β1 2 14 β8 β4
β3 6 3]
β²
= [β1 4 β32 β8 61 β4 3
]
Therefore, π΄π΅β² = [β1 4 β32 β8 61 β4 3
] β¦..(1) [π
π Mark]
Step 3:
π΅β² = [β1 2 1]β² = [β121
]
π΄β² = [1
β43
] = [1 β 4 3] [π
π Mark]
Step 4:
π΅β²π΄β² = [β121
] [1 β 4 3] = [β1 Γ 1 β1 Γ (β4) β1 Γ 32 Γ 1 2 Γ (β4) 2 Γ 31 Γ 1 1 Γ (β4) 1 Γ 3
]
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= [β1 4 β32 β8 61 β4 3
]
Therefore, π΅β²π΄β² = [β1 4 β32 β8 61 β4 3
]β― (2)
From the equation (1] and [2], we get
(π΄π΅) = π΅β²π΄β²
Hence proved.
(ii) Given:
π΄ = [012], π΅ = [1 5 7]
Step 1:
So, AB=[012] = [1 5 7]
= [0 Γ 1 0 Γ 5 0 Γ 71 Γ 1 1 Γ 5 1 Γ 72 Γ 1 2 Γ 5 2 Γ 7
] = [0 0 01 5 72 10 14
]
Step 2:
π΄π΅β² = [0 0 01 5 72 10 14
]
β²
= [0 1 20 5 100 7 14
]
Therefore π΄π΅β² = [0 1 20 5 100 7 14
] β― (1) [π
π Mark]
Step 3:
π΅β² = [1 5 7]β² = [157] and π΄β² = [
012] = [0 1 2] [
π
π Mark]
Step 4:
So, π΅β²π΄β² = [157] [0 1 2] = [
1 Γ 0 1 Γ 1 1 Γ 25 Γ 0 5 Γ 1 5 Γ 27 Γ 0 7 Γ 1 7 Γ 2
]
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= [0 1 20 5 100 7 14
] β― (2)
From the equation (1) and (2), we get
(π΄π΅)β = π΅β²π΄β²
Hence proved
6. (i) If π΄ = [cosπΌ sinπΌβsinπΌ cosπΌ
] , then verify that π΄β²π΄ = πΌ [2 Marks]
(ii) If π΄ = [sinπΌ cosπΌβ cosπΌ sinπΌ
], then verify that π΄β²π΄ = πΌ [2 marks]
Solution:
(i) Given:
π΄ = [πππ πΌ π πππΌβπ πππΌ πππ πΌ
]
Step 1:
β π΄β² = [πππ πΌ βπ πππΌπ πππΌ πππ πΌ
] [π
π Mark]
Step 2:
Therefore π΄β²π΄ = [πππ πΌ βπ πππΌπ οΏ½ΜοΏ½πΌ πππ πΌ
] [πππ πΌ π πππΌβπ πππΌ πππ πΌ
]
= β[πππ 2πΌ + π ππ2πΌ πππ πΌ π πππΌ β π πππΌ πππ πΌπ πππΌ πππ πΌ β πππ πΌ π πππΌ π ππ2πΌ + πππ 2πΌ
]
Step 3:
= [1 00 1
] = πΌ
β΄ π΄β²π΄ = πΌ [π
π Mark]
Hence proved.
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(ii) Given:
π΄ = [π πππΌ πππ πΌβπππ πΌ π πππΌ
]
Step 1:
β π΄β² = [π πππΌ βπππ πΌπππ πΌ π πππΌ
] [π
π Mark]
Step 2:
Therefore π΄β²π΄ = [π πππΌ βπππ πΌπππ πΌ π πππΌ
] [π πππΌ πππ πΌβπππ πΌ π πππΌ
]
= [π ππ2πΌ + πππ 2πΌ π πππΌ πππ πΌ β πππ πΌ π πππΌ
πππ πΌ π πππΌ β π πππΌ πππ πΌ πππ 2πΌ + π ππ2πΌ]
Step 3:
= [1 00 1
] = πΌ
β΄ π΄β²π΄ = πΌ [π
π Mark]
Hence proved.
7. i) Show that the matrix π΄ = [1 β1 5
β1 2 15 1 3
] is a symmetric matrix. [2 Marks]
(ii) Show that the matrix π΄ = [0 1 β1
β1 0 11 β1 0
] is a skew symmetric matrix. [2 Marks]
Solution:
(i) Given:
π΄ = [1 β1 5β1 2 15 1 3
]
Step 1:
β Aβ² = [1 β1 5
β1 2 15 1 3
]
β²
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= [1 β1 5
β1 2 15 1 3
]
Step 2:
β΄ π΄β² = π΄
Hence, the matrix π΄ = [1 β1 5
β1 2 15 1 3
] is a symmetric matrix.
(ii) π΄ = [0 1 β1
β1 0 11 β1 0
]
Step 1:
β π΄β² = [0 1 β1
β1 0 11 β1 0
]
β²
= [0 β1 11 0 β1
β1 1 0]
= β[0 1 β1
β1 0 11 β1 0
] = βπ΄
Step 2:
β π΄β² = βπ΄
Hence, the matrix A = [0 1 β1
β1 0 11 β1 0
] is a skew symmetric matrix.
8. For the matrix π΄ = [1 56 7
], verify that
(i) (π΄ + π΄β²) is a symmetric matrix [3 Marks]
(ii) (π΄ β π΄β²) is a skew symmetric matrix [3 Marks]
(i) Given:
π΄ = [1 56 7
]
Step 1:
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50 Practice more on Matrices www.embibe.com
β π΄β² = [1 56 7
]β²
= [1 65 7
]
Step 2:
Therefore, (π΄ + π΄β²) = [1 56 7
] + [1 65 7
] = [2 1111 14
]
Step 3:
(π΄ + π΄β²)β² = [2 1111 14
]β²
= [2 1111 14
]
β (π΄ + Aβ²)β² = (π΄ + π΄β²)
Hence the matrix (π΄ + π΄β²) is a symmetric matrix.
(ii) Given:
π΄ = [1 56 7
]
Step 1:
β π΄β² = [1 56 7
]β²
= [1 65 7
]
Step 2:
(π΄ β π΄β²) = [1 56 7
] β [1 65 7
] = [0 β11 0
]
Step 3:
β΄ (π΄ β π΄β²)β² = [0 β11 0
]β²
= [0 1
β1 0] = β [
0 β11 0
] = β(π΄ β π΄β²)
β (π΄ β π΄β²)β² = β(π΄ β π΄β²)
Hence the matrix (π΄ β π΄β²) is a skew symmetric matrix.
9. Find 1
2(π΄ + π΄β²) and
1
2(π΄ β π΄β²), when π΄ = [
0 π πβπ 0 πβπ βπ 0
] [3 Marks]
Solution:
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51 Practice more on Matrices www.embibe.com
Given that: π΄ = [0 π π
βπ 0 πβπ βπ 0
]
Step 1:
β π΄β² = [0 π π
βπ 0 πβπ βπ 0
]
β²
= [0 βπ βππ 0 βππ π 0
]
Step 2:
So, 1
2(π΄ + π΄β²) =
1
2([
0 π πβπ 0 πβπ βπ 0
] + [0 βπ βππ 0 βππ π 0
])
=1
2[0 0 00 0 00 0 0
] = [0 0 00 0 00 0 0
]
Step 3:
1
2(π΄ β π΄β²) =
1
2([
0 π πβπ 0 βπβπ βπ 0
] β [0 βπ βππ 0 βππ π 0
])
=1
2[
0 2π 2πβ2π 0 2πβ2π β2π 0
]=[0 π π
βπ 0 πβπ π 0
]
Hence, the required matrix of 1
2(π΄ + π΄β²) is [
0 0 00 0 00 0 0
] and 1
2(π΄ β π΄β²) is [
0 π πβπ 0 πβπ π 0
]
10. Express the following matrices as the sum of a symmetric and a skew symmetric matrix:
(i) [3 51 β1
] [4 Marks]
(ii) [6 β2 2β2 3 β12 β1 3
] [4 Marks]
(iii) [3 3 β1β2 β2 1β4 β5 2
] [4 Marks]
(iv) [1 5β1 2
] [4 Marks]
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52 Practice more on Matrices www.embibe.com
Solution:
(i) Given: π΄ = [3 51 β1
]
Step 1:
π΄ =1
2(π΄ + π΄β²) +
1
2(π΄ β π΄β²)
Consider, π =1
2(π΄ + π΄β²) and π =
1
2(π΄ β π΄β²) [
π
π Mark]
Step 2:
We have π΄β² = [3 15 β1
] [π
π Mark]
Step 3:
π =1
2(π΄ + π΄β²) =
1
2([
3 51 β1
] + [3 15 β1
])
=1
2[6 66 β2
] = [3 33 β1
] [π
π Mark]
Step 4:
πβ² = [3 33 β1
] = π
β π is a symmetric matrix. [π
π Mark]
Step 5:
π =1
2(π΄ β π΄β²) =
1
2([
3 51 β1
] β [3 15 β1
])
=1
2[0 4β4 0
] = [0 2β2 0
] [π
π Mark]
Step 6:
πβ² = [0 2β2 0
]β²
= [0 β22 0
] = β [0 2β2 0
] = βπ
β π is a skew symmetric matrix [π
π Mark]
Step 7:
Hence, π΄ = π + π = [3 33 β1
] + [0 2β2 0
]
π΄ = [3 51 β1
]
Thus, A is a sum of symmetric and skew symmetric matrix.
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53 Practice more on Matrices www.embibe.com
(ii) Given that: π΄ = [6 β2 2β2 3 β12 β1 3
]
Step 1:
π΄β² = [6 β2 2
β2 3 β12 β1 3
] [π
π Mark]
Step 2:
Therefore, π΄ =1
2(π΄ + π΄β²) +
1
2(π΄ β π΄β²)
πΏπt, π =1
2(π΄ + π΄β²) and π =
1
2(π΄ β π΄β²) [
π
π Mark]
Step 3:
π =1
2(π΄ + π΄β²) =
1
2([
6 β2 2β2 3 β12 β1 3
] + [6 β2 2
β2 3 β12 β1 3
])
=1
2[12 β4 4β4 6 β24 β2 6
] = [6 β2 2β2 3 β12 β1 3
] [π
π Mark]
Step 4:
πβ² = [6 β2 2
β2 3 β12 β1 3
]
β²
= [6 β2 2
β2 3 β12 β1 3
] = π [π
π Mark]
β π is a symmetric matrix
Step 5:
π =1
2(π΄ β π΄β²) =
1
2([
6 β2 2β2 3 β12 β1 3
] β [6 β2 2β2 3 β12 β1 3
])
=1
2[0 0 00 0 00 0 0
] = [0 0 00 0 00 0 0
] [π
π Mark]
Step 6:
πβ² = [0 0 00 0 00 0 0
]
β²
= [0 0 00 0 00 0 0
] = β [0 0 00 0 00 0 0
] = βπ [π
π Mark]
β π is a skew symmetric matrix.
Step 7:
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Hence, π΄ = π + π = [6 β2 2β2 3 β12 β1 3
] + [0 0 00 0 00 0 0
]
π΄ = [6 β2 2β2 3 β12 β1 3
]
Thus, A is a sum of symmetric and skew symmetric matrix.
(iii) Given: π΄ = [3 3 β1β2 β2 1β4 β5 2
]
Step 1:
π΄β² = [3 β2 β43 β2 β5
β1 1 2] [
π
π Mark]
Step 2:
Therefore, A =1
2(π΄ + π΄β²) +
1
2(π΄ β π΄β²)
Let, π =1
2(π΄ + π΄β²) and π =
1
2(π΄ β π΄β²) [
π
π Mark]
Step 3:
π =1
2(π΄ + π΄β²) =
1
2([
3 3 β1β2 β2 1β4 β5 2
] + [3 β2 β43 β2 β5β1 1 2
])
= β1
2[6 1 β51 β4 β4β5 β4 4
] =
[ 3
1
2β
5
21
2β2 β2
β5
2β2 2 ]
[π
π Mark]
Step 4:
πβ² =
[ 3
1
2β
5
21
2β2 β2
β5
2β2 2 ]
β²
=
[ 3
1
2β
5
21
2β2 β2
β5
2β2 2 ]
= π [π
π Mark]
β π is a symmetric matrix.
Step 5:
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π =1
2(π΄ β π΄β²) =
1
2([
3 3 β1β2 β2 1β4 β5 2
] β [3 β2 β43 β2 β5β1 1 2
]) [π
π Mark]
=1
2[0 5 3β5 0 6β3 β6 0
] =
[ 0
5
2
3
2
β5
20 3
β3
2β3 0]
Step 6:
πβ² =
[ 0
5
2
3
2
β5
20 3
β3
2β3 0]
β²
=
[ 0 β
5
2β
3
25
30 β3
3
23 0 ]
= β
[ 0 β
5
2β
3
25
20 β3
3
23 0 ]
= βπ
β π is a skew symmetric matrix. [π
π Mark]
Step 7:
Hence, π΄ = π + π =
[ 3
1
2β
5
21
2β2 β2
β5
2β2 2 ]
+
[
β
05
2
3
25
20 3
β3
2β3 0]
π΄ = [3 3 β1β2 β2 1β4 β5 2
]
Thus, A is a sum of symmetric and skew symmetric matrix.
(iv) Given that: π΄ = [1 5β1 2
]
Step 1:
π΄β² = [1 β15 2
] [π
π Mark]
Step 2:
Therefore, π΄ = β1
2(π΄ + π΄β²) +
1
2(π΄ β π΄β²)
Let, π =1
2(π΄ + π΄β²) and π =
1
2(π΄ β π΄β²) [
π
π Mark]
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Step 3:
π =1
2(π΄ + π΄β²)
=1
2([
1 5β1 2
] + [1 β15 2
])
=1
2[2 44 4
] = [1 22 2
] [π
π Mark]
Step 4:
πβ² = [1 22 2
] = [1 22 2
] = π
β π is a symmetric matrix [π
π Mark]
Step 5:
π =1
2(π΄ β π΄β²)
=1
2([
1 5β1 2
] β [1 β15 2
])
=1
2[0 6β6 0
] = [0 3β3 0
] [π
π Mark]
Step 6:
πβ² = [0 3β3 0
]β²
= [0 β33 0
] = β[0 3β3 0
] = βπ
β π is a skew symmetric matrix [π
π Mark]
Step 7:
Hence, π΄ = π + π = [1 22 2
] + [0 3β3 0
]
π΄ = [1 5β1 2
]
Thus, A is a sum of symmetric and skew symmetric matrix.
11. If π΄, π΅ are symmetric matrices of same order, then π΄π΅ β π΅π΄ is a [2 Marks]
(A) Skew symmetric matrix
(B) Symmetric matrix
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(C) Zero matrix
(D) Identity matrix
Solution:
Given:
A and B are symmetric matrices of same order
Step 1:
Let, (π΄π΅ β π΅π΄)β² = (π΄π΅)β² β (π΅π΄)β² [β΅ (π β π)β² = πβ² β πβ²] [π
π
Mark]
Step 2:
= π΅β²π΄β² β π΄β²π΅β² [β΅ (ππ)β = πβ²πβ²] [π
π Mark]
Step 3:
= π΅π΄ β AB [β΅Given: π΄β² = π΄, π΅β² = π΅] [π
π
Mark]
Step 4:
= β(π΄π΅ β π΅π΄)
β (π΄π΅ β π΅π΄)β² = β(π΄π΅ β π΅π΄)
Therefore, the matrix (AB β BA) is a skew symmetric matrix
Hence, the option (A) is correct. [π
π Mark]
12. If π΄ = [cosπΌ β sinπΌsinπΌ cosπΌ
], and π΄ + π΄β² = πΌ, then the value of πΌ is [2 marks]
(A) π
6
(B) π
3
(C) π
(D) 3π
2
Solution:
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(B)
Given that: π΄ + π΄β² = πΌ
π΄ = [cosπΌ β sinπΌsinπΌ cosπΌ
]
Step 1:
π΄β² = [cosπΌ sinπΌβsinπΌ cosπΌ
] [π
π Mark]
Step 2:
β [πππ πΌ βπ πππΌπ πππΌ πππ πΌ
] + [πππ πΌ π πππΌβπ πππΌ πππ πΌ
] = [1 00 1
]
β [2πππ πΌ 00 2πππ πΌ
] = [1 00 1
] [π
π Mark]
Step 3:
β 2 cos πΌ = 1
β cos πΌ =1
2
cos πΌ = cos 600 [β΅ cos 600 =1
2] [
π
π Mark]
Step 4:
On comparing angles, we get
πΌ = 60Β°
πΌ = 60Β° Γπ
180Β°
β πΌ =π
3
Hence, option (B) is correct. [π
π Mark]
EXERCISE 3.4
Using elementary transformations, find the inverse of each of the matrices, if it exists in Exercises 1 to 17.
1. [1 β12 3
] [3 Marks]
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Solution:
Let π΄ = [1 β12 3
],
Step 1:
We know that, π΄ = πΌπ΄, where πΌ = [1 00 1
].
β [1 β12 3
] = [1 00 1
]π΄ [π
π Mark]
Step 2:
β [1 β10 5
] = [1 0β2 1
] π΄ [applying π 2 βΆ π 2 β 2π 1] [π
π Mark]
Step 3:
β [1 β10 1
] = [1 0
β2
5
1
5
] π΄ [applying π 2 βΆ1
5π 2] [
π
π Mark]
Step 4:
β [1 00 1
] = [
3
5
1
5
β2
5
1
5
] π΄ [applying π 1 β π 1 + π 2] [π
π Mark]
Step 5:
I = [
3
5
1
5
β2
5
1
5
] π΄ [π
π Mark]
Step 6:
β΄ I = Aβ1A
Hence, π΄β1 = [
3
5
1
5
β2
5
1
5
] [π
π Mark]
2. [2 11 1
] [2 Marks]
Solution:
Let, π΄ = [2 11 1
],
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Step 1:
We know that, π΄ = πΌπ΄, where πΌ = [1 00 1
]
β [2 11 1
] [1 00 1
] π΄ [π
π Mark]
Step 2:
β [1 00 1
] = [1 β10 1
]π΄ [applying π 1 β π 1 β π 2] [π
π Mark]
Step 3:
β [1 00 1
] = [1 β1β1 2
]π΄ [applying π 2 β π 2 β π 1] [π
π Mark]
Step 4:
β΄ I = Aβ1A
Hence, π΄β1 = [1 β1
β1 2] [
π
π Mark]
3. [1 32 7
]
Solution:
Let, π΄ = [1 32 7
],
We know that, π΄ = πΌπ΄, where πΌ = [1 00 1
]
β [1 32 7
] = [1 00 1
] π΄
β [1 30 1
] = [1 0β2 1
]π΄ [applying π 2 β π 2 β 2π 1]
β [1 00 1
] = [7 β3β2 1
]π΄ [applying π 1 β π 1 β 3π 2]
β΄ I = Aβ1A
Hence, π΄β1 = [7 β3β2 1
]
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4. [2 35 7
] [3 Marks]
Solution:
Let, π΄ = [2 35 7
],
Step 1:
We know that, π΄ = πΌπ΄, where πΌ = [1 00 1
]
β [2 35 7
] = [1 00 1
] π΄ [π
π Mark]
Step 2:
β [1 25 7
] = [3 β10 1
]π΄ [applying π 1 β 3π 1 β π 2] [π
π Mark]
Step 3:
β [1 20 β3
] = [3 β1
β15 6]π΄ [applying π 2 β π 2 β 5π 1] [
π
π Mark]
Step 4:
β [1 20 1
] = [3 β15 β2
]π΄ [applying π 2 β β1
3π 2] [
π
π Mark]
Step 5:
β [1 00 1
] = [β7 35 β2
]π΄ [applying π 1 β π 1 β 2π 2] [π
π Mark]
Step 6:
β΄ I = Aβ1A
Hence, π΄β1 = [β7 35 β2
] [π
π Mark]
5. [2 17 4
] [3 Marks]
Solution:
Let, π΄ = [2 17 4
],
Step 1:
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We know that, π΄ = πΌπ΄, where πΌ = [1 00 1
]
β [2 17 4
] = [1 00 1
] π΄ [π
π Mark]
Step 2:
β [1 07 4
] = [4 β10 1
]π΄ [applying π 1 β 4π 1 β π 2] [π
π Mark]
Step 3:
β [1 00 4
] = [4 β1
β28 8]π΄ [applying π 2 β π 2 β 7π 1] [
π
π Mark]
Step 4:
β [1 00 1
] = [4 β1
β7 2]π΄ [applying π 2 β
1
4π 2] [
π
π Mark]
Step 5:
β΄ I = Aβ1A
Hence, π΄β1 = [4 β1
β7 2]1`
6. [2 51 3
] [3 marks]
Solution:
Let, π΄ = [2 51 3
],
Step 1:
We know that, π΄ = πΌπ΄, where πΌ = [1 00 1
]
β [2 51 3
] = [1 00 1
] π΄ [π
π Mark]
Step 2:
β [1 21 3
] = [1 β10 1
]π΄ [applying π 1 β π 1 β π 2] [π
π Mark]
Step 3:
β [1 20 1
] = [1 β1
β1 2]π΄ [applying π 2 β π 2 β π 1] [
π
π Mark]
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Step 4:
β [1 00 1
] = [3 β5
β1 2]π΄ [applying π 1 β π 1 β 2π 2] [
π
π Mark]
Step 5:
β΄ I = Aβ1A
Hence, π΄β1 = [3 β5
β1 2]
7. [3 15 2
] [3 Marks]
Solution:
Let, π΄ = [3 15 2
],
Step 1:
We know that, π΄ = πΌπ΄, where πΌ = [1 00 1
].
β [3 15 2
] = [1 00 1
] π΄ [π
π Mark]
Step 2:
β [1 05 2
] = [2 β10 1
]π΄ [Applying π 1 β 2π 1 β π 2] [π
π Mark]
Step 3:
β [1 00 2
] = [2 β1
β10 6]π΄ [Applying π 2 β 2π 2 β 5π 1] [
π
π Mark]
Step 4:
β [1 00 2
] = [2 β1
β5 3]π΄ [Applying π 2 β
1
2π 2] [
π
π Mark]
Step 5:
β΄ I = Aβ1A
Hence, π΄β1 = [2 β1
β5 3]
Class-XII-Maths Matrices
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8. [4 53 4
] [3 Marks]
Solution:
Let, π΄ = [4 53 4
],
Step 1:
We know that, π΄ = πΌπ΄, where πΌ = [1 00 1
].
β [4 53 4
] = [1 00 1
] π΄ [π
π Mark]
Step 2:
β [1 13 4
] = [1 β10 1
]π΄ [applying π 1 β π 1 β π 2] [π
π Mark]
Step 3:
β [1 10 1
] = [1 β1
β3 4]π΄ [applying π 2 β π 2 β 3π 1] [
π
π Mark]
Step 4:
β [1 00 1
] = [4 β5
β3 4]π΄ [applying R1 β π 1 β R2] [
π
π Mark]
Step 5:
β΄ I = Aβ1A
Hence, π΄β1 = [4 β5
β3 4]
9. [3 102 7
] [3 Marks]
Solution:
Let, π΄ = [3 102 7
],
Class-XII-Maths Matrices
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Step 1:
We know that, π΄ = πΌπ΄, where πΌ = [1 00 1
].
β [3 102 7
] = [1 00 1
] π΄ [π
π Mark]
Step 2:
β [1 32 7
] = [1 β10 1
]π΄ [applying π 1 β π 1 β π 2] [π
π Mark]
Step 3:
β [1 30 1
] = [1 β1
β2 3]π΄ [applying π 2 β π 2 β 2π 1] [
π
π Mark]
Step 4:
β [1 00 1
] = [7 β10
β2 3]π΄ [applying π 1 β π 1 β 3π 2] [
π
π Mark]
Step 5:
β΄ I = Aβ1A
Hence, π΄β1 = [7 β10
β2 3]
10. [3 β1β4 2
] [3 Marks]
Solution:
Let, π΄ = [3 β1β4 2
],
Step 1:
We know that, π΄ = πΌπ΄, where πΌ = [1 00 1
].
β [3 β1
β4 2] = [
1 00 1
] π΄ [π
π Mark]
Step 2:
β [1 1
β4 2] = [
3 20 1
]π΄ [applying π 1 β 3π 1 + 2π 2] [π
π Mark]
Step 3:
β [1 10 6
] = [3 212 9
] π΄ [applying π 2 β π 2 + 4π 1] [π
π Mark]
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Step 4:
β [1 10 1
] = [3 2
23
2
] π΄ [applying π 2 β1
6π 2] [
π
π Mark]
Step 5:
β [1 00 1
] = [1
1
2
23
2
] π΄ [applying π 1 β π 1 β π 2] [π
π Mark]
Step 6:
β΄ I = Aβ1A
Hence, π΄β1 = [1
1
2
23
2
]
11. [2 β61 β2
] [3 Marks]
Solution:
Let, π΄ = [2 β61 β2
],
Step 1:
We know that, π΄ = πΌπ΄, where πΌ = [1 00 1
].
β [2 β61 β2
] = [1 00 1
]π΄ [π
π Mark]
Step 2:
β [1 β41 β2
] = [1 β10 1
]π΄ [applying R1 β π 1-R2] [π
π Mark]
Step 3:
β [1 β40 2
] = [1 β1
β1 2]π΄ [applying R2 β π 2-R1] [
π
π Mark]
Step 4:
β [1 β40 1
] = [1 β1
β1
21 ]π΄ [applying π 2 β
1
2π 2] [
π
π Mark]
Step 5:
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β [1 00 1
] = [β1 3
β1
21]π΄ [applyingR1 β π 1-4R2] [
π
π Mark]
Step 6:
β΄ I = Aβ1A
Hence, π΄β1 = [β1 3
β1
21]
12. [6 β3β2 1
] [2 Marks]
Solution:
Let, π΄ = [6 β3β2 1
],
Step 1:
We know that, π΄ = πΌπ΄, where πΌ = [1 00 1
].
β [6 β3
β2 1] = [
1 00 1
] π΄ [π
π Mark]
Step 2:
β [6 β30 0
] = [1 03 1
]π΄ [applying π 2 β π 2 + 3π 1] [π
π Mark]
Step 3:
Since, on left hand side of the matrix all the elements of second row is zero.
Hence, π΄β1 does not exist.
13. [2 β3β1 2
] [3 marks]
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Solution:
Let, π΄ = [2 β3β1 2
],
Step 1:
We know that, π΄ = πΌπ΄ where πΌ = [1 00 1
].
β [2 β3
β1 2] = [
1 00 1
] π΄ [π
π Mark]
Step 2:
β [1 β1
β1 2] = [
1 10 1
] π΄ [applying π 1 β π 1 + π 2] [π
π Mark]
Step 3:
β [1 β10 1
] = [1 11 2
]π΄ [applying π 2 β π 2 + π 1] [π
π Mark]
Step 4:
β [1 00 1
] = [2 31 2
] π΄ [applying π 1 β π 1 + π 2] [π
π Mark]
Step 5:
β΄ I = Aβ1A
Hence, π΄β1 = [2 31 2
]
14. [2 14 2
] [2 marks]
Solution:
Let π΄ = [2 14 2
],
Step 1:
We know that, π΄ = πΌπ΄, where πΌ = [1 00 1
].
β [2 14 2
] = [1 00 1
] π΄ [π
π Mark]
Step 2:
Class-XII-Maths Matrices
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β [2 10 0
] = [1 0
β2 1]π΄ [ applying π 2 β π 2 β 2π 1] [
π
π Mark]
Step 3:
Since, on left hand side of the matrix all the elements of second row is zero.
Hence π΄β1 does not exist.
15. [2 β3 32 2 33 β2 2
] [5 Marks]
Solution:
Let, π΄ = [2 β3 32 2 33 β2 2
],
Step 1:
We know that, π΄ = πΌπ΄, where πΌ = [1 0 00 1 00 0 1
].
β [2 β3 32 2 33 β2 2
] = [1 0 00 1 00 0 1
]π΄ [π
π Mark]
Step 2:
β [1 1 42 2 33 β2 2
] = [1 1 β10 1 00 0 1
]π΄ [applying π 1 β π 1 + π 2 β π 3] [π
π Mark]
Step 3:
β [1 1 40 0 β53 β2 2
] = [1 1 β1
β2 β1 20 0 1
]π΄ [applying π 2 β π 2 β 2π 1] [π
π Mark]
Step 4:
β [1 1 40 0 β50 β5 β10
] = [1 1 β1
β2 β1 2β3 β3 4
]π΄ [applying π 3 β π 3 β 3π 1] [π
π Mark]
Step 5:
β [1 1 40 β5 β100 0 β5
] = [1 1 β1
β3 β3 4β2 β1 2
]π΄ [applying π 2 β π 3] [π
π Mark]
Step 6:
Class-XII-Maths Matrices
70 Practice more on Matrices www.embibe.com
β [1 1 40 1 20 0 1
] = [
1 1 β13
5
3
5β
4
52
5
1
5β
2
5
] π΄ [applying π 2 β β1
5π 2, π 3 β β
1
5π 3] [
π
π Mark]
Step 7:
β [1 1 40 1 00 0 1
] = [
1 1 β1
β1
5
1
50
2
5
1
5β
2
5
]π΄ [applying π 2 β π 2 β 2π 3] [π
π Mark]
Step 8:
β [1 0 40 1 00 0 1
] =
[
6
5
4
5β1
β1
5
1
50
2
5
1
5β
2
5]
π΄ [applying π 1 β π 1 β π 2] [π
π Mark]
Step 9:
β [1 0 00 1 00 0 1
] =
[ β
2
5 0
3
5
β1
5
1
50
2
5
1
5β
2
5]
π΄ [applying R1 β π 1 β 4R3] [π
π Mark]
Step 10:
β΄ I = Aβ1A
Hence,π΄β1 =
[ β
2
50
3
5
β1
5
1
50
2
5
1
5β
2
5]
[π
π Mark]
16. [1 3 β2
β3 0 β52 5 0
] [5 Marks]
Solution:
Let, π΄ = [1 3 β2
β3 0 β52 5 0
],
Class-XII-Maths Matrices
71 Practice more on Matrices www.embibe.com
Step 1:
We know that, π΄ = πΌπ΄, where πΌ = [1 0 00 1 00 0 1
]
β [1 3 β2
β3 0 β52 5 0
] = [1 0 00 1 00 0 1
]π΄ [π
π Mark]
Step 2:
β [1 3 β20 9 β112 5 0
] = [1 0 03 1 00 0 1
]π΄ [applying π 2 β π 2 + 3π 1] [π
π Mark]
Step 3:
β [1 3 β20 9 β110 β1 4
] = [1 0 03 1 0
β2 0 1]π΄ [applying R3 β π 3-2R1] [
π
π Mark]
Step 4:
β [1 3 β20 9 β110 0 25
] = [1 0 03 1 0
β15 1 9]π΄ [applying R3 β 9R3+R2] [
π
π Mark]
Step 5:
β [1 3 β20 9 β110 0 1
] = [
1 0 03 1 0
β3
5
1
25
9
25
] π΄ [applying π 3 β1
25π 3] [
π
π Mark]
Step 6:
β [1 3 β20 9 00 0 1
] = [
1 0 0
β18
5
36
25
99
25
β3
5
1
25
9
25
] π΄ [applying π 2 β π 2 + 11π 3] [π
π Mark]
Step 7:
β [1 3 β20 1 00 0 1
] = [
1 0 0
β2
5
4
25
11
25
β3
5
1
25
9
25
]π΄ [applying π 2 β1
9π 2] [
π
π Mark]
Step 8:
β [1 3 00 1 00 0 1
] =
[ β
1
5
2
25
18
25
β2
5
4
25
11
25
β3
5
1
25
9
25]
π΄ [applying R1 β π 1+2R3] [π
π Mark]
Step 9:
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β [1 0 00 1 00 0 1
] =
[ 1 β
2
5β
3
5
β2
5
4
25
11
25
β3
5
1
25
9
25 ]
π΄ [applying R1 β π 1-3R2] [π
π Mark]
Step 10:
β΄ I = Aβ1A
Hence, π΄β1 =
[ 1 β
2
5β
3
5
β2
5
4
25
11
25
β3
5
1
25
9
25 ]
[π
π Mark]
17. [2 0 β15 1 00 1 3
] [5 Marks]
Solution:
Let π΄ = [2 0 β15 1 00 1 3
],
Step 1:
We know that, π΄ = πΌπ΄, where πΌ = [1 0 00 1 00 0 1
].
β [2 0 β15 1 00 1 3
] = [1 0 00 1 00 0 1
]π΄ [π
π Mark]
Step 2:
β [1 β1 β35 1 00 1 3
] = [3 β1 00 1 00 0 1
]π΄ [applying R1 β 3R1-R2] [π
π Mark]
Step 3:
β [1 β1 β30 6 150 1 3
] = [3 β1 0
β15 6 00 0 1
]π΄ [applyingR2 β π 2-5R1] [π
π Mark]
Step 4:
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β [1 β1 β30 6 150 0 3
] = [3 β1 0
β15 6 015 β6 6
]π΄ [applyingR3 β 6R3-R2] [π
π Mark]
Step 5:
β [1 β1 β30 6 150 0 1
] = [3 β1 0
β15 6 05 β2 2
]π΄ [applying π 3 β1
3π 3] [
π
π Mark]
Step 6:
β [1 β1 β30 6 00 0 1
] = [3 β1 0
β90 36 β305 β2 2
]π΄ [applying R2 β π 2-15R3] [π
π Mark]
Step 7:
β [1 β1 β30 1 00 0 1
] = [3 β1 0
β15 6 β55 β2 2
]π΄ [applying π 2 β1
6π 2] [
π
π Mark]
Step 8:
β [1 β1 00 1 00 0 1
] = [18 β7 6
β15 6 β55 β2 2
]π΄ [applying R1 β π 1+3R3] [π
π Mark]
Step 9:
β [1 0 00 1 00 0 1
] = [3 β1 1
β15 6 β55 β2 2
]π΄ [applying R1 β π 1+R2] [π
π Mark]
Step 10:
β΄ I = Aβ1A
Hence, π΄β1 = [3 β1 1
β15 6 β55 β2 2
] [π
π Mark]
18. Matrices π΄ and π΅ will be inverse of each other only if [2 Marks]
(A) π΄π΅ = π΅π΄
(B) π΄π΅ = π΅π΄ = 0
(C) π΄π΅ = 0, π΅π΄ = πΌ
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(D) π΄π΅ = π΅π΄ = πΌ
Solution:
Given:
A and B will be inverse of each other.
π΄β1 = π΅ and π΅β1 = π΄
Step 1:
We know that,
π΄π΄β1 = πΌ [β΅ Aβ1 = B]
π΄π΅ = πΌ
Step 2:
π΅π΅β1 = πΌ [β΅ Bβ1 = A]
π΅π΄ = πΌ
So, π΄π΅ = π΅π΄ = πΌ.
Hence, option (D) is correct.
Miscellaneous Exercise on Chapter 3
1. Let π΄ = [0 10 0
], show that (ππΌ + ππ΄)π = πππΌ + πππβ1ππ΄, where πΌ is the identity matrix
of order 2 and π β π. [5 marks]
Solution:
Given: π΄ = [0 10 0
]
To Prove:
(ππΌ + ππ΄)π = πππΌ + πππβ1ππ΄
Proof: The proof is by mathematical induction [π
π Mark]
Step 1:
Consider, π(π): (ππ + ππ΄)π = πππΌ + πππβ1ππ΄ [π
π Mark]
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Step 2:
β΄ P(1): (ππΌ + ππ΄)1 = ππΌ + ππ΄
Hence, the result is true for π = 1. [π
π Mark]
Step 3:
Let the result be true for π = π,
So, π(π): (ππ + ππ΄)π = πππΌ + πππβ1ππ΄ [π
π Mark]
Step 4:
We must prove that it is true for π = π + 1 also.
(i.e.) π(π + 1): (ππ + ππ΄)π+1 = ππ+1πΌ + (π + 1)ππππ΄ [π
π Mark]
Step 5:
By taking LHS, we get
LHS= (ππΌ + ππ΄)π+1
= (ππΌ + ππ΄)π(ππΌ + ππ΄) [π
π Mark]
Step 6:
= (πππΌ + πππβ1ππ΄)(ππΌ + ππ΄) [β΅ (ππΌ + ππ΄)π+1 = πππΌ + πππβ1ππ΄]
[π
π Mark]
Step 7:
= ππ+1πΌ2 + πππΌππ΄ + ππππΌππ΄ + πππβ1π2π΄2 [π
π Mark]
Step 8:
= ππ+1πΌ + (π + 1)ππππ΄ [β΅ π΄2 = π΄ β π΄ = [0 10 0
] [0 10 0
] = [0 00 0
] = 0]
πΏπ»π = π π»π
Therefore, the result is true for π = π + 1.
Hence, by the Principle of Mathematical Induction, (ππΌ + ππ΄)π = πππΌ + πππβ1ππ΄ is true for all-natural numbers π.
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2. If π΄ = [1 1 11 1 11 1 1
], prove that π΄π = [3πβ1 3πβ1 3πβ1
3πβ1 3πβ1 3πβ1
3πβ1 3πβ1 3πβ1
], π β π. [5 Marks]
Solution:
Given:
π΄ = [1 1 11 1 11 1 1
]
To Prove:
π΄π = [3πβ1 3πβ1 3πβ1
3πβ1 3πβ1 3πβ1
3πβ1 3πβ1 3πβ1
], π β π
Proof: The proof is by mathematical induction. [π
π Mark]
Step 1:
Consider, π(π): π΄π = [3πβ1 3πβ1 3πβ1
3πβ1 3πβ1 3πβ1
3πβ1 3πβ1 3πβ1
], [π
π Mark]
Step 2:
Therefore, π(1): π΄1 = [30 30 30
30 30 30
30 30 30
] = [1 1 11 1 11 1 1
] = π΄
Hence, the result is true for π = 1. [π
π Mark]
Step 3:
Let the result is true for π = π,
So, π(π): π΄π = [3πβ1 3πβ1 3πβ1
3πβ1 3πβ1 3πβ1
3πβ1 3πβ1 3πβ1
] [π
π Mark]
Step 4:
Now,
We have to prove that it is true for π = π + 1 also.
(i.e.) π(π + 1) : π΄π+1 = [3π 3π 3π
3π 3π 3π
3π 3π 3π
] [π
π Mark]
Step 5:
By taking LHS, we get
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= π΄π+1 = π΄ππ΄
= [3πβ1 3πβ1 3πβ1
3πβ1 3πβ1 3πβ1
3πβ1 3πβ1 3πβ1
] [1 1 11 1 11 1 1
] [π
π Mark]
Step 6:
= [3πβ1 + 3πβ1 + 3πβ1 3πβ1 + 3πβ1 + 3πβ1 3πβ1 + 3πβ1 + 3πβ1
3πβ1 + 3πβ1 + 3πβ1 3πβ1 + 3πβ1 + 3πβ1 3πβ1 + 3πβ1 + 3πβ1
3πβ1 + 3πβ1 + 3πβ1 3πβ1 + 3πβ1 + 3πβ1 3πβ1 + 3πβ1 + 3πβ1
] [π
π Mark]
Step 7:
= [3. 3πβ1 3. 3πβ1 3. 3πβ1
3. 3πβ1 3. 3πβ1 3. 3πβ1
3. 3πβ1 3. 3πβ1 3. 3πβ1
] [π
π Mark]
Step 8:
= [3π 3π 3π
3π 3π 3π
3π 3π 3π
]
πΏπ»π = π π»π
Therefore, the result is true for π = π + 1.
Hence, by the Principle of Mathematical Induction, π΄π = [3πβ1 3πβ1 3πβ1
3πβ1 3πβ1 3πβ1
3πβ1 3πβ1 3πβ1
] is true
for all natural numbers π.
3. If π΄ = [3 β41 β1
], then prove that π΄π = [1 + 2π β4ππ 1 β 2π
], where π is any positive
integer. [5 Marks]
Solution:
Given:
π΄ = [3 β41 β1
]
To Prove:
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π΄π = [1 + 2π β4ππ 1 β 2π
],
Proof: The proof is by mathematical induction. [π
π Mark]
Step 1:
Consider, π(π) : π΄π = [1 + 2π β4π
π 1 β 2π] [
π
π Mark]
Step 2:
Therefore, π(1) : π΄1 = [1 + 2(1) β4(1)
1 1 β 2(1)] = [
3 β41 β1
] = π΄
Hence, the result is true for π = 1. [π
π Mark]
Step 3:
Let the result is true for π = π, therefore,
π(π) : π΄π = [1 + 2π β4π
π 1 β 2π] [
π
π Mark]
Step 4:
We have to prove that it is true for π = π + 1 also.
(i.e.) π(π + 1) : π΄π+1 = [1 + 2(π + 1) β4(π + 1)
π + 1 1 β 2(π + 1)] [
π
π Mark]
Step 5:
By taking LHS, we get
LHS= π΄π+1 = π΄ππ΄
= [1 + 2π β4π
π 1 β 2π] [
3 β41 β1
] [π
π Mark]
Step 6:
= [3 + 6π β 4π β4 β 8π + 4π3π + 1 β 2π β4π β 1 + 2π
] [π
π Mark]
Step 7:
= [1 + (2π + 2) β4π β 4
π + 1 1 β (2π + 2)] [
π
π Mark]
Step 8:
= [1 + 2(π + 1) β4(π + 1)
π + 1 1 β 2(π + 1)] =RHS
πΏπ»π = π π»π
Therefore, the result is true for π = π + 1.
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Hence, by the Principle of Mathematical Induction, π΄π = [1 + 2π β4π
π 1 β 2π] is true for
any positive integer numbers π.
4. If π΄ and π΅ are symmetric matrices, prove that π΄π΅ β π΅π΄ is a skew symmetric matrix. [2 Marks]
Solution:
Given: π΄ and π΅ are symmetric matrices
β΄ π΄β² = π΄ and π΅β² = π΅
Step 1:
Now,
(π΄π΅ β π΅π΄)β² = (π΄π΅)β² β (π΅π΄)β² [β΅ (π β π)β² = πβ² β πβ²] [π
π Mark]
Step 2:
= π΅β²π΄β² β π΄β²π΅β² [β΅ (π΄π΅)β² = π΅β²π΄β²] [π
π Mark]
Step 3:
= π΅π΄ β π΄π΅ [Given π΄β² = π΄,π΅β² = π΅] [π
π Mark]
Step 4:
= β(π΄π΅ β π΅π΄)
β (π΄π΅ β π΅π΄)β² = β(π΄π΅ β π΅π΄), [π
π Mark]
Therefore, π΄π΅βπ΅π΄ is a skew symmetric matrix.
Hence proved.
5. Show that the matrix π΅β²π΄π΅ is symmetric or skew symmetric according as π΄ is symmetric or skew symmetric. [π Marks]
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Solution:
Step 1:
If π΄ is a symmetric matrix, then Aβ² = π΄
Now, (π΅β²π΄π΅)β² = (π΄π΅)β²(π΅β²)β² [β΅ (π΄π΅)β² = π΅β²π΄β²] [π
π Mark]
Step 2:
= (π΄π΅)β²π΅ [β΅ (π΅β²)β² = π΅] [π
π Mark]
Step 3:
= π΅β²π΄β²π΅ [β΅ (π΄π΅)β² = π΅β²π΄β²] [π
π Mark]
Step 4:
= π΅β²π΄π΅ [Given Aβ² = π΄]
β (π΅β²π΄π΅)β² = Bβ²π΄π΅,
Therefore, the matrix π΅β²π΄π΅ is also symmetric matrix. [π
π Mark]
Step 5:
If π΄ is skew symmetric matrix, then π΄β² = βπ΄
Here, (π΅β²π΄π΅)β² = (π΄π΅)β²(π΅β²)β² [β΅ (π΄π΅)β² = π΅β²π΄β²] [π
π Mark]
Step 6:
= (π΄π΅)β²π΅ [β΅ (π΅β²)β² = B] [π
π Mark]
Step 7:
= Bβ²Aβ²B [β΅ (π΄π΅)β² = π΅β²π΄β²] [π
π Mark]
Step 8:
= βπ΅β²π΄π΅ [β΅ Given π΄β² = βπ΄]
β (π΅β²π΄π΅)β² = βBβ²π΄B,
Hence, the matrix π΅β²π΄π΅ is also a skewβsymmetric matrix. [π
π Mark]
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6. Find the values of π₯, π¦, π§ if the matrix π΄ = [0 2π¦ π§π₯ π¦ βπ§π₯ βπ¦ π§
] satisfy the equation π΄β²π΄ = πΌ.
[4 Marks]
Solution:
Given:
π΄ = [0 2π¦ π§π₯ π¦ βπ§π₯ βπ¦ π§
]
π΄β²π΄ = πΌ
Step 1:
β [0 2π¦ π§π₯ π¦ βπ§π₯ βπ¦ π§
] [0 π₯ π₯2π¦ π¦ βπ¦π§ βπ§ π§
] = [1 0 00 1 00 0 1
] [π
π Mark]
Step 2:
β [
0 + 4π¦2 + π§2 0 + 2π¦2 β π§2 0 β 2π¦2 + π§2
0 + 2π¦2 β π§2 π₯2 + π¦2 + π§2 π₯2 β π¦2 β π§2
0 β 2π¦2 + π§2 π₯2 β π¦2 β π§2 π₯2 + π¦2 + π§2
] = [1 0 00 1 00 0 1
]
Step 3:
If two matrices are equal, then their corresponding elements are also equal.
So, on comparing the corresponding elements, we get
4π¦2 + π§2 = 1
2π¦2 β π§2 = 0
π₯2 + π¦2 + π§2 = 1
Step 4:
On solving we get π₯ = Β±1
β2, π¦ = Β±
1
β6 and π§ = Β±
1
β3 [1
π
π Marks]
Hence, the values of π₯ = Β±1
β2, π¦ = Β±
1
β6 and π§ = Β±
1
β3
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7. For what values of π₯: [1 2 1] [1 2 02 0 11 0 2
] [02π₯] = π? [3 Marks]
Solution:
Given that:[1 2 1] [1 2 02 0 11 0 2
] [02π₯] = π
Step 1:
β [1 + 4 + 1 2 + 0 + 0 0 + 2 + 2] [02π₯] = π
Step 2:
β [6 2 4] [02π₯] = π [
π
π Mark]
Step 3:
[6(0) + 2(2) + 4(π₯)] = 0
β [0 + 4 + 4π₯] = [0] [π
π Mark]
Step 4:
β 4 + 4π₯ = 0 [π
π Mark]
Step 5:
Upon solving the above obtained equation, we get:
β π₯ = β1
Hence, the required value of π₯ is β 1 [π
π Mark]
8. If π΄ = [3 1β1 2
], show that π΄2 β 5π΄ + 7πΌ = 0. [3 Marks]
Solution:
Given:
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π΄ = [3 1β1 2
]
Step 1:
By taking LHS= π΄2 β 5π΄ + 7πΌ
= [3 1β1 2
] [3 1β1 2
] β 5 [3 1β1 2
] + 7 [1 00 1
]
Step 2:
= [9 β 1 3 + 2
β3 β 2 β1 + 4] β [
15 5β5 10
] + [7 00 7
]
= [8 5
β5 3] β [
15 5β5 10
] + [7 00 7
]
Step 3:
= [8 β 15 + 7 5 β 5 + 0β5 + 5 + 0 3 β 10 + 7
]
= [0 00 0
] = π
β΄ πΏπ»π = π π»π
π΄2 β 5π΄ + 7πΌ = 0
Hence it is proved.
9. Find π₯, if [π₯ β 5 β 1] [1 0 20 2 12 0 3
] [π₯41] = 0 [3 marks]
Solution:
Given: [π₯ β 5 β 1] [1 0 20 2 12 0 3
] [π₯41] = 0
Step 1:
[π₯ β 5 β 1] [1 0 20 2 12 0 3
] [π₯41] = 0
β [π₯ + 0 β 2 0 β 10 + 0 2π₯ β 5 β 3] [π₯41] = π
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β [π₯ β 2 β 10 2π₯ β 8] [π₯41] = π
Step 2:
β [π₯2 β 2π₯ β 40 + 2π₯ β 8] = [0]
Step 3:
β π₯2 = 48
β π₯ = Β±4β3
Hence, the required value of π₯ is Β± 4β3
10. A manufacturer produces three products π₯, π¦, π§ which he sells in two markets. Annual sales are indicated below:
Market Products
I 10,000 2,000 18,000
II 6,000 20,000 8,000
(a) If unit sale prices of π₯, π¦ and π§ are βΉ 2.50, βΉ 1.50 and βΉ 1.00, respectively, find the total revenue in each market with the help of matrix algebra. [3 Marks]
(b) If the unit costs of the above three commodities are βΉ 2.00, βΉ 1.00 and 50 paise respectively. Find the gross profit. [3 Marks]
Solution:
(a) Step 1:
If unit sale prices of π₯, π¦ and π§ are βΉ2.50, βΉ1.50 and βΉ1.00, then
Products Selling price
Market I
Market II
π₯ π¦ π§
[10000 2000 180006000 20000 8000
]
[βΉ2.50βΉ1.50βΉ1.00
]
Step 2:
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Total revenue of each market = Total product Γ Unit selling price
= [10000 2000 180006000 20000 8000
] [βΉ2.50βΉ1.50βΉ1.00
]
Step 3:
= [βΉ25000 + βΉ3000 + βΉ18000βΉ15000 + βΉ30000 + βΉ8000
] = [βΉ46000βΉ53000
]
Thus, the revenue of market I is βΉ46,000 and that of the market II is βΉ53,000.
(b) Step 1:
If the unit costs of the three commodities are βΉ2.00, βΉ1.00 and 50 paise, then
Products Selling price
Market I
Market II
π₯ π¦ π§
[10000 2000 180006000 20000 8000
]
[βΉ2.50βΉ1.50βΉ0.00
]
Step 2:
Gross profit from each market = Total product Γ Unit selling price
= [10000 2000 180006000 20000 8000
] [βΉ2.50βΉ1.50βΉ1.00
]
= [βΉ20000 + βΉ2000 + βΉ9000βΉ12000 + βΉ20000 + βΉ4000
] = [βΉ31000βΉ36000
]
Step 3:
Hence, the total revenue from market I is βΉ46,000 and that of the market II is βΉ31,000.
Therefore, the gross profit of market I=Revenue β Cost
= βΉ46000 β βΉ31000 = βΉ15000
Therefore, the gross profit of market II = βΉ53000 β βΉ36000 = βΉ17000
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Hence, the total gross profit from market I is βΉ15,000 and that of the market II is βΉ17,000.
11. ind the matrix π so that π [1 2 34 5 6
] = [β7 β8 β92 4 6
] [5 Marks]
Solution:
Given:
π [1 2 34 5 6
]2Γ3
= [β7 β8 β92 4 6
]2Γ3
Step 1:
Let, π = [π ππ π
] [π
π Mark]
Step 2:
Therefore, π [1 2 34 5 6
] = [β7 β8 β92 4 6
]
β [π ππ π
] [1 2 34 5 6
] = [β7 β8 β92 4 6
] [π
π Mark]
Step 3:
β [π + 4π 2π + 5π 3π + 6ππ + 4π 2π + 5π 3π + 6π
] = [β7 β8 β92 4 6
]
Step 4:
If two matrices are equal, then their corresponding elements are also equal.
So, on comparing the corresponding elements, we get
π + 4π = β7 [π
π Mark]
2π + 5π = β8 [π
π Mark]
π + 4π = 2 [π
π Mark]
2π + 5π = 4 [π
π Mark]
Step 5:
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On solving, we get π = 1, π = β2, π = 2 and π = 0
Hence, π = [1 β22 0
]
12. π΄ and π΅ are square matrices of the same order such that π΄π΅ = π΅π΄, then prove by induction that π΄π΅π = π΅ππ΄. Further, prove that (π΄π΅)π = π΄ππ΅π for all π β π. Choose the correct answer in the following questions: [5 marks]
Solution:
Given:
π΄π΅ = π΅π΄
To prove:
π΄π΅π = π΅ππ΄ and
(π΄π΅)π = π΄ππ΅π for all π β π
Proof: The proof is by mathematical induction.
Step 1:
Consider, π(π): π΄π΅π = π΅ππ΄, [π
π Mark]
Step 2:
Therefore, π(1): π΄π΅ = π΅π΄
Hence the result is true for π = 1. [π
π Mark]
Step 3:
Let it be true for π = π,
so, π(π): π΄π΅π = π΅ππ΄ [π
π Mark]
Step 4:
Now we have to prove that it is true for π = π + 1 also.
(i.e.) π(π + 1): π΄π΅π+1 = Bk+1π΄ [π
π Mark]
Step 5:
By taking LHS, we get
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LHS= π΄π΅π+1
= π΄π΅ππ΅
= π΅ππ΄π΅ [β΅ π΄π΅π = π΅ππ΄]
= π΅ππ΅π΄ [β΅ π΄π΅ = π΅π΄]
= π΅π+1π΄ =RHS
Hence, the result is true for π = π + 1, [π
π Mark]
Therefore, by the Principle of Mathematical Induction π΄π΅π = π΅ππ΄ is true for all natural numbers π.
Step 6:
If (π΄π΅)π = π΄ππ΅π
Now,
Consider π(π): (π΄π΅)n = π΄ππ΅π therefore, π(1): (π΄π΅)1 = π΄1π΅1
Hence the result is true for π = 1. [π
π Mark]
Step 7:
Let it is true for π = π,
so, π(π): (π΄π΅)π = π΄ππ΅π [π
π Mark]
Step 8:
Now we have to prove that it is true for π = π + 1 also.
(i.e.) π(π + 1): (π΄π΅)π+1 = π΄π+1π΅π+1 [π
π Mark]
Step 9:
By taking LHS, we get
LHS = (π΄π΅)π+1
= (π΄π΅)kπ΄π΅
= π΄ππ΅ππ΄π΅ [β΅ (π΄π΅)k = π΄ππ΅π] [π
π Mark]
Step 10:
= π΄ππ΄π΅ππ΅ [β΅ π΄π΅ = π΅π΄]
= π΄π+1π΅π+1 =RHS
Hence, the result is true for π = π + 1.
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Therefore, by the Principle of Mathematical Induction (π΄π΅)π = π΄ππ΅π is true for all
natural numbers π. [π
π Mark]
13. If π΄ = [πΌ π½πΎ βπΌ
] is such that π΄2 = πΌ, then [2 Marks]
(A) 1 + πΌ2 + π½πΎ = 0
(B) 1 β πΌ2 + π½πΎ = 0
(C) 1 β πΌ2 β π½πΎ = 0
(D) 1 + πΌ2 β π½πΎ = 0
Solution:
Given that: π΄2 = πΌ
Step 1:
β [πΌ π½πΎ βπΌ
] [πΌ π½πΎ βπΌ
] = [1 00 1
]
β [πΌ2 + π½πΎ πΌπ½ β π½πΌ
πΌπΎ β πΌπΎ π½πΎ + πΌ2 ] = [1 00 1
]
Step 2:
If two matrices are equal, then their corresponding elements are also equal.
So, on comparing the corresponding elements, we get
πΌ2 + π½πΎ = 1
1 β πΌ2 β π½πΎ = 0
Hence the option (C) is correct.
14. If the matrix π΄ is both symmetric and skew symmetric, then [2 Marks]
(A) π΄ is a diagonal matrix
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(B) π΄ is a zero matrix
(C) π΄ is a square matrix
(D) None of these
Solution:
Given:
Matrix π΄ is both symmetric and skew symmetric
Step 1:
We know that only a zero matrix is always both symmetric and skew symmetric.
Proof:
β΄ π΄β² = π΄ and π΄β² = βπ΄
Step 2:
On comparing both the equations, we get
β π΄ = βπ΄
π΄ + π΄ = π
2π΄ = π
π΄ = π
Hence, the option (B) is correct.
15. If π΄ is square matrix such that π΄2 = π΄, then (πΌ + π΄)3 β 7 π΄ is equal to [2 Marks]
(A) π΄
(B) πΌ β π΄
(C) πΌ
(D) 3π΄
Solution:
Given:
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π΄2 = π΄
Step 1:
(πΌ + π΄)3 β 7π΄
= πΌ3 + π΄3 + 3πΌ2π΄ + 3πΌπ΄2 β 7π΄ [π
π Mark]
Step 2:
= βπΌ + π΄2π΄ + 31π΄ + 31π΄2 β 7π΄ [β΅ πΌ3 = πΌ2 = πΌ] [π
π Mark]
Step 3:
= πΌ + π΄π΄ + 3π΄ + 3πΌπ΄ β 7π΄ [β΅ π΄2 = π΄] [π
π Mark]
Step 4:
= πΌ + π΄ + 3π΄ + 3π΄ β 7π΄
= πΌ + 7π΄ β 7π΄
= πΌ [β΅ πΌπ΄ = π΄] [π
π Mark]
Hence, the option (C) is correct