cbse ncert solutions for class 12 maths chapter 03 · back of chapter questions. cbse ncert...

Class-XII-Maths Matrices

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EXERCISE 3.1

1. In the matrix 𝐴 = [

2 5 19 −7

35 −25

√3 1 −5 17

], write:

(i) The order of the matrix,

(ii) The number of elements,

(iii) Write the elements 𝑎13, 𝑎21, 𝑎33, 𝑎24, 𝑎23.

Solution:

Given:

𝐴 = [

2 5 19 −7

35 −25

√3 1 −5 17

Number of rows = 3

Number of columns = 4

Step 1:

(i) The order of the matrix=number of rows × number of columns

Hence the order of the given matrix = 3 × 4

Step 2:

(𝑖𝑖) Since, the order of matrix is 3 × 4

So, the number of elements = 3 × 4 = 12

(𝑖𝑖𝑖) From the given matrix, we can observe the elements:

𝑎13 = 19

𝑎21 = 35

𝑎33 = −5

𝑎24 = 12

𝑎23 =5

Back of Chapter Questions

CBSE NCERT Solutions for Class 12 Maths Chapter 03

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 × 24, 2 × 12, 3 × 8, 4 × 6, 6 × 4,8 × 3, 12 × 2 and 24 × 1

Step 2:

Now, if it has 13 elements,

then the possible orders are 13 × 1 and 1 × 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 × 18, 2 × 9, 3 × 6, 6 × 3, 9 × 2 and 18 × 1

Step 2:

So, if it has 5 elements,

then the possible orders are 5 × 1 and 1 × 5

4. Construct a 2 × 2 matrix, 𝐴 = [𝑎𝑖𝑗], whose elements are given by:

(i) 𝑎𝑖𝑗 =(𝑖+𝑗)2

2 [2 marks]

(ii) 𝑎𝑖𝑗 =𝑖

𝑗 [2 marks]

(iii) 𝑎𝑖𝑗 =(𝑖+2𝑗)2

2 [2 marks]

Solution:

Given:

𝐴 = [𝑎𝑖𝑗]

Since, it is a 2 × 2 matrix

𝐴 = [𝑎11 𝑎12

𝑎21 𝑎22]

Step 1:

(i) Here, 𝑎𝑖𝑗 =(𝑖+𝑗)2

So, the elements of matrix are:

𝑎11 =(1+1)2

𝑎12 =(1+2)2

𝑎21 =(2+1)2

𝑎22 =(2+2)2

Therefore, the required matrix = [2

Step 2:

(ii) Here ,𝑎𝑖𝑗 =𝑖

𝑎11 =1

𝑎12 =1

𝑎21 =2

𝑎22 =2

Therefore, the required matrix=[1

2 1] [2 marks]

Step 3:

(𝑖𝑖𝑖) Here, 𝑎𝑖𝑗 =(𝑖+2𝑗)2

The elements of matrix are:

𝑎11 =(1+2(1))2

(1+2)2

𝑎12 =(1+2(2))2

(1+4)2

𝑎21 =(2+2(1))

(2+2)2

𝑎22 =(2+2(2))2

(2+4)2

Therefore, the required matrix = [9

8 18] [2 marks]

5. Construct a 3 × 4 matrix, whose elements are given by:

(i) 𝑎𝑖𝑗 =1

2| − 3𝑖 + 𝑗| [3 marks]

(ii) 𝑎𝑖𝑗 = 2𝑖 − 𝑗 [3 marks]

Solution:

Given:

Since, it is a 3 × 4 matrix

𝐴 = [

𝑎11 𝑎12 𝑎13 𝑎14

𝑎21 𝑎22 𝑎23 𝑎24

𝑎31 𝑎32 𝑎33 𝑎34

Step 1:

(i) Here, 𝑎𝑖𝑗 =1

2| − 3𝑖 + 𝑗|

𝑎11 =1

2|−3(1) + 1| =

2|−3 + 1| =

2|−2| =

2(2) = 1 [

𝟒 Mark]

𝑎12 =1

2| − 3(1) + 2| =

2| − 3 + 2| =

2| − 1| =

2(1) =

𝟒 Mark]

𝑎13 =1

2| − 3(1) + 3| =

2|0| =

2(0) = 0 [

𝟒 Mark]

𝑎14 =1

2| − 3(1) + 4| =

2| − 3 + 4| =

2|1| =

2(1) =

𝟒 Mark]

𝑎21 =1

2| − 3(2) + 1| =

2| − 6 + 1| =

2| − 5| =

2(5) =

𝟒 Mark]

𝑎22 =1

2| − 3 × 2 + 2| =

2| − 6 + 2| =

2| − 4| =

2(4) = 2 [

𝟒 Mark]

𝑎23 =1

2| − 3(2) + 1| =

2| − 6 + 3| =

2| − 3| =

2(3) =

𝟒 Mark]

𝑎24 =1

2| − 3 × 2 + 2| =

2| − 6 + 4| =

2| − 2| =

2(2) = 1 [

𝟒 Mark]

𝑎31 =1

2| − 3(3) + 1| =

2| − 9 + 1| =

2| − 8| =

2(8) = 4 [

𝟒 Mark]

𝑎32 =1

2| − 3(3) + 2| =

2| − 9 + 2| =

2| − 7| =

2(7) =

𝟒 Mark]

𝑎33 =1

2| − 3 × 3 + 3| =

2| − 9 + 3| =

2| − 6| =

2(6) = 3 [

𝟒 Mark]

𝑎34 ==1

2| − 3 × 3 + 4| =

2| − 9 + 4| =

2| − 5| =

2(5) =

𝟒 Mark]

Therefore, the required matrix A =

Step 2:

(ii) Here, 𝑎𝑖𝑗 = 2𝑖 − 𝑗,

𝑎11 = 2(1) − 1 = 2 − 1 = 1 [𝟏

𝟒 Mark]

𝑎12 = 2(1) − 2 = 2 − 2 = 0 [𝟏

𝟒 Mark]

𝑎13 = 2(1) − 3 = 2 − 3 = −1 [𝟏

𝟒 Mark]

𝑎14 = 2(1) − 4 = 2 − 4 = −2 [𝟏

𝟒 Mark]

𝑎21 = 2(2) − 1 = 4 − 1 = 3 [𝟏

𝟒 Mark]

𝑎22 = 2(2) − 2 = 4 − 2 = 2 [𝟏

𝟒 Mark]

𝑎23 = 2(2) − 3 = 4 − 3 = 1 [𝟏

𝟒 Mark]

𝑎24 = 2(2) − 4 = 4 − 4 = 0 [𝟏

𝟒 Mark]

𝑎31 = 2(3) − 1 = 6 − 1 = 5 [𝟏

𝟒 Mark]

𝑎32 = 2(3) − 2 = 6 − 2 = 4 [𝟏

𝟒 Mark]

𝑎33 = 2(3) − 3 = 6 − 3 = 3 [𝟏

𝟒 Mark]

𝑎34 = 2(3) − 4 = 6 − 4 = 2 [𝟏

𝟒 Mark]

Therefore, the required matrix A = [1 0 −1 −23 2 1 05 4 3 2

6. Find the values of 𝑥, 𝑦 and 𝑧 from the following equations:

(i) [4 3𝑥 5

] = [𝑦 𝑧1 5

] [2 Marks]

(ii) [𝑥 + 𝑦 25 + 𝑧 𝑥𝑦

] = [6 25 8

] [2 Marks]

(iii) [

𝑥 + 𝑦 + 𝑧𝑥 + 𝑧𝑦 + 𝑧

] = [957] [2 Marks]

Solution:

(i) Step 1:

Given: [4 3𝑥 5

] = [𝑦 𝑧1 5

If two matrices are equal, then their corresponding elements are also equal. [𝟏

𝟐 Mark]

Step 2:

Comparing the corresponding elements, we get

3 = 𝑧

𝑥 = 1

Therefore, 4 = y, 3 = 𝑧 and 𝑥 = 1 [𝟏𝟏

𝟐 Mark]

(ii)Step 1:

Given: [𝑥 + 𝑦 25 + 𝑧 𝑥𝑦

] = [6 25 8

𝟐 Mark]

Step 2:

𝑥 + 𝑦 = 6

5 + 𝑧 = 5

𝑥𝑦 = 8 [𝟏

𝟐 Mark]

Step 3:

By solving these above equations, we get

𝑥 = 2, 𝑦 = 4 and 𝑧 = 0 OR

𝑥 = 4, 𝑦 = 2 and 𝑧 = 0

Therefore, 𝑥 = 2, 𝑦 = 4 and 𝑧 = 0 or 𝑥 = 4, 𝑦 = 2 and 𝑧 = 0 [𝟏 Mark]

(iii) Step 1:

Given: [

𝑥 + 𝑦 + 𝑧𝑥 + 𝑧𝑦 + 𝑧

] = [957]

𝟐 Mark]

Step 2:

𝑥 + 𝑦 + 𝑧 = 9

𝑥 + 𝑧 = 5

𝑦 + 𝑧 = 7 [𝟏

𝟐 Mark]

Step 3:

𝑥 = 2,

𝑦 = 4

𝑧 = 3

Therefore, 𝑥 = 2, 𝑦 = 4 and 𝑧 = 3 [𝟏 Mark]

7. Find the value of 𝑎, 𝑏, 𝑐 and 𝑑 from the equation:

[𝑎 − 𝑏 2𝑎 + 𝑐2𝑎 − 𝑏 3𝑐 + 𝑑

] = [−1 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

𝑎 − 𝑏 = −1… (1)

2𝑎 − 𝑏 = 0… (2)

2𝑎 + 𝑐 = 5… (3)

3𝑐 + 𝑑 = 13… (4) [1𝟏

𝟐 Marks]

Step 2:

By solving equation (1) and (2), we get

𝑎 = 1, 𝑏 = 2

Putting the value of 𝑎 in equation (3), we get

𝑐 = 3

Putting the value of 𝑐 in the equation (4), we get

𝑑 = 4

Hence, the required values are 𝑎 = 1, 𝑏 = 2, 𝑐 = 3 and 𝑑 = 4 [1𝟏

𝟐 Marks]

8. 𝐴 = [𝑎𝑖𝑗]𝑚×𝑛 is a square matrix, if

(A) 𝑚 < 𝑛

(B) 𝑚 > 𝑛

(C) 𝑚 = 𝑛

(D) None of these

Solution:

Step 1:

𝐴 = [𝑎𝑖𝑗]𝑚×𝑛 means matrix A is of order m × n

In a square matrix the number of column is same as the number of rows.

Number of rows = Number of columns

𝑚 = 𝑛

Hence, the option (𝐶) is correct.

9. Which of the given values of 𝑥 and 𝑦 make the following pair of matrices equal

[3𝑥 + 7 5𝑦 + 1 2 − 3𝑥

] , [0 𝑦 − 28 4

] [2 marks]

(A) 𝑥 =−1

3, 𝑦 = 7

(B) Not possible to find

(C) 𝑦 = 7, 𝑥 =−2

(D) 𝑥 =−1

3, 𝑦 =

Solution:

Step 1:

Here [3𝑥 + 7 5𝑦 + 1 2 − 3𝑥

] = [0 𝑦 − 28 4

𝟐 Mark]

Step 2:

3𝑥 + 7 = 0

5 = 𝑦 − 2

𝑦 − 1 = 8

2 − 3𝑥 = 4 [𝟏

𝟐 Mark]

Step 3:

𝑦 = 7, 𝑥 = −7

3 and 𝑥 = −

Here, the value of 𝑥 is not unique.

Therefore, the option (𝐵) is correct.

10. The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is:

(A) 27

(B) 18

(C) 81

(D) 512 [2 Marks]

Solution:

Given:

It is a 3 × 3 matrix.

Step 1:

𝑎11 𝑎12 𝑎13

𝑎21 𝑎22 𝑎23

𝑎31 𝑎32 𝑎33

The total number of elements in a matrix of order 3 × 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.

EXERCISE 3.2

1. Let 𝐴 = [2 43 2

], 𝐵 = [1 3−2 5

], 𝐶 = [−2 5 3 4

Find each of the following:

(i) 𝐴 + 𝐵 [2 marks]

(ii) 𝐴 − 𝐵 [2 marks]

(iii) 3𝐴 − 𝐶 [3 marks]

(iv) AB [3 marks]

(v) BA [3 marks]

Solution:

Given:

𝐴 = [2 43 2

], 𝐵 = [1 3−2 5

], 𝐶 = [−2 5 3 4

(i) 𝐴 + 𝐵

Step 1:

𝐴 + 𝐵 = [2 43 2

] + [1 3−2 5

Step 2:

= [2 + 1 4 + 33 − 2 2 + 5

] = [3 71 7

Hence, 𝐴 + 𝐵 = [3 71 7

(ii) A − B

Step 1:

A − B = [2 43 2

] − [1 3

−2 5]

Step 2:

= [2 − 1 4 − 33 + 2 2 − 5

] = [1 15 −3

Hence, A − B = [1 15 −3

(iii) 3𝐴 − 𝐶

Step 1:

3𝐴 − 𝐶 = 3 [2 43 2

] − [−2 53 4

Step 2:

= [6 129 6

] − [−2 53 4

Step 3:

= [6 + 2 12 − 59 − 3 6 − 4

] = [8 76 2

Hence, 3𝐴 − 𝐶 = [8 76 2

(iv) 𝐴𝐵

Step 1:

𝐴𝐵 = [2 43 2

] [1 3−2 5

= [2 × 1 + 4 × −2 2 × 3 + 4 × 53 × 1 + 2 × −2 3 × 3 + 2 × 5

Step 2:

= [2 − 8 6 + 203 − 4 9 + 10

Step 3:

= [−6 26−1 19

Hence, 𝐴𝐵 = [−6 26−1 19

(v) 𝐵𝐴

Step 1:

𝐵𝐴 = [1 3−2 5

] [2 43 2

= [1 × 2 + 3 × 3 1 × 4 + 3 × 2−2 × 2 + 5 × 3 −2 × 4 + 5 × 2

Step 2:

= [2 + 9 4 + 6

−4 + 15 −8 + 10]

Step 3:

= [11 1011 2

Hence, 𝐵𝐴 = [11 1011 2

2. Compute the following:

(i) [𝑎 𝑏−𝑏 𝑎

] + [𝑎 𝑏𝑏 𝑎

] [2 marks]

(ii) [𝑎2 + 𝑏2 𝑏2 + 𝑐2

𝑎2 + 𝑐2 𝑎2 + 𝑏2] + [2𝑎𝑏 2𝑏𝑐−2𝑎𝑐 −2𝑎𝑏

] [2 marks]

(iii) [−1 4 −68 5 162 8 5

] + [12 7 68 0 53 2 4

] [2 marks]

(iv) [cos2𝑥 sin2𝑥sin2𝑥 cos2𝑥

] + [sin2𝑥 cos2𝑥

cos2𝑥 sin2𝑥] [2 marks]

Solution:

(i) Given:

[𝑎 𝑏−𝑏 𝑎

] + [𝑎 𝑏𝑏 𝑎

Step 1:

= [𝑎 + 𝑎 𝑏 + 𝑏−𝑏 + 𝑏 𝑎 + 𝑎

Step 2:

= [2𝑎 2𝑏0 2𝑎

Therefore, [𝑎 𝑏−𝑏 𝑎

] + [𝑎 𝑏𝑏 𝑎

] = [2𝑎 2𝑏0 2𝑎

(ii) Given:

[𝑎2 + 𝑏2 𝑏2 + 𝑐2

𝑎2 + 𝑐2 𝑎2 + 𝑏2] + [2𝑎𝑏 2𝑏𝑐−2𝑎𝑐 −2𝑎𝑏

Step 1:

= [𝑎2 + 𝑏2 + 2𝑎𝑏 𝑏2 + 𝑐2 + 2𝑏𝑐

𝑎2 + 𝑐2 − 2𝑎𝑐 𝑎2 + 𝑏2 − 2𝑎𝑏]

Step 2:

= [(𝑎 + 𝑏)2 (𝑏 + 𝑐)2

(𝑎 − 𝑐)2 (𝑎 − 𝑏)2]

Therefore, [𝑎2 + 𝑏2 𝑏2 + 𝑐2

𝑎2 + 𝑐2 𝑎2 + 𝑏2] + [2𝑎𝑏 2𝑏𝑐−2𝑎𝑐 −2𝑎𝑏

] = [(𝑎 + 𝑏)2 (𝑏 + 𝑐)2

(𝑎 − 𝑐)2 (𝑎 − 𝑏)2]

(iii) Given:

[−1 4 −68 5 162 8 5

] + [12 7 68 0 53 2 4

Step 1:

= [−1 + 12 4 + 7 −6 + 68 + 8 5 + 0 16 + 52 + 3 8 + 2 5 + 4

Step 2:

= [11 11 016 5 215 10 9

Therefore, [−1 4 −68 5 162 8 5

] + [12 7 68 0 53 2 4

] = [11 11 016 5 215 10 9

(iv) Given:

[𝑐𝑜𝑠2𝑥 𝑠𝑖𝑛2𝑥𝑠𝑖𝑛2𝑥 𝑐𝑜𝑠2𝑋

] + [𝑠𝑖𝑛2𝑋 𝑐𝑜𝑠2𝑋

𝑐𝑜𝑠2𝑥 𝑠𝑖𝑛2𝑥]

Step 1:

= [𝑐𝑜𝑠2𝑥 + 𝑠𝑖𝑛2𝑋 𝑠𝑖𝑛2𝑥 + 𝑐𝑜𝑠2𝑥𝑠𝑖𝑛2𝑥 + 𝑐𝑜𝑠2𝑋 𝑐𝑜𝑠2𝑥 + 𝑠𝑖𝑛2𝑥

Step 2:

= [1 11 1

Therefore, [𝑐𝑜𝑠2𝑥 𝑠𝑖𝑛2𝑥𝑠𝑖𝑛2𝑥 𝑐𝑜𝑠2𝑋

] + [𝑠𝑖𝑛2𝑋 𝑐𝑜𝑠2𝑋

𝑐𝑜𝑠2𝑥 𝑠𝑖𝑛2𝑥] = [

1 11 1

3. Compute the indicated products.

(i) [𝑎 𝑏−𝑏 𝑎

] [𝑎 −𝑏𝑏 𝑎

] [2 marks]

(ii) [123] [2 3 4] [2 marks]

(iii) [1 −22 3

] [1 2 32 3 1

] [2 marks]

(iv) [2 3 43 4 54 5 6

] [1 −3 50 2 43 0 5

] [2 marks]

(v) [2 13 2

−1 1] [

1 0 1−1 2 1

] [2 marks]

(vi) [3 −1 3−1 0 2

] [2 −31 03 1

] [2 marks]

Solution:

(i) Given:

[𝑎 𝑏−𝑏 𝑎

Step 1:

= [𝑎 × 𝑎 + 𝑏 × 𝑏 𝑎 × (−𝑏) + 𝑏 × 𝑎

−𝑏 × 𝑎 + 𝑏 × 𝑏 (−𝑏) × (−𝑏) + 𝑎 × 𝑎]

Step 2:

= [ 𝑎2 + 𝑏2 −𝑎𝑏 + 𝑎𝑏−𝑎𝑏 + 𝑏2 𝑏2 + 𝑎2 ]

= [𝑎2 + 𝑏2 0

0 𝑏2 + 𝑎2]

Hence, [𝑎 𝑏−𝑏 𝑎

] = [𝑎2 + 𝑏2 0

0 𝑏2 + 𝑎2]

(ii) Given:

[123] [2 3 4]

Step 1:

= [1 × 2 1 × 3 1 × 42 × 2 2 × 3 2 × 43 × 2 3 × 3 3 × 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 −22 3

] [1 2 32 3 1

Step 1:

= [1 × 1 + (−2) × 2 1 × 2 + (−2) × 3 1 × 3 + (−2) × 1

2 × 1 + 3 × 2 2 × 2 + 3 × 3 2 × 3 + 3 × 1]

= [1 − 4 2 − 6 3 − 22 + 6 4 + 9 6 + 3

Step 2:

= [−3 −4 18 13 9

Hence, [1 −22 3

] [1 2 32 3 1

] = [−3 −4 18 13 9

(iv) Given:

[2 3 43 4 54 5 6

] [1 −3 50 2 43 0 5

Step 1:

= [2 × 1 + 3 × 0 + 4 × 3 2 × (−3) + 3 × 2 + 4 × 0 2 × 5 + 3 × 4 + 4 × 53 × 1 + 4 × 0 + 5 × 3 3 × (−3) + 4 × 2 + 5 × 0 3 × 5 + 4 × 4 + 5 × 54 × 1 + 5 × 0 + 6 × 3 4 × (−3) + 5 × 2 + 6 × 0 4 × 5 + 5 × 4 + 6 × 5

Step 2:

= [2 + 0 + 12 −6 + 6 + 0 10 + 12 + 203 + 0 + 15 −9 + 0 + 0 15 + 16 + 254 + 0 + 18 −12 + 10 + 0 20 + 20 + 30

= [14 0 4218 −1 5622 −2 70

Hence, [2 3 43 4 54 5 6

] [1 −3 50 2 43 0 5

] = [14 0 4218 −1 5622 −2 70

(v) Given:

[2 13 2−1 1

] [1 0 1

−1 2 1]

Step 1:

2 × 1 + 1 × (−1) 2 × 0 + 1 × 2 2 × 1 + 1 × 13 × 1 + 2 × (−1) 3 × 0 + 2 × 2 3 × 1 + 2 × 1−1 × 1 + 1 × (−1) −1 × 0 + 1 × 2 −1 × 1 + 1 × 1

Step 2:

= [1 2 31 4 5−2 2 0

Hence, [2 13 2−1 1

] [1 0 1

−1 2 1] = [

1 2 31 4 5−2 2 0

(vi) Given:

[3 −1 3−1 0 2

] [2 −31 03 1

Step 1:

= [3 × 2 + (−1) × 1 + 3 × 3 3 × (−3) + (−1) × 0 + 3 × 1−1 × 2 + 0 × 1 + 2 × 3 −1 × (−3) + 0 × 0 + 2 × 1

Step 2:

= [6 − 1 + 9 −9 − 0 + 3

−2 + 0 + 6 3 + 0 + 2]

= [14 −64 5

Hence, [3 −1 3−1 0 2

] [2 −31 03 1

] = [14 −64 5

4. If 𝐴 = [1 2 −35 0 21 −1 1

], 𝐵 = [3 −1 24 2 52 0 3

] and 𝐶 = [4 1 20 3 21 −2 3

], then compute (𝐴 + 𝐵)

and (𝐵—𝐶). Also, verify that 𝐴 + (𝐵 − 𝐶) = (𝐴 + 𝐵) − 𝐶. [3 marks]

Solution:

Given:

𝐴 = [1 2 −35 0 21 −1 1

], 𝐵 = [3 −1 24 2 52 0 3

] and 𝐶 = [4 1 20 3 21 −2 3

Step 1:

So, 𝐴 + 𝐵 = [1 2 −35 0 21 −1 1

] + [3 −1 24 2 52 0 3

= [1 + 3 2 − 1 −3 + 25 + 4 0 + 2 2 + 51 + 2 −1 + 0 1 + 3

∴ 𝐴 + 𝐵 = [4 1 −19 2 73 −1 4

Step 2:

Now, 𝐵 − 𝐶 = [3 −1 24 2 52 0 3

] − [4 1 20 3 21 −2 3

= [3 − 4 −1 − 1 2 − 24 − 0 2 − 3 5 − 22 − 1 0 − (−2) 3 − 3

∴ 𝐵 − 𝐶 = [−1 −2 04 −1 31 2 0

Step 3:

𝐿𝐻𝑆 = 𝐴 + (𝐵 − 𝐶) = [1 2 −35 0 21 −1 1

] + [−1 −2 04 −1 31 2 0

] = [0 0 −39 −1 52 1 1

𝑅𝐻𝑆 = (𝐴 + 𝐵) − 𝐶 = [4 1 −19 2 73 −1 4

] − [4 1 20 3 21 −2 3

] = [0 0 −39 −1 52 1 1

∴ 𝐿𝐻𝑆 = 𝑅𝐻𝑆

Hence, 𝐴 + (𝐵 − 𝐶) = (𝐴 + 𝐵) − 𝐶 = [0 0 −39 −1 52 1 1

5. If 𝐴 =

, then compute 3𝐴 − 5𝐵. [3 Marks]

Solution:

Given:

𝐴 =

and 𝐵 =

Step 1:

3𝐴 − 5𝐵 = 3

Step 2:

3× 3 1 × 3

3× 3 2 × 3

3× 3]

5× 5 1 × 5

5× 5]

Step 3:

= [2 3 51 2 47 6 2

] − [2 3 51 2 47 6 2

= [2 − 2 3 − 3 5 − 51 − 1 2 − 2 4 − 47 − 7 6 − 6 2 − 2

] = [0 0 00 0 00 0 0

Therefore, 3𝐴 − 5𝐵 = [0 0 00 0 00 0 0

6. Simplify cos𝜃 [cos𝜃 sin𝜃−sin𝜃 cos𝜃

] + sin𝜃 [sin𝜃 −cos𝜃cos𝜃 sin𝜃

] [3 Marks]

Solution:

Given:

𝑐𝑜𝑠𝜃 [𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃−𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃

] + 𝑠𝑖𝑛𝜃 [𝑠𝑖𝑛𝜃 −𝑐𝑜𝑠𝜃𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃

Step 1:

= [cos𝜃(cos𝜃) cos𝜃(sin𝜃)

cos𝜃(−sin𝜃) cos𝜃(cos𝜃)] + [

sin𝜃(sin𝜃) sin𝜃(−cos𝜃)

sin𝜃(cos𝜃) sin𝜃(sin𝜃)]

= [𝑐𝑜𝑠2𝜃 𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃−𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 𝑐𝑜𝑠2𝜃

] + [𝑠𝑖𝑛2𝜃 −𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃

𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 𝑠𝑖𝑛2𝜃]

Step 2:

= [𝑐𝑜𝑠2𝜃 + 𝑠𝑖𝑛2𝜃 𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 − 𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃−𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 + 𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 𝑐𝑜𝑠2𝜃 + 𝑠𝑖𝑛2𝜃

= [cos2𝜃 + sin2𝜃 00 cos2𝜃 + sin2𝜃

] (∵ cos2𝜃 + sin2𝜃 = 1)

Step 3:

= [1 00 1

] = 𝐼

Hence, cos𝜃 [cos𝜃 sin𝜃−sin𝜃 cos𝜃

] + sin𝜃 [sin𝜃 −cos𝜃cos𝜃 sin𝜃

] = [1 00 1

7. Find 𝑋 and 𝑌, if

(i) 𝑋 + 𝑌 = [7 02 5

] and 𝑋 − 𝑌 = [3 00 3

] [3 Marks]

(ii) 2𝑋 + 3𝑌 = [2 34 0

] and 3𝑋 + 2𝑌 = [2 −2−1 5

] [3 Marks]

Solution:

(i) Given:

𝑋 + 𝑌 = [7 02 5

] ⋯ (1)

and 𝑋 − 𝑌 = [3 00 3

] …. (2)

Step 1:

Adding equation (1) and (2), we get

𝑋 + 𝑌 + 𝑋 − 𝑌 = [7 02 5

] + [3 00 3

2𝑋 = [10 02 8

⇒ 𝑋 = [5 01 4

] [1𝟏

𝟐 Mark]

Step 2:

Putting the value of 𝑋 in equation (1), we get

[5 01 4

] + 𝑌 = [7 02 5

𝑌 = [7 02 5

] − [5 01 4

⇒ 𝑌 = [2 01 1

Therefore, 𝑋 = [5 01 4

] and 𝑌 = [2 01 1

] [1𝟏

𝟐 Mark]

(ii) Given:

2𝑋 + 3𝑌 = [2 34 0

]… (1)

3𝑋 + 2Y = [2 −2−1 5

] ⋯(2)

Step 1:

Multiply equation (1) by 3 and equation (2) by 2, on subtracting, we get

3 (2𝑋 + 3𝑌) − 2(3𝑋 + 2𝑌) = 3 [2 34 0

] − 2 [2 −2−1 5

⇒ 6𝑋 + 9𝑌 − 6𝑋 − 4𝑌 = [6 912 0

] − [4 −4−2 10

⇒ 5𝑌 = [2 1314 −10

⇒ 𝑌 = [

] [1𝟏

𝟐 Mark]

Step 2:

Putting the value of 𝑌 in equation (1), we get

2𝑋 + 3 [

] = [2 34 0

⇒ 2𝑋 = [2 34 0

] − [

= [2 −

53 −

4 −42

50 + 6

⇒ 𝑋 = [

Hence, 𝑋 = [

] and 𝑌 = [

] [1𝟏

𝟐 Mark]

8. Find X, if 𝑌 = [3 21 4

] and 2𝑋 + 𝑌 = [1 0−3 2

] [3 Marks]

Solution:

Given:

𝑌 = [3 21 4

Step 1:

2𝑋 + 𝑌 = [1 0−3 2

⇒ 2𝑋 + [3 21 4

] = [1 0−3 2

] [⋅: 𝑌 = [3 21 4

[1𝟏

𝟐 Mark]

Step 3:

⇒ 2𝑋 = [1 0−3 2

] − [3 21 4

] = [−2 − 2−4 − 2

⇒ 𝑋 = [−1 − 1−2 − 1

Hence, the value of 𝑋 = [−1 − 1−2 − 1

][1𝟏

𝟐 Mark]

9. Find 𝑥 and 𝑦, if 2 [1 30 𝑥

] + [𝑦 01 2

] = [5 61 8

] [3 Marks]

Solution:

Given:

2 [1 30 𝑥

] + [𝑦 01 2

] = [5 61 8

Step 1:

⇒ [2 60 2𝑥

] + [𝑦 01 2

] = [5 61 8

⇒ [2 + 𝑦 6 + 00 + 1 2𝑥 + 2

] = [5 61 8

Step 2:

2 + 𝑦 = 5 and 2𝑥 + 2 = 8

Step 3:

By solving these equations, we get

𝑦 = 3 and 𝑥 = 3

Therefore, the values of 𝑦 = 3 and 𝑥 = 3.

10. Solve the equation for 𝑥, 𝑦, 𝑧 and 𝑡, if 2 [𝑥 𝑧𝑦 𝑡] + 3 [

1 −10 2

] = 3 [3 54 6

] [3 Marks]

Solution:

Given:

2 [𝑥 𝑧𝑦 𝑡] + 3 [

1 −10 2

] = 3 [3 54 6

Step 1:

[𝑥 × 2 𝑧 × 2𝑦 × 2 𝑡 × 2

] + [1 × 3 −1 × 30 × 3 2 × 3

] = [3 × 3 5 × 34 × 3 6 × 3

⇒ [2𝑥 2𝑧2𝑦 2𝑡

] + [3 −30 6

] = [9 1512 18

Step 2:

2𝑥 + 3 = 9

2𝑧 − 3 = 15

2𝑦 = 12

2𝑡 + 6 = 18

Step 3:

𝑥 = 3, 𝑧 = 9, 𝑦 = 6 and 𝑡 = 6

Hence, the values of 𝑥 = 3, 𝑧 = 9, 𝑦 = 6 and 𝑡 = 6

11. If 𝑥 [23] + 𝑦 [

] = [105

], find the values of 𝑥 and 𝑦. [3 Marks]

Solution:

Given:

𝑥 [23] + 𝑦 [

] = [105

Step 1:

⇒ [2𝑥3𝑥

] + [−𝑦𝑦 ] = [

⇒ [2𝑥 − 𝑦3𝑥 + 𝑦

] = [105

Step 2:

2𝑥 − 𝑦 = 10

3𝑥 + 𝑦 = 5

Step 3:

Adding both the equations, we get

5𝑥 = 15

⇒ 𝑥 = 3

Putting the value of 𝑥 in the equation 3𝑥 + 𝑦 = 5, we get

3(3) + 𝑦 = 5

⇒ 𝑦 = −4

Therefore, the values x and y are 3 and -4 respectively.

12. Given 3 [𝑥 𝑦𝑧 𝑤

] = [𝑥 6−1 2𝑤

] + [4 𝑥 + 𝑦𝑧 + 𝑤 3

], find the values of 𝑥, 𝑦, 𝑧 and 𝑤. [3

Marks]

Solution:

Given:

3 [𝑥 𝑦𝑧 𝑤

] = [𝑥 6−1 2𝑤

] + [4 𝑥 + 𝑦𝑧 + 𝑤 3

Step 1:

[3 × 𝑥 3 × 𝑦3 × 𝑧 3 × 𝑤

] = [𝑥 + 4 6 + 𝑥 + 𝑦

−1 + 𝑧 + 𝑤 2𝑤 + 3]

⇒ [3𝑥 3𝑦3𝑧 3𝑤

] = [𝑥 + 4 6 + 𝑥 + 𝑦1 + 𝑧 + 𝑤 2𝑤 + 3

Step 2:

3𝑥 = 𝑥 + 4

3𝑦 = 6 + 𝑥 + 𝑦

3𝑧 = −1 + 𝑧 + 𝑤

3𝑤 = 2𝑤 + 3

Step 3:

⇒ 𝑥 = 2, 𝑦 = 4, 𝑧 = 1 and 𝑤 = 3

Therefore, the values of 𝑥 = 2, 𝑦 = 4, 𝑧 = 1 and 𝑤 = 3

13. If 𝐹(𝑥) = [cos 𝑥 − sin𝑥 0sin𝑥 cos𝑥 00 0 1

], show that 𝐹(𝑥)𝐹(𝑦) = 𝐹(𝑥 + 𝑦). [3 Marks]

Solution:

Given:

𝐹(𝑥) = [cos 𝑥 − sin𝑥 0sin𝑥 cos 𝑥 00 0 1

Step 1:

For 𝐹(𝑦), replace x by y in 𝐹(𝑥)

𝐹(𝑦) = [cos𝑦 − sin 𝑦 0sin 𝑦 cos𝑦 00 0 1

] [𝟏

𝟐 Mark]

Step 2:

By taking LHS = 𝐹(𝑥)𝐹(𝑦)

= [𝑐𝑜𝑠𝑥 −𝑠𝑖𝑛𝑥 0𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥 00 0 1

] [cos 𝑦 − sin𝑦 0 0sin𝑦 cos 𝑦 0

0 0 1] [

𝟐 Mark]

Step 3:

cos𝑥 cos𝑦 + (−sin𝑥)sin𝑦 + 0 cos𝑥(−sin𝑦) + (−sin𝑥)cos𝑦 + 0 0 + 0 + 0 × 1

sin𝑥 cos𝑦 + cos𝑥 sin𝑦 + 0 sin𝑥(−sin𝑦) + cos𝑥 cos𝑦 + 0 0 + 0 + 0 × 1

0 × cos𝑦 + 0 × sin𝑦 + 0 × 1 0 × (−sin𝑦) + 0 × cos𝑦 + 0 0 + 0 + 1 × 1

= [cos(𝑥 + 𝑦) −[cos𝑥 sin𝑦 + sin𝑥 cos𝑦] 0

sin(𝑥 + 𝑦) cos𝑥 cos𝑦 − sin𝑥 sin𝑦 00 0 1

∵ cos𝑥 cos𝑦 − sin𝑥 sin𝑦 = cos(𝑥 + 𝑦)sin𝑥 cos𝑦 + cos𝑥 sin𝑦 = sin(𝑥 + 𝑦)

[1Mark]

Step 4:

= [𝑐𝑜𝑠(𝑥 + 𝑦) −𝑠𝑖𝑛(𝑥 + 𝑦) 0𝑠𝑖𝑛(𝑥 + 𝑦) 𝑐𝑜𝑠(𝑥 + 𝑦) 00 0 1

] = 𝐹(𝑥 + 𝑦) = 𝑅𝐻𝑆

𝐿𝐻𝑆 = 𝑅𝐻𝑆

Hence it is proved 𝐹(𝑥)𝐹(𝑦) = 𝐹(𝑥 + 𝑦).

14. Show that

(i) [5 −16 7

] [2 13 4

] ≠ [2 13 4

] [5 −16 7

] [3 Marks]

(ii) [1 2 30 1 01 1 0

] [−1 1 00 −1 12 3 4

] ≠ [−1 1 00 −1 12 3 4

] [1 2 30 1 01 1 0

] [4 Marks]

Solution:

(i)Step 1:

By taking 𝐿𝐻𝑆 = [5 −16 7

] [2 13 4

= [5 × 2 + (−1) × 3 5 × 1 + (−1) × 4

6 × 2 + 7 × 3 6 × 1 + 7 × 4] [

𝟐 Mark]

Step 2:

= [10 − 3 5 − 412 + 21 6 + 28

= [7 133 34

Step 3:

By taking 𝑅𝐻𝑆 = [2 13 4

] [5 −16 7

= [2 × 5 + 1 × 6 2 × (−1) + 1 × 73 × 5 + 4 × 6 6 × 1 + 7 × 4

] [𝟏

𝟐 Mark]

Step 4:

= [10 + 6 −2 + 715 + 24 −3 + 28

] = [16 539 25

Hence, [5 −16 7

] [2 13 4

] ≠ [2 13 4

] [5 −16 7

(ii)Step 1:

By taking 𝐿𝐻𝑆 = [1 2 30 1 01 1 0

] [−1 1 00 −1 12 3 4

= [1 × (−1) + 2 × 0 + 3 × 2 1 × 1 + 2 × (−1) + 3 × 3 1 × 0 + 2 × 1 + 3 × 40 × (−1) + 1 × 0 + 0 × 2 0 × 1 + 1 × (−1) + 0 × 3 0 × 0 + 1 × 1 + 0 × 41 × (−1) + 1 × 0 + 0 × 2 1 × 1 + 1 × (−1) + 0 × 3 1 × 0 + 1 × 1 + 0 × 4

Step 2:

= [−1 + 0 + 6 1 − 0 + 9 0 + 2 + 120 + 0 + 0 0 − 1 + 0 0 + 1 + 0

−1 + 0 + 0 1 − 1 + 0 0 + 1 + 0]

𝐿𝐻𝑆 = [5 8 140 −1 1−1 0 1

Step 3:

By taking 𝑅𝐻𝑆 = [−1 1 00 −1 12 3 4

] [1 2 30 1 01 1 0

[−1 × 1 + 1 × 0 + 0 × 1 −1 × 2 + 1 × 1 + 0 × 1 −1 × 3 + 1 × 0 + 0 × 00 × 1 + (−1) × 0 + 1 × 1 0 × 2 + (−1) × 1 + 1 × 1 0 × 3 + (−1) × 0 + 1 × 02 × 1 + 3 × 0 + 4 × 1 2 × 2 + 3 × 1 + 4 × 1 2 × 3 + 3 × 0 + 4 × 0

Step 4:

= [−1 + 0 + 0 −2 + 1 + 0 −3 + 0 + 00 + 0 + 1 0 − 1 + 1 0 + 0 + 02 + 0 + 4 4 + 3 + 4 6 + 0 + 0

𝑅𝐻𝑆 = [−1 −1 −31 0 06 11 6

∴ 𝐿𝐻𝑆 ≠ 𝑅𝐻𝑆

Hence, [1 2 30 1 01 1 0

] [−1 1 00 −1 12 3 4

] ≠ [−1 1 00 −1 32 3 4

] [1 2 30 1 01 1 0

15. Find 𝐴2 − 5𝐴 + 6𝐼, if 𝐴 = [2 0 12 1 31 −1 0

] [4 marks]

Solution:

Given:

𝐴 = [2 0 12 1 31 −1 0

Step 1:

We know that, 𝐴2 = 𝐴. 𝐴

⇒ 𝐴2 = [2 0 12 1 31 −1 0

] [2 0 12 1 31 −1 0

2 × 2 + 0 × 2 + 1 × 1 2 × 0 + 0 × 1 + 1 × (−1) 2 × 1 + 0 × 3 + 1 × 02 × 2 + 1 × 2 + 3 × 1 2 × 0 + 1 × 1 + 3 × (−1) 2 × 1 + 1 × 3 + 3 × 01 × 2 + (−1) × 2 + 0 × 1 1 × 0 + (−1) × 1 + 0 × (−1) 1 × 1 + (−1) × 3 + 0 × 0

Step 2:

= [4 + 0 + 1 0 + 0 − 1 2 + 0 + 04 + 2 + 3 0 + 1 − 3 2 + 3 + 02 − 2 + 0 0 − 1 + 0 1 − 3 + 0

𝐴2 = [5 −1 29 −2 50 −1 −2

Step 3:

Therefore, 𝐴2 − 5𝐴 + 6𝐼

= [5 −1 29 −2 50 −1 −2

] − 5 [2 0 12 1 31 −1 0

] + 6 [1 0 00 1 00 0 1

Step 4:

= [5 −1 29 −2 50 −1 −2

] − [10 0 510 5 155 −5 0

] + [6 0 00 6 00 0 6

= [5 − 10 + 6 −1 − 0 + 0 2 − 5 + 09 − 10 + 0 −2 − 5 + 6 5 − 15 + 00 − 5 + 0 −1 + 5 + 0 −2 − 0 + 6

= [1 −1 −3

−1 −1 −10−5 4 4

Hence, 𝐴2 − 5𝐴 + 6𝐼 = [1 −1 −3

−1 −1 −10−5 4 4

16. If 𝐴 = [1 0 20 2 12 0 3

], prove that 𝐴3 − 6𝐴2 + 7𝐴 + 2𝐼 = 0 [5 Marks]

Solution:

Given:

𝐴 = [1 0 20 2 12 0 3

Step 1:

⇒ 𝐴2 = [1 0 20 2 12 0 3

] [1 0 20 2 12 0 3

= [1 × 1 + 0 × 0 + 2 × 2 1 × 0 + 0 × 2 + 2 × 0 1 × 2 + 0 × 1 + 2 × 30 × 1 + 2 × 0 + 1 × 2 0 × 0 + 2 × 2 + 1 × 0 0 × 2 + 2 × 1 + 1 × 32 × 1 + 0 × 0 + 3 × 2 2 × 0 + 0 × 2 + 3 × 0 2 × 2 + 0 × 1 + 3 × 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

𝐴2 = [5 0 82 4 58 0 13

Step 3:

𝐴3 = 𝐴2𝐴 = [5 0 82 4 58 0 13

] [1 0 20 2 12 0 3

= [5 × 1 + 0 × 0 + 8 × 2 5 × 0 + 0 × 2 + 8 × 0 5 × 2 + 0 × 1 + 8 × 3 2 × 1 + 4 × 0 + 5 × 2 2 × 0 + 4 × 2 + 5 × 0 2 × 2 + 4 × 1 + 5 + 3 8 × 1 + 0 × 0 + 13 × 2 8 × 0 + 0 × 2 + 13 × 0 8 × 2 + 0 × 1 + 13 × 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

𝐴3 = [21 0 3412 8 2334 0 55

Step 4:

Therefore, 𝐴3 − 6𝐴2 + 7𝐴 + 2𝐼

𝐿𝐻𝑆 = [21 0 3412 8 2334 0 55

] − 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

] − [

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 − 30 + 7 + 2 0 − 0 + 0 + 0 34 − 48 + 14 + 012 − 12 + 0 + 0 8 − 24 + 14 + 2 23 − 30 + 7 + 034 − 48 + 14 + 0 0 + 0 + 0 + 0 55 − 78 + 21 + 2

= [0 0 00 0 00 0 0

∵ 𝐿𝐻𝑆 = 𝑅𝐻𝑆

Hence proved.

17. If 𝐴 = [3 −24 −2

] and 𝐼 = [1 00 1

], find 𝑘 so that 𝐴2 = 𝑘𝐴 − 2𝐼 [3 marks]

Solution:

Given:

𝐴 = [3 −24 −2

] and 𝐼 = [1 00 1

Step 1:

𝐴2 = 𝑘𝐴 − 2𝐼

⇒ [3 −24 −2

] [3 −24 −2

] = 𝑘 [3 −24 −2

] − 2 [1 00 1

Step 2:

⇒ [3 × 3 + (−2) × 4 3 × (−2) + (−2) × (−2)

4 × 3 + (−2) × 4 4 × (−2) + (−2) × (−2)] = [

3𝑘 −2𝑘4𝑘 −2𝑘

] − [2 00 2

⇒ [1 −24 −4

] = [3𝑘 − 2 −2𝑘

4𝑘 −2𝑘 − 2]

Step 3:

Here, matrices are equal.

⇒ 4𝑘 = 4

∴ 𝑘 = 1

Hence, the value of 𝑘 is 1.

18. If 𝐴 = [0 − tan

tan𝛼

] and 𝐼 is the identity matrix of order 2, show that

𝐼 + 𝐴 = (𝐼 − 𝐴) [cos 𝛼 − sin𝛼sin𝛼 cos𝛼

] [5 Marks]

Solution:

Given:

𝐴 = [0 − tan

tan𝛼

Step 1:

By considering LHS = 𝐼 + 𝐴

= [1 00 1

] + [0 − tan

tan𝛼

] = [1 − tan

tan𝛼

Step 2:

By taking RHS = (𝐼 − 𝐴) [𝑐𝑜𝑠𝛼 −𝑠𝑖𝑛𝛼𝑠𝑖𝑛𝛼 𝑐𝑜𝑠𝛼

= ([1 00 1

] − [0 − tan

tan𝛼

]) [𝑐𝑜𝑠𝛼 −𝑠𝑖𝑛𝛼𝑠𝑖𝑛𝛼 𝑐𝑜𝑠𝛼

= [1 − tan

tan𝛼

] [𝑐𝑜𝑠𝛼 −𝑠𝑖𝑛𝛼𝑠𝑖𝑛𝛼 𝑐𝑜𝑠𝛼

Step 3:

= [cos𝛼 + tan

2sin𝛼 −sin𝛼 + tan

2cos 𝛼

− tan𝛼

2cos 𝛼 + sin𝛼 tan

2sin 𝛼 + cos𝛼

Step 4:

[ 𝑐𝑜𝑠𝛼 +

sin𝛼

𝑐𝑜𝑠𝛼

𝑠𝑖𝑛𝛼 −𝑠𝑖𝑛𝛼 +𝑠𝑖𝑛

𝑐𝑜𝑠𝛼

−𝑠𝑖𝑛

𝑐𝑜𝑠𝛼

𝑐𝑜𝑠𝛼 + 𝑠𝑖𝑛𝛼𝑠𝑖𝑛

𝑐𝑜𝑠𝛼

𝑠𝑖𝑛𝛼 + 𝑐𝑜𝑠𝛼]

𝟐 Mark]

Step 5:

[ 𝑐𝑜𝑠𝛼 𝑐𝑜𝑠

2+𝑠𝑖𝑛

2𝑠𝑖𝑛𝛼

𝑐𝑜𝑠𝛼

−𝑠𝑖𝑛𝛼 𝑐𝑜𝑠𝛼

2+𝑠𝑖𝑛

2𝑐𝑜𝑠𝛼

𝑐𝑜𝑠𝛼

−𝑐𝑜𝑠𝛼 𝑠𝑖𝑛𝛼

2+𝑐𝑜𝑠

2𝑠𝑖𝑛𝛼

𝑐𝑜𝑠𝛼

𝑠𝑖𝑛𝛼 𝑠𝑖𝑛𝛼

2+𝑐𝑜𝑠

2𝑐𝑜𝑠𝛼

𝑐𝑜𝑠𝛼

𝟐 Mark]

Step 6:

[ cos(𝛼−

𝑐𝑜𝑠𝛼

−sin(𝛼−𝛼

𝑐𝑜𝑠𝛼

sin(𝛼−𝛼

𝑐𝑜𝑠𝛼

cos(𝛼−𝛼

𝑐𝑜𝑠𝛼

𝟐 Mark]

Step 7:

= [1 − tan

tan𝛼

] = 𝐿𝐻𝑆 [𝟏

𝟐 Mark]

∵ 𝐿𝐻𝑆 = 𝑅𝐻𝑆

Hence, it is proved.

19. A trust fund has ₹ 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 ₹ 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of: [5 Marks]

(a) ₹1800

(b) ₹2000

Solution:

Given:

Trust fund= ₹ 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 = ₹ 𝑥

The amount invested in second bond = ₹(30000 − 𝑥)

(𝑎) If the total annual interest is ₹1800, then

Investment in Bonds (in ₹) Annual Interest Rate Interest (in ≤]

30,000 − 𝑥] [5%7%

] [1800]

Step 2:

By solving, we get

𝑥 × 5% + (30000 − 𝑥) × 7%=1800

⇒5𝑥

100(30000 − 𝑥) = 1800

⇒ 5𝑥 + 210000 − 7𝑥 = 180000

⇒ −2𝑥 = −30000

𝑥 = 15000

Step 3:

So, amount investment at 5% = ₹15000

The amount investment at 7% = ₹(30000 − 𝑥)

= ₹(30000 − 15000) = ₹15000

Therefore, the amount invested in first bond is ₹15000 and second bond is ₹15000 [𝟏

Step 4:

(b) If the total annual interest is ₹2000, then

Investment in Bonds (in ₹] Annual Interest Rate Interest (in₹)

30,000 − 𝑥] [5%7%

] [2000]

Step 5:

By solving, we get

𝑥 × 5% + (30000 − 𝑥) × 7% = 2000

⇒5𝑥

100(30000 − 𝑥) = 2000

⇒ 5𝑥 + 210000 − 7𝑥 = 200000

⇒ −2𝑥 = −10000

⇒ 𝑥 = 5000

Step 6:

So, amount investment at 5% = ₹5000

The amount investment at 7% = ₹(30000 − 𝑥)

= ₹(30000 − 5000) = ₹25000

Therefore, the amount invested in first bond is ₹ 5000 and second bond is ₹25000. [𝟏

20. The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are ₹ 80, ₹ 60 and ₹ 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

[3 Marks]

Solution:

Given:

Step 1:

Number of chemistry books = 10 dozen = 10 × 12 = 120

Number of physics books = 8 dozen = 8 × 12 = 96

Number of economics books = 10 dozen = 10 × 12 = 120 [𝟏

𝟐 Mark]

Step 2:

Number of books

Chemistry Physics Economics

Selling Price per book Total amount (in ₹)

[120 96 120] [806040

] [𝑥]

[1Mark]

Step 3:

The shopkeeper receives the total amount of = Number of books ×Selling price per book

= [120 96 120] [806040

= [180 × 80 + 96 × 60 + 120 × 40]

= 120 × 80 + 96 × 60 + 120 × 40

= 9600 + 5760 + 4800

= ₹ 20160 [1𝟏

𝟐 Marks]

Hence, the bookshop will receive ₹20160 from selling all the books.

21. Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k,

respectively.

The restriction on 𝑛, 𝑘 and 𝑝 so that 𝑃𝑌 + 𝑊𝑌 will be defined are: [2 Marks]

(A) 𝑘 = 3, 𝑝 = 𝑛

(B) 𝑘 is arbitrary, 𝑝 = 2

(C) 𝑝 is arbitrary, 𝑘 = 3

(D) 𝑘 = 2, 𝑝 = 3

Solution:

Given:

The order of 𝑃 = 𝑝 × 𝑘

The order of 𝑌 = 3 × 𝑘.

Step 1:

So, 𝑃𝑌 will be defined if, 𝑘 = 3.

Hence, the order of 𝑃𝑌 is 𝑝 × 𝑘.

Step 2:

Order of 𝑊 = 𝑛 × 3 and order of 𝑌 = 3 × 𝑘

According to order, WY is defined and its order is 𝑛 × 𝑘.

𝑃𝑌 + 𝑊𝑌 is defined, if the order of PY and WY are equal.

⇒ 𝑝 × 𝑘 = 𝑛 × 𝑘

⇒ 𝑝 = 𝑛,

Therefore, the option (A) is correct.

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: [2 marks]

(A) 𝑝 × 2

(B) 2 × 𝑛

(C) 𝑛 × 3

(D) 𝑝 × 𝑛

Solution:

Given:

The order of 𝑋 = 2 × 𝑛

The order of 𝑍 = 2 × 𝑝

𝑛 = 𝑝

Step 1:

During the addition or subtraction, the order of matrix doesn’t change.

Therefore, the order of 7𝑋 − 5𝑍 = order of 𝑋 = order of 𝑍

Step 2:

⇒ 2 × 𝑛 = 2 × 𝑝

= 2 × 𝑛

Hence, the option (𝐵) is correct

EXERCISE 3.3

1. Find the transpose of each of the following matrices:

(ii) [1 −12 3

(iii) [−1 5 6

√3 5 62 3 −1

Solution:

(i) Let 𝐴 = [

𝐴′ = [

2−1]

Hence the transpose of the given matrix is [5 1

2 − 1]

(𝑖𝑖) Let 𝐵 = [1 −12 3

𝐵′ = [1 −12 3

= [1 2

−1 3]

Hence the transpose of the given matrix is [1 2−1 3

(iii) Let 𝐶 = [−1 5 6

√3 5 62 3 −1

𝐶′ = [−1 5 6

√3 5 62 3 −1

= [−1 √3 25 5 36 6 −1

Hence the transpose of the given matrix is [−1 √3 25 5 36 6 −1

2. If 𝐴 = [−1 2 35 7 9−2 1 1

] and 𝐵 = [−4 1 −51 2 01 3 1

], then verify that

(i) (𝐴 + 𝐵)′ = 𝐴′ + 𝐵′, [3 Marks]

(ii) (𝐴 − 𝐵)′ = 𝐴′ − 𝐵′ [3 Marks]

Solution:

Given:

𝐴 = [−1 2 35 7 9−2 1 1

] and 𝐵 = [−4 1 −51 2 01 3 1

(i) (𝐴 + 𝐵)′ = 𝐴′ + 𝐵′ Step 1:

(𝐴 + 𝐵)

= [−1 2 35 7 9

−2 1 1] + [

−4 1 −51 2 01 3 1

] = [−1 − 4 2 + 1 3 − 55 + 1 7 + 2 9 + 0

−2 + 1 1 + 3 1 + 1]

(𝐴 + 𝐵) = [−5 3 −26 9 9

−1 4 2]

Step 2:

Hence (𝐴 + 𝐵)’ = [−5 −6 −13 9 4

−2 9 2] ….(1) [

𝟐 Mark]

Step 3:

𝐴′ + 𝐵′ = [−1 2 35 7 9

−2 1 1]

+ [−4 1 −51 2 01 3 1

𝟐 Mark]

Step 4:

= [−1 5 −22 7 13 9 1

] + [−4 1 11 2 3

−5 0 1] = [

−1 − 4 5 + 1 −2 + 12 + 1 7 + 2 1 + 33 − 5 9 + 0 1 + 1

] = [−5 6 −13 9 4

−2 9 2] ….(2)

From the equation (1) and (2], we get

(𝐴 + 𝐵)’ = 𝐴′ + 𝐵′ [1Mark]

Hence it is proved.

(ii) (𝐴 − 𝐵)′ = 𝐴′ − 𝐵′

Step 1:

(𝐴 − 𝐵)

= [−1 2 35 7 9−2 1 1

] − [−4 1 −51 2 01 3 1

] = [−1 + 4 2 − 1 3 + 55 − 1 7 − 2 9 − 0

−2 − 1 1 − 3 1 − 1]

(𝐴 − 𝐵) = [3 1 84 5 9

−3 −2 0] [1Mark]

Step 2:

Thus, (𝐴 − 𝐵)’ = [3 4 −31 5 −28 9 0

] ⋯ (1) [𝟏

𝟐 Mark]

Step 3:

A′ − 𝐵′ = [−1 2 35 7 9

−2 1 1]

− [−4 1 −51 2 01 3 1

𝟐 Mark]

Step 4:

= [−1 5 −22 7 13 9 1

] − [−4 1 11 2 3

−5 0 1] = [

−1 + 4 5 − 1 −2 − 12 − 1 7 − 2 1 − 33 + 5 9 − 0 1 − 1

] = [3 4 −31 5 −28 9 0

] ⋯ (2)

From the equation (1) and (2), we get

(𝐴 + 𝐵)′ = 𝐴′ + 𝐵′ [1Mark]

Hence it is proved.

3. If 𝐴′ = [3 4−1 20 1

] and 𝐵 = [−1 2 11 2 3

], then verify that

(i) (𝐴 + 𝐵)′ = 𝐴′ + 𝐵′ [3 Marks]

(ii) (𝐴 − 𝐵)′ = 𝐴′ − 𝐵′ [3 Marks]

Solution:

Given:

𝐴′ = [3 4−1 20 1

] and 𝐵 = [−1 2 11 2 3

(i) (𝐴 + 𝐵)′ = 𝐴′ + 𝐵′ Step 1:

𝐴′ = [3 4

−1 20 1

⇒ 𝐴 = [3 −1 04 2 1

] [∵ (𝐴′)′ = 𝐴] [𝟏

𝟐 Mark]

Step 2:

(𝐴 + 𝐵) = [3 −1 04 2 1

] + [−1 2 11 2 3

= [3 − 1 −1 + 2 0 + 14 + 1 2 + 2 1 + 3

] = [2 1 15 4 4

Step 3:

Therefore (𝐴 + 𝐵)′ = [2 51 41 4

] …. (1) [𝟏

𝟐 Mark]

Step 4:

𝐴′ + 𝐵’ = [3 −1 04 2 1

+ [−1 2 11 2 3

= [3 4

−1 20 1

] + [−1 12 21 3

] = [3 − 1 4 + 1

−1 + 2 2 + 20 + 1 1 + 3

𝐴′ + 𝐵’ = [2 51 41 4

] ⋯ (2)

(𝐴 + 𝐵)’ = 𝐴′ + 𝐵′

(ii) (𝐴 − 𝐵)′ = 𝐴′ − 𝐵′

Step 1:

𝐴′ = [3 4−1 20 1

⇒ 𝐴 = [3 −𝑙 04 2 1

] [∵ (𝐴′)′ = 𝐴] [𝟏

𝟐 Mark]

Step 2:

(A − B) = [3 −1 04 2 1

]—[−1 2 11 2 3

= [3 + 1 −1 − 2 0 − 14 − 1 2 − 2 1 − 3

] = [4 −3 −13 0 −2

Step 3:

Therefore (A − B)’ = [4 3−3 0−1 −2

] ⋯ (1) [𝟏

𝟐 Mark]

Step 4:

𝐴′ − 𝐵′ = [3 −1 04 2 1

− [−1 2 11 2 3

= [3 4

−1 20 1

] − [−1 12 21 3

] = [3 + 1 4 − 1

−1 − 2 2 − 20 − 1 1 − 3

= [4 3

−3 0−1 −2

]⋯ (2)

From the equation (1) and (2], we get

(𝐴 − 𝐵)’ = 𝐴′ − 𝐵′

4. If 𝐴 = [−2 31 2

] and 𝐵 = [−1 01 2

], then find (𝐴 + 2𝐵)′ [2 Marks]

Solution:

Given:

A = [−2 31 2

] and 𝐵 = [−1 01 2

Step 1:

(𝐴 + 2𝐵) = [−2 31 2

] + 2 [−1 01 2

= [−2 31 2

] + [−2 02 4

Step 2:

= [−2 − 2 3 + 01 + 2 2 + 4

] = [−4 33 6

] [𝟏

𝟐 Mark]

Step 3:

(𝐴 + 2𝐵)’ = [−4 33 6

= [−4 33 6

] [𝟏

𝟐 Mark]

Therefore, (𝐴 + 2𝐵)’ = [−4 33 6

5. For the matrices 𝐴 and 𝐵, verify that (𝐴𝐵)′ = 𝐵′𝐴′, where

(i) 𝐴 = [1−43

], 𝐵 = [−1 2 1] [3 marks]

(ii) 𝐴 = [012], 𝐵 = [1 5 7] [3 marks]

Solution:

(i) Given:

𝐴 = [1−43

] , 𝐵 = [−1 2 1]

Step 1:

So, 𝐴𝐵 = [1−43

] [−1 2 1]

= [1 × (−1) 1 × 2 1 × 1−4 × (−1) −4 × 2 −4 × 13 × (−1) 3 × 2 3 × 1

] = [−1 2 14 −8 −4

−3 6 3]

Step 2:

𝐴𝐵′ = [−1 2 14 −8 −4

−3 6 3]

= [−1 4 −32 −8 61 −4 3

Therefore, 𝐴𝐵′ = [−1 4 −32 −8 61 −4 3

] …..(1) [𝟏

𝟐 Mark]

Step 3:

𝐵′ = [−1 2 1]′ = [−121

𝐴′ = [1

] = [1 − 4 3] [𝟏

𝟐 Mark]

Step 4:

𝐵′𝐴′ = [−121

] [1 − 4 3] = [−1 × 1 −1 × (−4) −1 × 32 × 1 2 × (−4) 2 × 31 × 1 1 × (−4) 1 × 3

= [−1 4 −32 −8 61 −4 3

Therefore, 𝐵′𝐴′ = [−1 4 −32 −8 61 −4 3

]⋯ (2)

From the equation (1] and [2], we get

(𝐴𝐵) = 𝐵′𝐴′

Hence proved.

(ii) Given:

𝐴 = [012], 𝐵 = [1 5 7]

Step 1:

So, AB=[012] = [1 5 7]

= [0 × 1 0 × 5 0 × 71 × 1 1 × 5 1 × 72 × 1 2 × 5 2 × 7

] = [0 0 01 5 72 10 14

Step 2:

𝐴𝐵′ = [0 0 01 5 72 10 14

= [0 1 20 5 100 7 14

Therefore 𝐴𝐵′ = [0 1 20 5 100 7 14

] ⋯ (1) [𝟏

𝟐 Mark]

Step 3:

𝐵′ = [1 5 7]′ = [157] and 𝐴′ = [

012] = [0 1 2] [

𝟐 Mark]

Step 4:

So, 𝐵′𝐴′ = [157] [0 1 2] = [

1 × 0 1 × 1 1 × 25 × 0 5 × 1 5 × 27 × 0 7 × 1 7 × 2

= [0 1 20 5 100 7 14

] ⋯ (2)

(𝐴𝐵)’ = 𝐵′𝐴′

Hence proved

6. (i) If 𝐴 = [cos𝛼 sin𝛼−sin𝛼 cos𝛼

] , then verify that 𝐴′𝐴 = 𝐼 [2 Marks]

(ii) If 𝐴 = [sin𝛼 cos𝛼− cos𝛼 sin𝛼

], then verify that 𝐴′𝐴 = 𝐼 [2 marks]

Solution:

(i) Given:

𝐴 = [𝑐𝑜𝑠𝛼 𝑠𝑖𝑛𝛼−𝑠𝑖𝑛𝛼 𝑐𝑜𝑠𝛼

Step 1:

⇒ 𝐴′ = [𝑐𝑜𝑠𝛼 −𝑠𝑖𝑛𝛼𝑠𝑖𝑛𝛼 𝑐𝑜𝑠𝛼

] [𝟏

𝟐 Mark]

Step 2:

Therefore 𝐴′𝐴 = [𝑐𝑜𝑠𝛼 −𝑠𝑖𝑛𝛼𝑠�̇�𝛼 𝑐𝑜𝑠𝛼

] [𝑐𝑜𝑠𝛼 𝑠𝑖𝑛𝛼−𝑠𝑖𝑛𝛼 𝑐𝑜𝑠𝛼

= −[𝑐𝑜𝑠2𝛼 + 𝑠𝑖𝑛2𝛼 𝑐𝑜𝑠𝛼 𝑠𝑖𝑛𝛼 − 𝑠𝑖𝑛𝛼 𝑐𝑜𝑠𝛼𝑠𝑖𝑛𝛼 𝑐𝑜𝑠𝛼 − 𝑐𝑜𝑠𝛼 𝑠𝑖𝑛𝛼 𝑠𝑖𝑛2𝛼 + 𝑐𝑜𝑠2𝛼

Step 3:

= [1 00 1

] = 𝐼

∴ 𝐴′𝐴 = 𝐼 [𝟏

𝟐 Mark]

Hence proved.

(ii) Given:

𝐴 = [𝑠𝑖𝑛𝛼 𝑐𝑜𝑠𝛼−𝑐𝑜𝑠𝛼 𝑠𝑖𝑛𝛼

Step 1:

⇒ 𝐴′ = [𝑠𝑖𝑛𝛼 −𝑐𝑜𝑠𝛼𝑐𝑜𝑠𝛼 𝑠𝑖𝑛𝛼

] [𝟏

𝟐 Mark]

Step 2:

Therefore 𝐴′𝐴 = [𝑠𝑖𝑛𝛼 −𝑐𝑜𝑠𝛼𝑐𝑜𝑠𝛼 𝑠𝑖𝑛𝛼

] [𝑠𝑖𝑛𝛼 𝑐𝑜𝑠𝛼−𝑐𝑜𝑠𝛼 𝑠𝑖𝑛𝛼

= [𝑠𝑖𝑛2𝛼 + 𝑐𝑜𝑠2𝛼 𝑠𝑖𝑛𝛼 𝑐𝑜𝑠𝛼 − 𝑐𝑜𝑠𝛼 𝑠𝑖𝑛𝛼

𝑐𝑜𝑠𝛼 𝑠𝑖𝑛𝛼 − 𝑠𝑖𝑛𝛼 𝑐𝑜𝑠𝛼 𝑐𝑜𝑠2𝛼 + 𝑠𝑖𝑛2𝛼]

Step 3:

= [1 00 1

] = 𝐼

∴ 𝐴′𝐴 = 𝐼 [𝟏

𝟐 Mark]

Hence proved.

7. i) Show that the matrix 𝐴 = [1 −1 5

−1 2 15 1 3

] is a symmetric matrix. [2 Marks]

(ii) Show that the matrix 𝐴 = [0 1 −1

−1 0 11 −1 0

] is a skew symmetric matrix. [2 Marks]

Solution:

(i) Given:

𝐴 = [1 −1 5−1 2 15 1 3

Step 1:

⇒ A′ = [1 −1 5

−1 2 15 1 3

= [1 −1 5

−1 2 15 1 3

Step 2:

∴ 𝐴′ = 𝐴

Hence, the matrix 𝐴 = [1 −1 5

−1 2 15 1 3

] is a symmetric matrix.

(ii) 𝐴 = [0 1 −1

−1 0 11 −1 0

Step 1:

⇒ 𝐴′ = [0 1 −1

−1 0 11 −1 0

= [0 −1 11 0 −1

−1 1 0]

= −[0 1 −1

−1 0 11 −1 0

] = −𝐴

Step 2:

⇒ 𝐴′ = −𝐴

Hence, the matrix A = [0 1 −1

−1 0 11 −1 0

] is a skew symmetric matrix.

8. For the matrix 𝐴 = [1 56 7

], verify that

(i) (𝐴 + 𝐴′) is a symmetric matrix [3 Marks]

(ii) (𝐴 − 𝐴′) is a skew symmetric matrix [3 Marks]

(i) Given:

𝐴 = [1 56 7

Step 1:

⇒ 𝐴′ = [1 56 7

= [1 65 7

Step 2:

Therefore, (𝐴 + 𝐴′) = [1 56 7

] + [1 65 7

] = [2 1111 14

Step 3:

(𝐴 + 𝐴′)′ = [2 1111 14

= [2 1111 14

⇒ (𝐴 + A′)′ = (𝐴 + 𝐴′)

Hence the matrix (𝐴 + 𝐴′) is a symmetric matrix.

(ii) Given:

𝐴 = [1 56 7

Step 1:

⇒ 𝐴′ = [1 56 7

= [1 65 7

Step 2:

(𝐴 − 𝐴′) = [1 56 7

] − [1 65 7

] = [0 −11 0

Step 3:

∴ (𝐴 − 𝐴′)′ = [0 −11 0

= [0 1

−1 0] = − [

0 −11 0

] = −(𝐴 − 𝐴′)

⇒ (𝐴 − 𝐴′)′ = −(𝐴 − 𝐴′)

Hence the matrix (𝐴 − 𝐴′) is a skew symmetric matrix.

9. Find 1

2(𝐴 + 𝐴′) and

2(𝐴 − 𝐴′), when 𝐴 = [

0 𝑎 𝑏−𝑎 0 𝑐−𝑏 −𝑐 0

] [3 Marks]

Solution:

Given that: 𝐴 = [0 𝑎 𝑏

−𝑎 0 𝑐−𝑏 −𝑐 0

Step 1:

⇒ 𝐴′ = [0 𝑎 𝑏

−𝑎 0 𝑐−𝑏 −𝑐 0

= [0 −𝑎 −𝑏𝑎 0 −𝑐𝑏 𝑐 0

Step 2:

2(𝐴 + 𝐴′) =

0 𝑎 𝑏−𝑎 0 𝑐−𝑏 −𝑐 0

] + [0 −𝑎 −𝑏𝑎 0 −𝑐𝑏 𝑐 0

2[0 0 00 0 00 0 0

] = [0 0 00 0 00 0 0

Step 3:

2(𝐴 − 𝐴′) =

0 𝑎 𝑏−𝑎 0 −𝑐−𝑏 −𝑐 0

] − [0 −𝑎 −𝑏𝑎 0 −𝑐𝑏 𝑐 0

0 2𝑎 2𝑏−2𝑎 0 2𝑐−2𝑏 −2𝑐 0

]=[0 𝑎 𝑏

−𝑎 0 𝑐−𝑏 𝑐 0

Hence, the required matrix of 1

2(𝐴 + 𝐴′) is [

0 0 00 0 00 0 0

] and 1

2(𝐴 − 𝐴′) is [

0 𝑎 𝑏−𝑎 0 𝑐−𝑏 𝑐 0

10. Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

(i) [3 51 −1

] [4 Marks]

(ii) [6 −2 2−2 3 −12 −1 3

] [4 Marks]

(iii) [3 3 −1−2 −2 1−4 −5 2

] [4 Marks]

(iv) [1 5−1 2

] [4 Marks]

Solution:

(i) Given: 𝐴 = [3 51 −1

Step 1:

𝐴 =1

2(𝐴 + 𝐴′) +

2(𝐴 − 𝐴′)

Consider, 𝑃 =1

2(𝐴 + 𝐴′) and 𝑄 =

2(𝐴 − 𝐴′) [

𝟐 Mark]

Step 2:

We have 𝐴′ = [3 15 −1

] [𝟏

𝟐 Mark]

Step 3:

𝑃 =1

2(𝐴 + 𝐴′) =

3 51 −1

] + [3 15 −1

2[6 66 −2

] = [3 33 −1

] [𝟏

𝟐 Mark]

Step 4:

𝑃′ = [3 33 −1

] = 𝑃

⇒ 𝑃 is a symmetric matrix. [𝟏

𝟐 Mark]

Step 5:

𝑄 =1

2(𝐴 − 𝐴′) =

3 51 −1

] − [3 15 −1

2[0 4−4 0

] = [0 2−2 0

] [𝟏

𝟐 Mark]

Step 6:

𝑄′ = [0 2−2 0

= [0 −22 0

] = − [0 2−2 0

] = −𝑄

⇒ 𝑄 is a skew symmetric matrix [𝟏

𝟐 Mark]

Step 7:

Hence, 𝐴 = 𝑃 + 𝑄 = [3 33 −1

] + [0 2−2 0

𝐴 = [3 51 −1

Thus, A is a sum of symmetric and skew symmetric matrix.

(ii) Given that: 𝐴 = [6 −2 2−2 3 −12 −1 3

Step 1:

𝐴′ = [6 −2 2

−2 3 −12 −1 3

] [𝟏

𝟐 Mark]

Step 2:

Therefore, 𝐴 =1

2(𝐴 + 𝐴′) +

2(𝐴 − 𝐴′)

𝐿𝑒t, 𝑃 =1

2(𝐴 + 𝐴′) and 𝑄 =

2(𝐴 − 𝐴′) [

𝟐 Mark]

Step 3:

𝑃 =1

2(𝐴 + 𝐴′) =

6 −2 2−2 3 −12 −1 3

] + [6 −2 2

−2 3 −12 −1 3

2[12 −4 4−4 6 −24 −2 6

] = [6 −2 2−2 3 −12 −1 3

] [𝟏

𝟐 Mark]

Step 4:

𝑃′ = [6 −2 2

−2 3 −12 −1 3

= [6 −2 2

−2 3 −12 −1 3

] = 𝑃 [𝟏

𝟐 Mark]

⇒ 𝑃 is a symmetric matrix

Step 5:

𝑄 =1

2(𝐴 − 𝐴′) =

6 −2 2−2 3 −12 −1 3

] − [6 −2 2−2 3 −12 −1 3

2[0 0 00 0 00 0 0

] = [0 0 00 0 00 0 0

] [𝟏

𝟐 Mark]

Step 6:

𝑄′ = [0 0 00 0 00 0 0

= [0 0 00 0 00 0 0

] = − [0 0 00 0 00 0 0

] = −𝑄 [𝟏

𝟐 Mark]

⇒ 𝑄 is a skew symmetric matrix.

Step 7:

Hence, 𝐴 = 𝑃 + 𝑄 = [6 −2 2−2 3 −12 −1 3

] + [0 0 00 0 00 0 0

𝐴 = [6 −2 2−2 3 −12 −1 3

(iii) Given: 𝐴 = [3 3 −1−2 −2 1−4 −5 2

Step 1:

𝐴′ = [3 −2 −43 −2 −5

−1 1 2] [

𝟐 Mark]

Step 2:

Therefore, A =1

2(𝐴 + 𝐴′) +

2(𝐴 − 𝐴′)

Let, 𝑃 =1

2(𝐴 + 𝐴′) and 𝑄 =

2(𝐴 − 𝐴′) [

𝟐 Mark]

Step 3:

𝑃 =1

2(𝐴 + 𝐴′) =

3 3 −1−2 −2 1−4 −5 2

] + [3 −2 −43 −2 −5−1 1 2

= −1

2[6 1 −51 −4 −4−5 −4 4

2−2 −2

2−2 2 ]

𝟐 Mark]

Step 4:

𝑃′ =

2−2 −2

2−2 2 ]

2−2 −2

2−2 2 ]

= 𝑃 [𝟏

𝟐 Mark]

⇒ 𝑃 is a symmetric matrix.

Step 5:

𝑄 =1

2(𝐴 − 𝐴′) =

3 3 −1−2 −2 1−4 −5 2

] − [3 −2 −43 −2 −5−1 1 2

]) [𝟏

𝟐 Mark]

2[0 5 3−5 0 6−3 −6 0

2−3 0]

Step 6:

𝑄′ =

2−3 0]

[ 0 −

30 −3

23 0 ]

[ 0 −

20 −3

23 0 ]

= −𝑄

⇒ 𝑄 is a skew symmetric matrix. [𝟏

𝟐 Mark]

Step 7:

Hence, 𝐴 = 𝑃 + 𝑄 =

2−2 −2

2−2 2 ]

2−3 0]

𝐴 = [3 3 −1−2 −2 1−4 −5 2

(iv) Given that: 𝐴 = [1 5−1 2

Step 1:

𝐴′ = [1 −15 2

] [𝟏

𝟐 Mark]

Step 2:

Therefore, 𝐴 = −1

2(𝐴 + 𝐴′) +

2(𝐴 − 𝐴′)

Let, 𝑃 =1

2(𝐴 + 𝐴′) and 𝑄 =

2(𝐴 − 𝐴′) [

𝟐 Mark]

Step 3:

𝑃 =1

2(𝐴 + 𝐴′)

1 5−1 2

] + [1 −15 2

2[2 44 4

] = [1 22 2

] [𝟏

𝟐 Mark]

Step 4:

𝑃′ = [1 22 2

] = [1 22 2

] = 𝑃

⇒ 𝑃 is a symmetric matrix [𝟏

𝟐 Mark]

Step 5:

𝑄 =1

2(𝐴 − 𝐴′)

1 5−1 2

] − [1 −15 2

2[0 6−6 0

] = [0 3−3 0

] [𝟏

𝟐 Mark]

Step 6:

𝑄′ = [0 3−3 0

= [0 −33 0

] = −[0 3−3 0

] = −𝑄

⇒ 𝑄 is a skew symmetric matrix [𝟏

𝟐 Mark]

Step 7:

Hence, 𝐴 = 𝑃 + 𝑄 = [1 22 2

] + [0 3−3 0

𝐴 = [1 5−1 2

11. If 𝐴, 𝐵 are symmetric matrices of same order, then 𝐴𝐵 − 𝐵𝐴 is a [2 Marks]

(A) Skew symmetric matrix

(B) Symmetric matrix

(C) Zero matrix

(D) Identity matrix

Solution:

Given:

A and B are symmetric matrices of same order

Step 1:

Let, (𝐴𝐵 − 𝐵𝐴)′ = (𝐴𝐵)′ − (𝐵𝐴)′ [∵ (𝑋 ‐ 𝑌)′ = 𝑋′ − 𝑌′] [𝟏

Step 2:

= 𝐵′𝐴′ − 𝐴′𝐵′ [∵ (𝑋𝑌)’ = 𝑌′𝑋′] [𝟏

𝟐 Mark]

Step 3:

= 𝐵𝐴 − AB [∵Given: 𝐴′ = 𝐴, 𝐵′ = 𝐵] [𝟏

Step 4:

= −(𝐴𝐵 − 𝐵𝐴)

⇒ (𝐴𝐵 − 𝐵𝐴)′ = −(𝐴𝐵 − 𝐵𝐴)

Therefore, the matrix (AB — BA) is a skew symmetric matrix

Hence, the option (A) is correct. [𝟏

𝟐 Mark]

12. If 𝐴 = [cos𝛼 − sin𝛼sin𝛼 cos𝛼

], and 𝐴 + 𝐴′ = 𝐼, then the value of 𝛼 is [2 marks]

(A) 𝜋

(B) 𝜋

(C) 𝜋

(D) 3𝜋

Solution:

Given that: 𝐴 + 𝐴′ = 𝐼

𝐴 = [cos𝛼 − sin𝛼sin𝛼 cos𝛼

Step 1:

𝐴′ = [cos𝛼 sin𝛼−sin𝛼 cos𝛼

] [𝟏

𝟐 Mark]

Step 2:

⇒ [𝑐𝑜𝑠𝛼 −𝑠𝑖𝑛𝛼𝑠𝑖𝑛𝛼 𝑐𝑜𝑠𝛼

] + [𝑐𝑜𝑠𝛼 𝑠𝑖𝑛𝛼−𝑠𝑖𝑛𝛼 𝑐𝑜𝑠𝛼

] = [1 00 1

⇒ [2𝑐𝑜𝑠𝛼 00 2𝑐𝑜𝑠𝛼

] = [1 00 1

] [𝟏

𝟐 Mark]

Step 3:

⇒ 2 cos 𝛼 = 1

⇒ cos 𝛼 =1

cos 𝛼 = cos 600 [∵ cos 600 =1

𝟐 Mark]

Step 4:

On comparing angles, we get

𝛼 = 60°

𝛼 = 60° ×𝜋

⇒ 𝛼 =𝜋

Hence, option (B) is correct. [𝟏

𝟐 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 −12 3

] [3 Marks]

Solution:

Let 𝐴 = [1 −12 3

Step 1:

We know that, 𝐴 = 𝐼𝐴, where 𝐼 = [1 00 1

⇒ [1 −12 3

] = [1 00 1

]𝐴 [𝟏

𝟐 Mark]

Step 2:

⇒ [1 −10 5

] = [1 0−2 1

] 𝐴 [applying 𝑅2 ⟶ 𝑅2 − 2𝑅1] [𝟏

𝟐 Mark]

Step 3:

⇒ [1 −10 1

] = [1 0

] 𝐴 [applying 𝑅2 ⟶1

5𝑅2] [

𝟐 Mark]

Step 4:

⇒ [1 00 1

] 𝐴 [applying 𝑅1 → 𝑅1 + 𝑅2] [𝟏

𝟐 Mark]

Step 5:

] 𝐴 [𝟏

𝟐 Mark]

Step 6:

∴ I = A−1A

Hence, 𝐴−1 = [

] [𝟏

𝟐 Mark]

2. [2 11 1

] [2 Marks]

Solution:

Let, 𝐴 = [2 11 1

Step 1:

⇒ [2 11 1

] [1 00 1

] 𝐴 [𝟏

𝟐 Mark]

Step 2:

⇒ [1 00 1

] = [1 −10 1

]𝐴 [applying 𝑅1 → 𝑅1 − 𝑅2] [𝟏

𝟐 Mark]

Step 3:

⇒ [1 00 1

] = [1 −1−1 2

𝟐 Mark]

Step 4:

∴ I = A−1A

Hence, 𝐴−1 = [1 −1

−1 2] [

𝟐 Mark]

3. [1 32 7

Solution:

Let, 𝐴 = [1 32 7

⇒ [1 32 7

] = [1 00 1

] 𝐴

⇒ [1 30 1

] = [1 0−2 1

]𝐴 [applying 𝑅2 → 𝑅2 − 2𝑅1]

⇒ [1 00 1

] = [7 −3−2 1

]𝐴 [applying 𝑅1 → 𝑅1 − 3𝑅2]

∴ I = A−1A

Hence, 𝐴−1 = [7 −3−2 1

4. [2 35 7

] [3 Marks]

Solution:

Let, 𝐴 = [2 35 7

Step 1:

⇒ [2 35 7

] = [1 00 1

] 𝐴 [𝟏

𝟐 Mark]

Step 2:

⇒ [1 25 7

] = [3 −10 1

]𝐴 [applying 𝑅1 → 3𝑅1 − 𝑅2] [𝟏

𝟐 Mark]

Step 3:

⇒ [1 20 −3

] = [3 −1

−15 6]𝐴 [applying 𝑅2 → 𝑅2 − 5𝑅1] [

𝟐 Mark]

Step 4:

⇒ [1 20 1

] = [3 −15 −2

]𝐴 [applying 𝑅2 → −1

3𝑅2] [

𝟐 Mark]

Step 5:

⇒ [1 00 1

] = [−7 35 −2

]𝐴 [applying 𝑅1 → 𝑅1 − 2𝑅2] [𝟏

𝟐 Mark]

Step 6:

∴ I = A−1A

Hence, 𝐴−1 = [−7 35 −2

] [𝟏

𝟐 Mark]

5. [2 17 4

] [3 Marks]

Solution:

Let, 𝐴 = [2 17 4

Step 1:

⇒ [2 17 4

] = [1 00 1

] 𝐴 [𝟏

𝟐 Mark]

Step 2:

⇒ [1 07 4

] = [4 −10 1

]𝐴 [applying 𝑅1 → 4𝑅1 − 𝑅2] [𝟏

𝟐 Mark]

Step 3:

⇒ [1 00 4

] = [4 −1

−28 8]𝐴 [applying 𝑅2 → 𝑅2 − 7𝑅1] [

𝟐 Mark]

Step 4:

⇒ [1 00 1

] = [4 −1

−7 2]𝐴 [applying 𝑅2 →

4𝑅2] [

𝟐 Mark]

Step 5:

∴ I = A−1A

Hence, 𝐴−1 = [4 −1

−7 2]1`

6. [2 51 3

] [3 marks]

Solution:

Let, 𝐴 = [2 51 3

Step 1:

⇒ [2 51 3

] = [1 00 1

] 𝐴 [𝟏

𝟐 Mark]

Step 2:

⇒ [1 21 3

] = [1 −10 1

𝟐 Mark]

Step 3:

⇒ [1 20 1

] = [1 −1

−1 2]𝐴 [applying 𝑅2 → 𝑅2 − 𝑅1] [

𝟐 Mark]

Step 4:

⇒ [1 00 1

] = [3 −5

−1 2]𝐴 [applying 𝑅1 → 𝑅1 − 2𝑅2] [

𝟐 Mark]

Step 5:

∴ I = A−1A

Hence, 𝐴−1 = [3 −5

−1 2]

7. [3 15 2

] [3 Marks]

Solution:

Let, 𝐴 = [3 15 2

Step 1:

⇒ [3 15 2

] = [1 00 1

] 𝐴 [𝟏

𝟐 Mark]

Step 2:

⇒ [1 05 2

] = [2 −10 1

]𝐴 [Applying 𝑅1 → 2𝑅1 − 𝑅2] [𝟏

𝟐 Mark]

Step 3:

⇒ [1 00 2

] = [2 −1

−10 6]𝐴 [Applying 𝑅2 → 2𝑅2 − 5𝑅1] [

𝟐 Mark]

Step 4:

⇒ [1 00 2

] = [2 −1

−5 3]𝐴 [Applying 𝑅2 →

2𝑅2] [

𝟐 Mark]

Step 5:

∴ I = A−1A

Hence, 𝐴−1 = [2 −1

−5 3]

8. [4 53 4

] [3 Marks]

Solution:

Let, 𝐴 = [4 53 4

Step 1:

⇒ [4 53 4

] = [1 00 1

] 𝐴 [𝟏

𝟐 Mark]

Step 2:

⇒ [1 13 4

] = [1 −10 1

𝟐 Mark]

Step 3:

⇒ [1 10 1

] = [1 −1

−3 4]𝐴 [applying 𝑅2 → 𝑅2 − 3𝑅1] [

𝟐 Mark]

Step 4:

⇒ [1 00 1

] = [4 −5

−3 4]𝐴 [applying R1 → 𝑅1 − R2] [

𝟐 Mark]

Step 5:

∴ I = A−1A

Hence, 𝐴−1 = [4 −5

−3 4]

9. [3 102 7

] [3 Marks]

Solution:

Let, 𝐴 = [3 102 7

Step 1:

⇒ [3 102 7

] = [1 00 1

] 𝐴 [𝟏

𝟐 Mark]

Step 2:

⇒ [1 32 7

] = [1 −10 1

𝟐 Mark]

Step 3:

⇒ [1 30 1

] = [1 −1

−2 3]𝐴 [applying 𝑅2 → 𝑅2 − 2𝑅1] [

𝟐 Mark]

Step 4:

⇒ [1 00 1

] = [7 −10

−2 3]𝐴 [applying 𝑅1 → 𝑅1 − 3𝑅2] [

𝟐 Mark]

Step 5:

∴ I = A−1A

Hence, 𝐴−1 = [7 −10

−2 3]

10. [3 −1−4 2

] [3 Marks]

Solution:

Let, 𝐴 = [3 −1−4 2

Step 1:

⇒ [3 −1

−4 2] = [

1 00 1

] 𝐴 [𝟏

𝟐 Mark]

Step 2:

⇒ [1 1

−4 2] = [

3 20 1

]𝐴 [applying 𝑅1 → 3𝑅1 + 2𝑅2] [𝟏

𝟐 Mark]

Step 3:

⇒ [1 10 6

] = [3 212 9

] 𝐴 [applying 𝑅2 → 𝑅2 + 4𝑅1] [𝟏

𝟐 Mark]

Step 4:

⇒ [1 10 1

] = [3 2

] 𝐴 [applying 𝑅2 →1

6𝑅2] [

𝟐 Mark]

Step 5:

⇒ [1 00 1

] = [1

] 𝐴 [applying 𝑅1 → 𝑅1 − 𝑅2] [𝟏

𝟐 Mark]

Step 6:

∴ I = A−1A

Hence, 𝐴−1 = [1

11. [2 −61 −2

] [3 Marks]

Solution:

Let, 𝐴 = [2 −61 −2

Step 1:

⇒ [2 −61 −2

] = [1 00 1

]𝐴 [𝟏

𝟐 Mark]

Step 2:

⇒ [1 −41 −2

] = [1 −10 1

]𝐴 [applying R1 → 𝑅1-R2] [𝟏

𝟐 Mark]

Step 3:

⇒ [1 −40 2

] = [1 −1

−1 2]𝐴 [applying R2 → 𝑅2-R1] [

𝟐 Mark]

Step 4:

⇒ [1 −40 1

] = [1 −1

21 ]𝐴 [applying 𝑅2 →

2𝑅2] [

𝟐 Mark]

Step 5:

⇒ [1 00 1

] = [−1 3

21]𝐴 [applyingR1 → 𝑅1-4R2] [

𝟐 Mark]

Step 6:

∴ I = A−1A

Hence, 𝐴−1 = [−1 3

12. [6 −3−2 1

] [2 Marks]

Solution:

Let, 𝐴 = [6 −3−2 1

Step 1:

⇒ [6 −3

−2 1] = [

1 00 1

] 𝐴 [𝟏

𝟐 Mark]

Step 2:

⇒ [6 −30 0

] = [1 03 1

]𝐴 [applying 𝑅2 → 𝑅2 + 3𝑅1] [𝟏

𝟐 Mark]

Step 3:

Since, on left hand side of the matrix all the elements of second row is zero.

Hence, 𝐴−1 does not exist.

13. [2 −3−1 2

] [3 marks]

Solution:

Let, 𝐴 = [2 −3−1 2

Step 1:

We know that, 𝐴 = 𝐼𝐴 where 𝐼 = [1 00 1

⇒ [2 −3

−1 2] = [

1 00 1

] 𝐴 [𝟏

𝟐 Mark]

Step 2:

⇒ [1 −1

−1 2] = [

1 10 1

𝟐 Mark]

Step 3:

⇒ [1 −10 1

] = [1 11 2

]𝐴 [applying 𝑅2 → 𝑅2 + 𝑅1] [𝟏

𝟐 Mark]

Step 4:

⇒ [1 00 1

] = [2 31 2

𝟐 Mark]

Step 5:

∴ I = A−1A

Hence, 𝐴−1 = [2 31 2

14. [2 14 2

] [2 marks]

Solution:

Let 𝐴 = [2 14 2

Step 1:

⇒ [2 14 2

] = [1 00 1

] 𝐴 [𝟏

𝟐 Mark]

Step 2:

⇒ [2 10 0

] = [1 0

−2 1]𝐴 [ applying 𝑅2 → 𝑅2 − 2𝑅1] [

𝟐 Mark]

Step 3:

Since, on left hand side of the matrix all the elements of second row is zero.

Hence 𝐴−1 does not exist.

15. [2 −3 32 2 33 −2 2

] [5 Marks]

Solution:

Let, 𝐴 = [2 −3 32 2 33 −2 2

Step 1:

We know that, 𝐴 = 𝐼𝐴, where 𝐼 = [1 0 00 1 00 0 1

⇒ [2 −3 32 2 33 −2 2

] = [1 0 00 1 00 0 1

]𝐴 [𝟏

𝟐 Mark]

Step 2:

⇒ [1 1 42 2 33 −2 2

] = [1 1 −10 1 00 0 1

]𝐴 [applying 𝑅1 → 𝑅1 + 𝑅2 − 𝑅3] [𝟏

𝟐 Mark]

Step 3:

⇒ [1 1 40 0 −53 −2 2

] = [1 1 −1

−2 −1 20 0 1

𝟐 Mark]

Step 4:

⇒ [1 1 40 0 −50 −5 −10

] = [1 1 −1

−2 −1 2−3 −3 4

𝟐 Mark]

Step 5:

⇒ [1 1 40 −5 −100 0 −5

] = [1 1 −1

−3 −3 4−2 −1 2

]𝐴 [applying 𝑅2 ↔ 𝑅3] [𝟏

𝟐 Mark]

Step 6:

⇒ [1 1 40 1 20 0 1

1 1 −13

] 𝐴 [applying 𝑅2 → −1

5𝑅2, 𝑅3 → −

5𝑅3] [

𝟐 Mark]

Step 7:

⇒ [1 1 40 1 00 0 1

1 1 −1

𝟐 Mark]

Step 8:

⇒ [1 0 40 1 00 0 1

𝐴 [applying 𝑅1 → 𝑅1 − 𝑅2] [𝟏

𝟐 Mark]

Step 9:

⇒ [1 0 00 1 00 0 1

𝐴 [applying R1 → 𝑅1 − 4R3] [𝟏

𝟐 Mark]

Step 10:

∴ I = A−1A

Hence,𝐴−1 =

𝟐 Mark]

16. [1 3 −2

−3 0 −52 5 0

] [5 Marks]

Solution:

Let, 𝐴 = [1 3 −2

−3 0 −52 5 0

Step 1:

⇒ [1 3 −2

−3 0 −52 5 0

] = [1 0 00 1 00 0 1

]𝐴 [𝟏

𝟐 Mark]

Step 2:

⇒ [1 3 −20 9 −112 5 0

] = [1 0 03 1 00 0 1

]𝐴 [applying 𝑅2 → 𝑅2 + 3𝑅1] [𝟏

𝟐 Mark]

Step 3:

⇒ [1 3 −20 9 −110 −1 4

] = [1 0 03 1 0

−2 0 1]𝐴 [applying R3 → 𝑅3-2R1] [

𝟐 Mark]

Step 4:

⇒ [1 3 −20 9 −110 0 25

] = [1 0 03 1 0

−15 1 9]𝐴 [applying R3 → 9R3+R2] [

𝟐 Mark]

Step 5:

⇒ [1 3 −20 9 −110 0 1

1 0 03 1 0

] 𝐴 [applying 𝑅3 →1

25𝑅3] [

𝟐 Mark]

Step 6:

⇒ [1 3 −20 9 00 0 1

] 𝐴 [applying 𝑅2 → 𝑅2 + 11𝑅3] [𝟏

𝟐 Mark]

Step 7:

⇒ [1 3 −20 1 00 0 1

]𝐴 [applying 𝑅2 →1

9𝑅2] [

𝟐 Mark]

Step 8:

⇒ [1 3 00 1 00 0 1

𝐴 [applying R1 → 𝑅1+2R3] [𝟏

𝟐 Mark]

Step 9:

⇒ [1 0 00 1 00 0 1

[ 1 −

𝐴 [applying R1 → 𝑅1-3R2] [𝟏

𝟐 Mark]

Step 10:

∴ I = A−1A

Hence, 𝐴−1 =

[ 1 −

𝟐 Mark]

17. [2 0 −15 1 00 1 3

] [5 Marks]

Solution:

Let 𝐴 = [2 0 −15 1 00 1 3

Step 1:

⇒ [2 0 −15 1 00 1 3

] = [1 0 00 1 00 0 1

]𝐴 [𝟏

𝟐 Mark]

Step 2:

⇒ [1 −1 −35 1 00 1 3

] = [3 −1 00 1 00 0 1

]𝐴 [applying R1 → 3R1-R2] [𝟏

𝟐 Mark]

Step 3:

⇒ [1 −1 −30 6 150 1 3

] = [3 −1 0

−15 6 00 0 1

]𝐴 [applyingR2 → 𝑅2-5R1] [𝟏

𝟐 Mark]

Step 4:

⇒ [1 −1 −30 6 150 0 3

] = [3 −1 0

−15 6 015 −6 6

]𝐴 [applyingR3 → 6R3-R2] [𝟏

𝟐 Mark]

Step 5:

⇒ [1 −1 −30 6 150 0 1

] = [3 −1 0

−15 6 05 −2 2

3𝑅3] [

𝟐 Mark]

Step 6:

⇒ [1 −1 −30 6 00 0 1

] = [3 −1 0

−90 36 −305 −2 2

]𝐴 [applying R2 → 𝑅2-15R3] [𝟏

𝟐 Mark]

Step 7:

⇒ [1 −1 −30 1 00 0 1

] = [3 −1 0

−15 6 −55 −2 2

6𝑅2] [

𝟐 Mark]

Step 8:

⇒ [1 −1 00 1 00 0 1

] = [18 −7 6

−15 6 −55 −2 2

]𝐴 [applying R1 → 𝑅1+3R3] [𝟏

𝟐 Mark]

Step 9:

⇒ [1 0 00 1 00 0 1

] = [3 −1 1

−15 6 −55 −2 2

]𝐴 [applying R1 → 𝑅1+R2] [𝟏

𝟐 Mark]

Step 10:

∴ I = A−1A

Hence, 𝐴−1 = [3 −1 1

−15 6 −55 −2 2

] [𝟏

𝟐 Mark]

18. Matrices 𝐴 and 𝐵 will be inverse of each other only if [2 Marks]

(A) 𝐴𝐵 = 𝐵𝐴

(B) 𝐴𝐵 = 𝐵𝐴 = 0

(C) 𝐴𝐵 = 0, 𝐵𝐴 = 𝐼

(D) 𝐴𝐵 = 𝐵𝐴 = 𝐼

Solution:

Given:

A and B will be inverse of each other.

𝐴−1 = 𝐵 and 𝐵−1 = 𝐴

Step 1:

We know that,

𝐴𝐴−1 = 𝐼 [∵ A−1 = B]

𝐴𝐵 = 𝐼

Step 2:

𝐵𝐵−1 = 𝐼 [∵ B−1 = A]

𝐵𝐴 = 𝐼

So, 𝐴𝐵 = 𝐵𝐴 = 𝐼.

Hence, option (D) is correct.

Miscellaneous Exercise on Chapter 3

1. Let 𝐴 = [0 10 0

], show that (𝑎𝐼 + 𝑏𝐴)𝑛 = 𝑎𝑛𝐼 + 𝑛𝑎𝑛−1𝑏𝐴, where 𝐼 is the identity matrix

of order 2 and 𝑛 ∈ 𝑁. [5 marks]

Solution:

Given: 𝐴 = [0 10 0

To Prove:

(𝑎𝐼 + 𝑏𝐴)𝑛 = 𝑎𝑛𝐼 + 𝑛𝑎𝑛−1𝑏𝐴

Proof: The proof is by mathematical induction [𝟏

𝟐 Mark]

Step 1:

Consider, 𝑃(𝑛): (𝑎𝑙 + 𝑏𝐴)𝑛 = 𝑎𝑛𝐼 + 𝑛𝑎𝑛−1𝑏𝐴 [𝟏

𝟐 Mark]

Step 2:

∴ P(1): (𝑎𝐼 + 𝑏𝐴)1 = 𝑎𝐼 + 𝑏𝐴

Hence, the result is true for 𝑛 = 1. [𝟏

𝟐 Mark]

Step 3:

Let the result be true for 𝑛 = 𝑘,

So, 𝑃(𝑘): (𝑎𝑙 + 𝑏𝐴)𝑘 = 𝑎𝑘𝐼 + 𝑛𝑎𝑘−1𝑏𝐴 [𝟏

𝟐 Mark]

Step 4:

We must prove that it is true for 𝑛 = 𝑘 + 1 also.

(i.e.) 𝑃(𝑘 + 1): (𝑎𝑙 + 𝑏𝐴)𝑘+1 = 𝑎𝑘+1𝐼 + (𝑘 + 1)𝑎𝑘𝑏𝐴 [𝟏

𝟐 Mark]

Step 5:

By taking LHS, we get

LHS= (𝑎𝐼 + 𝑏𝐴)𝑘+1

= (𝑎𝐼 + 𝑏𝐴)𝑘(𝑎𝐼 + 𝑏𝐴) [𝟏

𝟐 Mark]

Step 6:

= (𝑎𝑘𝐼 + 𝑛𝑎𝑘−1𝑏𝐴)(𝑎𝐼 + 𝑏𝐴) [∵ (𝑎𝐼 + 𝑏𝐴)𝑘+1 = 𝑎𝑘𝐼 + 𝑛𝑎𝑘−1𝑏𝐴]

𝟐 Mark]

Step 7:

= 𝑎𝑘+1𝐼2 + 𝑎𝑘𝐼𝑏𝐴 + 𝑛𝑎𝑘𝐼𝑏𝐴 + 𝑛𝑎𝑘−1𝑏2𝐴2 [𝟏

𝟐 Mark]

Step 8:

= 𝑎𝑘+1𝐼 + (𝑘 + 1)𝑎𝑘𝑏𝐴 [∵ 𝐴2 = 𝐴 ⋅ 𝐴 = [0 10 0

] [0 10 0

] = [0 00 0

] = 0]

Therefore, the result is true for 𝑛 = 𝑘 + 1.

Hence, by the Principle of Mathematical Induction, (𝑎𝐼 + 𝑏𝐴)𝑛 = 𝑎𝑛𝐼 + 𝑛𝑎𝑛−1𝑏𝐴 is true for all-natural numbers 𝑛.

2. If 𝐴 = [1 1 11 1 11 1 1

], prove that 𝐴𝑛 = [3𝑛−1 3𝑛−1 3𝑛−1

3𝑛−1 3𝑛−1 3𝑛−1

], 𝑛 ∈ 𝑁. [5 Marks]

Solution:

Given:

𝐴 = [1 1 11 1 11 1 1

To Prove:

𝐴𝑛 = [3𝑛−1 3𝑛−1 3𝑛−1

3𝑛−1 3𝑛−1 3𝑛−1

], 𝑛 ∈ 𝑁

Proof: The proof is by mathematical induction. [𝟏

𝟐 Mark]

Step 1:

Consider, 𝑝(𝑛): 𝐴𝑛 = [3𝑛−1 3𝑛−1 3𝑛−1

3𝑛−1 3𝑛−1 3𝑛−1

], [𝟏

𝟐 Mark]

Step 2:

Therefore, 𝑃(1): 𝐴1 = [30 30 30

30 30 30

] = [1 1 11 1 11 1 1

] = 𝐴

𝟐 Mark]

Step 3:

Let the result is true for 𝑛 = 𝑘,

So, 𝑃(𝑘): 𝐴𝑘 = [3𝑘−1 3𝑘−1 3𝑘−1

3𝑘−1 3𝑘−1 3𝑘−1

] [𝟏

𝟐 Mark]

Step 4:

We have to prove that it is true for 𝑛 = 𝑘 + 1 also.

(i.e.) 𝑃(𝑘 + 1) : 𝐴𝑘+1 = [3𝑘 3𝑘 3𝑘

3𝑘 3𝑘 3𝑘

] [𝟏

𝟐 Mark]

Step 5:

= 𝐴𝑘+1 = 𝐴𝑘𝐴

= [3𝑘−1 3𝑘−1 3𝑘−1

3𝑘−1 3𝑘−1 3𝑘−1

] [1 1 11 1 11 1 1

] [𝟏

𝟐 Mark]

Step 6:

= [3𝑘−1 + 3𝑘−1 + 3𝑘−1 3𝑘−1 + 3𝑘−1 + 3𝑘−1 3𝑘−1 + 3𝑘−1 + 3𝑘−1

3𝑘−1 + 3𝑘−1 + 3𝑘−1 3𝑘−1 + 3𝑘−1 + 3𝑘−1 3𝑘−1 + 3𝑘−1 + 3𝑘−1

] [𝟏

𝟐 Mark]

Step 7:

= [3. 3𝑘−1 3. 3𝑘−1 3. 3𝑘−1

3. 3𝑘−1 3. 3𝑘−1 3. 3𝑘−1

] [𝟏

𝟐 Mark]

Step 8:

= [3𝑘 3𝑘 3𝑘

3𝑘 3𝑘 3𝑘

Hence, by the Principle of Mathematical Induction, 𝐴𝑛 = [3𝑛−1 3𝑛−1 3𝑛−1

3𝑛−1 3𝑛−1 3𝑛−1

] is true

for all natural numbers 𝑛.

3. If 𝐴 = [3 −41 −1

], then prove that 𝐴𝑛 = [1 + 2𝑛 −4𝑛𝑛 1 − 2𝑛

], where 𝑛 is any positive

integer. [5 Marks]

Solution:

Given:

𝐴 = [3 −41 −1

To Prove:

𝐴𝑛 = [1 + 2𝑛 −4𝑛𝑛 1 − 2𝑛

Proof: The proof is by mathematical induction. [𝟏

𝟐 Mark]

Step 1:

Consider, 𝑃(𝑛) : 𝐴𝑛 = [1 + 2𝑛 −4𝑛

𝑛 1 − 2𝑛] [

𝟐 Mark]

Step 2:

Therefore, 𝑃(1) : 𝐴1 = [1 + 2(1) −4(1)

1 1 − 2(1)] = [

3 −41 −1

] = 𝐴

𝟐 Mark]

Step 3:

Let the result is true for 𝑛 = 𝑘, therefore,

𝑃(𝑘) : 𝐴𝑘 = [1 + 2𝑘 −4𝑘

𝑘 1 − 2𝑘] [

𝟐 Mark]

Step 4:

We have to prove that it is true for 𝑛 = 𝑘 + 1 also.

(i.e.) 𝑃(𝑘 + 1) : 𝐴𝑘+1 = [1 + 2(𝑘 + 1) −4(𝑘 + 1)

𝑘 + 1 1 − 2(𝑘 + 1)] [

𝟐 Mark]

Step 5:

LHS= 𝐴𝑘+1 = 𝐴𝑘𝐴

= [1 + 2𝑘 −4𝑘

𝑘 1 − 2𝑘] [

3 −41 −1

] [𝟏

𝟐 Mark]

Step 6:

= [3 + 6𝑘 − 4𝑘 −4 − 8𝑘 + 4𝑘3𝑘 + 1 − 2𝑘 −4𝑘 − 1 + 2𝑘

] [𝟏

𝟐 Mark]

Step 7:

= [1 + (2𝑘 + 2) −4𝑘 − 4

𝑘 + 1 1 − (2𝑘 + 2)] [

𝟐 Mark]

Step 8:

= [1 + 2(𝑘 + 1) −4(𝑘 + 1)

𝑘 + 1 1 − 2(𝑘 + 1)] =RHS

Hence, by the Principle of Mathematical Induction, 𝐴𝑛 = [1 + 2𝑛 −4𝑛

𝑛 1 − 2𝑛] is true for

any positive integer numbers 𝑛.

4. If 𝐴 and 𝐵 are symmetric matrices, prove that 𝐴𝐵 − 𝐵𝐴 is a skew symmetric matrix. [2 Marks]

Solution:

Given: 𝐴 and 𝐵 are symmetric matrices

∴ 𝐴′ = 𝐴 and 𝐵′ = 𝐵

Step 1:

(𝐴𝐵 − 𝐵𝐴)′ = (𝐴𝐵)′ − (𝐵𝐴)′ [∵ (𝑋 ‐ 𝑌)′ = 𝑋′ − 𝑌′] [𝟏

𝟐 Mark]

Step 2:

= 𝐵′𝐴′ − 𝐴′𝐵′ [∵ (𝐴𝐵)′ = 𝐵′𝐴′] [𝟏

𝟐 Mark]

Step 3:

= 𝐵𝐴 − 𝐴𝐵 [Given 𝐴′ = 𝐴,𝐵′ = 𝐵] [𝟏

𝟐 Mark]

Step 4:

= −(𝐴𝐵 − 𝐵𝐴)

⇒ (𝐴𝐵 − 𝐵𝐴)′ = −(𝐴𝐵 − 𝐵𝐴), [𝟏

𝟐 Mark]

Therefore, 𝐴𝐵‐𝐵𝐴 is a skew symmetric matrix.

Hence proved.

5. Show that the matrix 𝐵′𝐴𝐵 is symmetric or skew symmetric according as 𝐴 is symmetric or skew symmetric. [𝟒 Marks]

Solution:

Step 1:

If 𝐴 is a symmetric matrix, then A′ = 𝐴

Now, (𝐵′𝐴𝐵)′ = (𝐴𝐵)′(𝐵′)′ [∵ (𝐴𝐵)′ = 𝐵′𝐴′] [𝟏

𝟐 Mark]

Step 2:

= (𝐴𝐵)′𝐵 [∵ (𝐵′)′ = 𝐵] [𝟏

𝟐 Mark]

Step 3:

= 𝐵′𝐴′𝐵 [∵ (𝐴𝐵)′ = 𝐵′𝐴′] [𝟏

𝟐 Mark]

Step 4:

= 𝐵′𝐴𝐵 [Given A′ = 𝐴]

⇒ (𝐵′𝐴𝐵)′ = B′𝐴𝐵,

Therefore, the matrix 𝐵′𝐴𝐵 is also symmetric matrix. [𝟏

𝟐 Mark]

Step 5:

If 𝐴 is skew symmetric matrix, then 𝐴′ = −𝐴

Here, (𝐵′𝐴𝐵)′ = (𝐴𝐵)′(𝐵′)′ [∵ (𝐴𝐵)′ = 𝐵′𝐴′] [𝟏

𝟐 Mark]

Step 6:

= (𝐴𝐵)′𝐵 [∵ (𝐵′)′ = B] [𝟏

𝟐 Mark]

Step 7:

= B′A′B [∵ (𝐴𝐵)′ = 𝐵′𝐴′] [𝟏

𝟐 Mark]

Step 8:

= −𝐵′𝐴𝐵 [∵ Given 𝐴′ = −𝐴]

⇒ (𝐵′𝐴𝐵)′ = −B′𝐴B,

Hence, the matrix 𝐵′𝐴𝐵 is also a skew‐symmetric matrix. [𝟏

𝟐 Mark]

6. Find the values of 𝑥, 𝑦, 𝑧 if the matrix 𝐴 = [0 2𝑦 𝑧𝑥 𝑦 −𝑧𝑥 −𝑦 𝑧

] satisfy the equation 𝐴′𝐴 = 𝐼.

[4 Marks]

Solution:

Given:

𝐴 = [0 2𝑦 𝑧𝑥 𝑦 −𝑧𝑥 −𝑦 𝑧

𝐴′𝐴 = 𝐼

Step 1:

⇒ [0 2𝑦 𝑧𝑥 𝑦 −𝑧𝑥 −𝑦 𝑧

] [0 𝑥 𝑥2𝑦 𝑦 −𝑦𝑧 −𝑧 𝑧

] = [1 0 00 1 00 0 1

] [𝟏

𝟐 Mark]

Step 2:

0 + 4𝑦2 + 𝑧2 0 + 2𝑦2 − 𝑧2 0 − 2𝑦2 + 𝑧2

0 + 2𝑦2 − 𝑧2 𝑥2 + 𝑦2 + 𝑧2 𝑥2 − 𝑦2 − 𝑧2

0 − 2𝑦2 + 𝑧2 𝑥2 − 𝑦2 − 𝑧2 𝑥2 + 𝑦2 + 𝑧2

] = [1 0 00 1 00 0 1

Step 3:

So, on comparing the corresponding elements, we get

4𝑦2 + 𝑧2 = 1

2𝑦2 − 𝑧2 = 0

𝑥2 + 𝑦2 + 𝑧2 = 1

Step 4:

On solving we get 𝑥 = ±1

√2, 𝑦 = ±

√6 and 𝑧 = ±

√3 [1

𝟐 Marks]

Hence, the values of 𝑥 = ±1

√2, 𝑦 = ±

√6 and 𝑧 = ±

7. For what values of 𝑥: [1 2 1] [1 2 02 0 11 0 2

] [02𝑥] = 𝑂? [3 Marks]

Solution:

Given that:[1 2 1] [1 2 02 0 11 0 2

] [02𝑥] = 𝑂

Step 1:

⇒ [1 + 4 + 1 2 + 0 + 0 0 + 2 + 2] [02𝑥] = 𝑂

Step 2:

⇒ [6 2 4] [02𝑥] = 𝑂 [

𝟐 Mark]

Step 3:

[6(0) + 2(2) + 4(𝑥)] = 0

⇒ [0 + 4 + 4𝑥] = [0] [𝟏

𝟐 Mark]

Step 4:

⇒ 4 + 4𝑥 = 0 [𝟏

𝟐 Mark]

Step 5:

Upon solving the above obtained equation, we get:

⇒ 𝑥 = −1

Hence, the required value of 𝑥 is − 1 [𝟏

𝟐 Mark]

8. If 𝐴 = [3 1−1 2

], show that 𝐴2 − 5𝐴 + 7𝐼 = 0. [3 Marks]

Solution:

Given:

𝐴 = [3 1−1 2

Step 1:

By taking LHS= 𝐴2 − 5𝐴 + 7𝐼

= [3 1−1 2

] [3 1−1 2

] − 5 [3 1−1 2

] + 7 [1 00 1

Step 2:

= [9 − 1 3 + 2

−3 − 2 −1 + 4] − [

15 5−5 10

] + [7 00 7

= [8 5

−5 3] − [

15 5−5 10

] + [7 00 7

Step 3:

= [8 − 15 + 7 5 − 5 + 0−5 + 5 + 0 3 − 10 + 7

= [0 00 0

] = 𝑂

∴ 𝐿𝐻𝑆 = 𝑅𝐻𝑆

𝐴2 − 5𝐴 + 7𝐼 = 0

Hence it is proved.

9. Find 𝑥, if [𝑥 − 5 − 1] [1 0 20 2 12 0 3

] [𝑥41] = 0 [3 marks]

Solution:

Given: [𝑥 − 5 − 1] [1 0 20 2 12 0 3

] [𝑥41] = 0

Step 1:

[𝑥 − 5 − 1] [1 0 20 2 12 0 3

] [𝑥41] = 0

⇒ [𝑥 + 0 − 2 0 − 10 + 0 2𝑥 − 5 − 3] [𝑥41] = 𝑂

⇒ [𝑥 − 2 − 10 2𝑥 − 8] [𝑥41] = 𝑂

Step 2:

⇒ [𝑥2 − 2𝑥 − 40 + 2𝑥 − 8] = [0]

Step 3:

⇒ 𝑥2 = 48

⇒ 𝑥 = ±4√3

Hence, the required value of 𝑥 is ± 4√3

10. A manufacturer produces three products 𝑥, 𝑦, 𝑧 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 𝑥, 𝑦 and 𝑧 are ₹ 2.50, ₹ 1.50 and ₹ 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 ₹ 2.00, ₹ 1.00 and 50 paise respectively. Find the gross profit. [3 Marks]

Solution:

(a) Step 1:

If unit sale prices of 𝑥, 𝑦 and 𝑧 are ₹2.50, ₹1.50 and ₹1.00, then

Products Selling price

Market I

Market II

𝑥 𝑦 𝑧

[10000 2000 180006000 20000 8000

[₹2.50₹1.50₹1.00

Step 2:

Total revenue of each market = Total product × Unit selling price

= [10000 2000 180006000 20000 8000

] [₹2.50₹1.50₹1.00

Step 3:

= [₹25000 + ₹3000 + ₹18000₹15000 + ₹30000 + ₹8000

] = [₹46000₹53000

Thus, the revenue of market I is ₹46,000 and that of the market II is ₹53,000.

(b) Step 1:

If the unit costs of the three commodities are ₹2.00, ₹1.00 and 50 paise, then

Products Selling price

Market I

Market II

𝑥 𝑦 𝑧

[10000 2000 180006000 20000 8000

[₹2.50₹1.50₹0.00

Step 2:

Gross profit from each market = Total product × Unit selling price

= [10000 2000 180006000 20000 8000

] [₹2.50₹1.50₹1.00

= [₹20000 + ₹2000 + ₹9000₹12000 + ₹20000 + ₹4000

] = [₹31000₹36000

Step 3:

Hence, the total revenue from market I is ₹46,000 and that of the market II is ₹31,000.

Therefore, the gross profit of market I=Revenue – Cost

= ₹46000 − ₹31000 = ₹15000

Therefore, the gross profit of market II = ₹53000 − ₹36000 = ₹17000

Hence, the total gross profit from market I is ₹15,000 and that of the market II is ₹17,000.

11. ind the matrix 𝑋 so that 𝑋 [1 2 34 5 6

] = [−7 −8 −92 4 6

] [5 Marks]

Solution:

Given:

𝑋 [1 2 34 5 6

= [−7 −8 −92 4 6

Step 1:

Let, 𝑋 = [𝑎 𝑏𝑐 𝑑

] [𝟏

𝟐 Mark]

Step 2:

Therefore, 𝑋 [1 2 34 5 6

] = [−7 −8 −92 4 6

⇒ [𝑎 𝑏𝑐 𝑑

] [1 2 34 5 6

] = [−7 −8 −92 4 6

] [𝟏

𝟐 Mark]

Step 3:

⇒ [𝑎 + 4𝑏 2𝑎 + 5𝑏 3𝑎 + 6𝑏𝑐 + 4𝑑 2𝑐 + 5𝑑 3𝑐 + 6𝑑

] = [−7 −8 −92 4 6

Step 4:

𝑎 + 4𝑏 = −7 [𝟏

𝟐 Mark]

2𝑎 + 5𝑏 = −8 [𝟏

𝟐 Mark]

𝑐 + 4𝑑 = 2 [𝟏

𝟐 Mark]

2𝑐 + 5𝑑 = 4 [𝟏

𝟐 Mark]

Step 5:

On solving, we get 𝑎 = 1, 𝑏 = −2, 𝑐 = 2 and 𝑑 = 0

Hence, 𝑋 = [1 −22 0

12. 𝐴 and 𝐵 are square matrices of the same order such that 𝐴𝐵 = 𝐵𝐴, then prove by induction that 𝐴𝐵𝑛 = 𝐵𝑛𝐴. Further, prove that (𝐴𝐵)𝑛 = 𝐴𝑛𝐵𝑛 for all 𝑛 ∈ 𝑁. Choose the correct answer in the following questions: [5 marks]

Solution:

Given:

𝐴𝐵 = 𝐵𝐴

To prove:

𝐴𝐵𝑛 = 𝐵𝑛𝐴 and

(𝐴𝐵)𝑛 = 𝐴𝑛𝐵𝑛 for all 𝑛 ∈ 𝑁

Proof: The proof is by mathematical induction.

Step 1:

Consider, 𝑃(𝑛): 𝐴𝐵𝑛 = 𝐵𝑛𝐴, [𝟏

𝟐 Mark]

Step 2:

Therefore, 𝑃(1): 𝐴𝐵 = 𝐵𝐴

Hence the result is true for 𝑛 = 1. [𝟏

𝟐 Mark]

Step 3:

Let it be true for 𝑛 = 𝑘,

so, 𝑃(𝑘): 𝐴𝐵𝑘 = 𝐵𝑘𝐴 [𝟏

𝟐 Mark]

Step 4:

Now we have to prove that it is true for 𝑛 = 𝑘 + 1 also.

(i.e.) 𝑃(𝑘 + 1): 𝐴𝐵𝑘+1 = Bk+1𝐴 [𝟏

𝟐 Mark]

Step 5:

LHS= 𝐴𝐵𝑘+1

= 𝐴𝐵𝑘𝐵

= 𝐵𝑘𝐴𝐵 [∵ 𝐴𝐵𝑘 = 𝐵𝑘𝐴]

= 𝐵𝑘𝐵𝐴 [∵ 𝐴𝐵 = 𝐵𝐴]

= 𝐵𝑘+1𝐴 =RHS

Hence, the result is true for 𝑛 = 𝑘 + 1, [𝟏

𝟐 Mark]

Therefore, by the Principle of Mathematical Induction 𝐴𝐵𝑛 = 𝐵𝑛𝐴 is true for all natural numbers 𝑛.

Step 6:

If (𝐴𝐵)𝑛 = 𝐴𝑛𝐵𝑛

Consider 𝑃(𝑛): (𝐴𝐵)n = 𝐴𝑛𝐵𝑛 therefore, 𝑃(1): (𝐴𝐵)1 = 𝐴1𝐵1

Hence the result is true for 𝑛 = 1. [𝟏

𝟐 Mark]

Step 7:

Let it is true for 𝑛 = 𝑘,

so, 𝑃(𝑘): (𝐴𝐵)𝑘 = 𝐴𝑘𝐵𝑘 [𝟏

𝟐 Mark]

Step 8:

Now we have to prove that it is true for 𝑛 = 𝑘 + 1 also.

(i.e.) 𝑃(𝑘 + 1): (𝐴𝐵)𝑘+1 = 𝐴𝑘+1𝐵𝑘+1 [𝟏

𝟐 Mark]

Step 9:

LHS = (𝐴𝐵)𝑘+1

= (𝐴𝐵)k𝐴𝐵

= 𝐴𝑘𝐵𝑘𝐴𝐵 [∵ (𝐴𝐵)k = 𝐴𝑘𝐵𝑘] [𝟏

𝟐 Mark]

Step 10:

= 𝐴𝑘𝐴𝐵𝑘𝐵 [∵ 𝐴𝐵 = 𝐵𝐴]

= 𝐴𝑘+1𝐵𝑘+1 =RHS

Hence, the result is true for 𝑛 = 𝑘 + 1.

Therefore, by the Principle of Mathematical Induction (𝐴𝐵)𝑛 = 𝐴𝑛𝐵𝑛 is true for all

natural numbers 𝑛. [𝟏

𝟐 Mark]

13. If 𝐴 = [𝛼 𝛽𝛾 −𝛼

] is such that 𝐴2 = 𝐼, then [2 Marks]

(A) 1 + 𝛼2 + 𝛽𝛾 = 0

(B) 1 − 𝛼2 + 𝛽𝛾 = 0

(C) 1 − 𝛼2 − 𝛽𝛾 = 0

(D) 1 + 𝛼2 − 𝛽𝛾 = 0

Solution:

Given that: 𝐴2 = 𝐼

Step 1:

⇒ [𝛼 𝛽𝛾 −𝛼

] [𝛼 𝛽𝛾 −𝛼

] = [1 00 1

⇒ [𝛼2 + 𝛽𝛾 𝛼𝛽 − 𝛽𝛼

𝛼𝛾 − 𝛼𝛾 𝛽𝛾 + 𝛼2 ] = [1 00 1

Step 2:

𝛼2 + 𝛽𝛾 = 1

1 − 𝛼2 − 𝛽𝛾 = 0

Hence the option (C) is correct.

14. If the matrix 𝐴 is both symmetric and skew symmetric, then [2 Marks]

(A) 𝐴 is a diagonal matrix

(B) 𝐴 is a zero matrix

(C) 𝐴 is a square matrix

(D) None of these

Solution:

Given:

Matrix 𝐴 is both symmetric and skew symmetric

Step 1:

We know that only a zero matrix is always both symmetric and skew symmetric.

Proof:

∴ 𝐴′ = 𝐴 and 𝐴′ = −𝐴

Step 2:

On comparing both the equations, we get

⇒ 𝐴 = −𝐴

𝐴 + 𝐴 = 𝑂

2𝐴 = 𝑂

𝐴 = 𝑂

Hence, the option (B) is correct.

15. If 𝐴 is square matrix such that 𝐴2 = 𝐴, then (𝐼 + 𝐴)3 − 7 𝐴 is equal to [2 Marks]

(A) 𝐴

(B) 𝐼 − 𝐴

(C) 𝐼

(D) 3𝐴

Solution:

Given:

𝐴2 = 𝐴

Step 1:

(𝐼 + 𝐴)3 − 7𝐴

= 𝐼3 + 𝐴3 + 3𝐼2𝐴 + 3𝐼𝐴2 − 7𝐴 [𝟏

𝟐 Mark]

Step 2:

= −𝐼 + 𝐴2𝐴 + 31𝐴 + 31𝐴2 − 7𝐴 [∵ 𝐼3 = 𝐼2 = 𝐼] [𝟏

𝟐 Mark]

Step 3:

= 𝐼 + 𝐴𝐴 + 3𝐴 + 3𝐼𝐴 − 7𝐴 [∵ 𝐴2 = 𝐴] [𝟏

𝟐 Mark]

Step 4:

= 𝐼 + 𝐴 + 3𝐴 + 3𝐴 − 7𝐴

= 𝐼 + 7𝐴 − 7𝐴

= 𝐼 [∵ 𝐼𝐴 = 𝐴] [𝟏

𝟐 Mark]

Hence, the option (C) is correct

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