matrices - kar

Class XII, Mathematics, Chapter Matrices, Module-3

Distance Learning Programme: An AEES Initiative module3/1

Matrices

Problem 1. Construct a matrix A = [aij]2×2 whose elements aij are given by aij = e2ix sinjx .

For i = 1, j = 1, a11 = e2x sin x For i = 1, j = 2, a12 = e2x sin 2x For i = 2, j = 1, a21 = e4x sin x For i = 2, j = 2, a22 = e4x sin 2x

Thus A = [

]

Problem 2. If A =[

] , B = [

], C = [ ], D = [

],

, thenwhich of the sums A + B, B + C, C + D and B + D is defined? Solution Only B + D is defined since matrices of the same order can only be added.

Problem 3. Express the matrix A as the sum of a symmetric and a skew symmetric matrix, where

A =[

]

Solution

A =[

]

A’ = [

]

Thus A + A’ = [

], and A-A’ = [

]

So A =

( )

( )

=

[

] +

[

]

= [

] + [

]

Problem 4 If A, B are symmetric matrices of same order, then AB – BA is a (A) Skew symmetric matrix (B) Symmetric matrix (C) Zero matrix (D) Identity matrix

Solution:

Given A, B are symmetric matrices



Implies A’ = A and B’ = B

Now t check AB – BA

Lets take its transpose

(AB-BA)’ = (AB)’ – (BA)’

= B’A’- A’B’

= BA – AB

= - (AB – BA)

Thus AB – BA is a skew symmetric matrix.

Problem 5: Find the inverse of the matrix

2 -3 3

2 2 3

3 -2 2

Answer:

Let A = 2 -3 3

2 2 3

3 -2 2

We know that A = IA

=> 2 -3 3 1 0 0 A

2 2 3 = 0 1 0

3 -2 2 0 0 1

=> 2 -3 3 1 1 0 A

0 5 0 = -1 1 0 [R2 -> R2 – R1]

3 -2 2 0 0 1

=> 2 -3 3 1 0 -1 A

0 5 0 = -1/5 1/5 0 [R2 -> R2/5]

3 -2 2 0 0 1

=> -1 -1 1 1 0 -1 A

0 1 0 = -1/5 1/5 0 [R1 -> R1 - R3]

3 -2 2 0 0 1



=> -1 0 1 4/5 1/5 -1 A

0 1 0 = -1/5 1/5 0 [R1 -> R1 + R2 and R3 -> R3 + 2R2]

3 0 2 -2/5 2/5 1

=> -1 0 1 4/5 1/5 -1 A

0 1 0 = -1/5 1/5 0 [R3 -> R3 + 3R1]

0 0 5 2 1 -2

=> -1 0 1 4/5 1/5 -1 A

0 1 0 = -1/5 1/5 0 [R3 -> R3/5]

0 0 1 2/5 1/5 -2/5

=> -1 0 0 2/5 0 3/5 A

0 1 0 = -1/5 1/5 0 [R1 -> R1 – R3]

0 0 1 2/5 1/5 -2/5

=> 1 0 0 -2/5 0 3/5 A

0 1 0 = -1/5 1/5 0 [R1 -> (-1)R1]

0 0 1 2/5 1/5 -2/5

So, A-1 = -2/5 0 3/5

-1/5 1/5 0

2/5 1/5 -2/5

Problem 6:

If A = 0 -tan α/2

tan α/2 0 and I is the identity matrix of order 2, show that

I + A = (I - A) cos α -sin α

sin α cos α

Answer:



LHS:

I + A = 1 0 0 - tan α/2

0 1 + tan α/2 0

= 1 -tan α/2

tan α/2 1 ………………..1

RHS:

(I - A) cos α -sin α

sin α cos α

= 1 0 - 0 - tan α/2 cos α -sin α

0 1 tan α/2 0 sin α cos α

= 1 tan α/2 cos α -sin α

- tan α/2 1 sin α cos α

= cos α + sin α * tan α/2 -sin α + cos α * tan α/2

- cos α * tan α/2 + sin α sin α * tan α/2 + cos α …………..2

= 1 – 2sin2 α/2 + 2 sin α/2 sin α/2 tan α/2 -2 sin α/2 cos α/2 + (2cos2 α/2 – 1) * tanα/2

- (2cos2 α/2 - 1) tan α/2 + 2sin α/2 cos α/2 2 sin α/2 cos α/2 tan α/2 + 1 – 2 sin2 α/2 = 1 – 2sin2 α/2 + 2sin2 α/2 -2 sin α/2 cos α/2 + 2 sin α/2 sin α/2 tanα/2 - 2sin α/2 cos α/2 + tan α/2 + 2sin α/2 cos α/2 2 sin2 α/2 + 1 – 2 sin2 α/2

= 1 -tan α/2

tan α/2 1

Thus, from equation 1 and 2, we get

LHS = RHS

So, I + A = (I - A) cos α -sin α

sin α cos α



Problem 7: A trust fund has Rs 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 Rs 30,000 among the two

types of bonds. If the trust fund must obtain an annual total interest of:

(a) Rs 1,800 (b) Rs 2,000

Answer:

(a) Let Rs x be invested in the first bond. Then, the sum of money invested in the second bond

will be Rs (30000 − x).

It is given that the first bond pays 5% interest per year and the second bond pays 7% interest

per year.

Therefore, in order to obtain an annual total interest of Rs 1800, we have:

[x (30000 - x)][5/100] = 1800 [SI for 1 year = (Principal * Rate)/100]

[7/100]

=> 5x/100 + 7(30000 - x)/100 = 1800

=> 5x + 210000 – 7x = 180000

=> 210000 – 2x = 180000

=> 2x = 210000 – 180000

=> 2x = 30000

=> x = 15000

Thus, in order to obtain an annual total interest of Rs 1800, the trust fund should invest Rs

15000 in the first bond and the remaining Rs 15000 in the second bond.

(b) Let Rs x be invested in the first bond. Then, the sum of money invested in the second bond

will be Rs (30000 − x).



Therefore, in order to obtain an annual total interest of Rs 2000, we have:

[x (30000 - x)][5/100] = 2000 [SI for 1 year = (Principal * Rate)/100]

[7/100]

=> 5x/100 + 7(30000 - x)/100 = 2000

=> 5x + 210000 – 7x = 200000

=> 210000 – 2x = 200000

=> 2x = 210000 – 200000

=> 2x = 10000

=> x = 5000

Thus, in order to obtain an annual total interest of Rs 2000, the trust fund should invest Rs 5000 in the first bond and the remaining Rs 25000 in the second bond.

Problem 8:

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

Answer:

The bookshop has 10 dozen chemistry books, 8 dozen physics books, and 10 dozen economics books.

The selling prices of a chemistry book, a physics book, and an economics book are respectively given as Rs 80, Rs 60, and Rs 40.

The total amount of money that will be received from the sale of all these books can be

represented in the form of a matrix as:

12[10 8 10] [80]

[60]

[40]

= 12[10 * 80 + 8 * 60 + 10 * 40]



= 12[800 + 480 + 400]

= 12 * 1680

= 20160

Thus, the bookshop will receive Rs 20160 from the sale of all these books.

Problem 9:

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:

A.k = 3, p = n B. k is arbitrary, p = 2 C. p is arbitrary, k = 3 D. k = 2, p = 3

Answer:

Matrices P and Y are of the orders p * k and 3 * k respectively.

Therefore, matrix PY will be defined if k = 3.

Consequently, PY will be of the order p * k.

Matrices W and Y are of the orders n * 3 and 3 * k respectively.

Since the number of columns in W is equal to the number of rows in Y, matrix WY is well-

defined and is of the order n * k.

Matrices PY and WY can be added only when their orders are the same.

However, PY is of the order p * k and WY is of the order n * k.

Therefore, we must have p = n

Thus, k = 3 and p = n are the restrictions on n, k, and p so that will be defined.

Problem 10:

Let A = 0 1 ,show that (aI + bA)n = anI + nan-1bA, where I is the identity matrix of order 3

0 0

and n є N.



Answer:

Given, A = 0 0

0 1

We shall prove the result by using the principle of mathematical induction.

For n = 1, we have:

P(1): (aI + bA)1 = a1I + 1 * a1-1bA = aI + bA, n є N

Therefore, the result is true for n = 1.

Let the result be true for n = k.

That is,

P(k): (aI + bA)k = akI + kak-1bA

Now, we prove that the result is true for n = k + 1.

Consider (aI + bA)k+1 = (aI + bA)k * (aI + bA)

= (akI + kak-1bA) * (aI + bA)

= ak-1I + kakbAI + akbIA + kak-1b2A2

= ak-1I + (k + 1)akbA + kak-1b2A2 ………….1

Now,

A2 = 0 1 0 1 = 0 0 = O

0 0 0 0 0 0

From Equation 1, we have:

(aI + bA)k+1 = ak-1I + (k + 1)akbA + O

=> (aI + bA)k+1 = ak-1I + (k + 1)akbA

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

Thus, by the principle of mathematical induction, we have:



(aI + bA)n = anI + nan-1bA, where A = 0 1 , n є N

Problem 11:Show that the matrix B’AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.

Answer:

We suppose that A is a symmetric matrix,

then A’ = A .............1

Consider (B’AB)’ = {B’(AB)}’

= (AB)’(B’)’ [Since (AB)’ = B’A’]

= B’A’(B) [Since (B’)’ = B]

= B’(A’B)

= B’AB [From equation 1]

So, (B’AB)’ = B’AB

Thus, if A is a symmetric matrix, then B’AB is a symmetric matrix.

Now, we suppose that A is a skew-symmetric matrix.

Consider (B’AB)’ = {B’(AB)}’

= (AB)’(B’)’ [Since (AB)’ = B’A’]

= B’A’(B) [Since (B’)’ = B]

= B’(A’B)

= B’(-A)B [From equation 1]

= - B’AB

So, (B’AB)’ = -B’AB

Thus, if A is a skew-symmetric matrix, then B’AB is a skew-symmetric matrix.

Hence, if A is a symmetric or skew-symmetric matrix, then B’AB is a symmetric or skew



symmetric matrix accordingly.

Problem 12: If 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 є N.

Answer:

A and B are square matrices of the same order such that AB = BA.

For n = 1, we have:

P(1): AB = BA [Given]

=> AB1 = B1A



P(k): ABk = BkA ………….1


ABk+1 = ABk.B

= (BkA)B [From equation 1]

= Bk.(AB) [Associative rule]

= Bk.(BA) [Since AB = BA]

= (BkB)A

= Bk+1A


Thus, by the principle of mathematical induction, we have ABn = BnA, n є N.

Now, we prove that (AB)n = AnBn for all n є N

For n = 1, we have:

(AB)1 = A1B1 = AB





(AB)k = AkBk …………..2


(AB)k+1 = (AB)k .(AB)

= (AkBk).(AB) [From equation 2]

= Ak(Bk A)B [Associative rule]

= Ak(ABk)B [(AB)n = AnBn for all n є N]

= (Ak A)( Bk B) [Associative rule]

= Ak+1 Bk+1


Thus, by the principle of mathematical induction, we get (AB)n = An Bn, for all natural numbers.

Problem 13: If A is square matrix such that A2 = A then (I + A)3 – 7A is equal to

1. A B. I − A C. I D. 3A

Answer: C

(I + A)3 – 7A = I3 + A3 + 3I2A + 3IA2 – 7A

= I3 + A.A2 + 3I2A + 3IA2 – 7A

= I + A.A + 3A + 3A – 7A [Since A2 = A and I3 = I]

= I + A2 + 6A – 7A

= I + A – A [Since A2 = A]

= I

So, (I + A)3 – 7A = I



Problem 14: If

A = 1 0 2

0 2 1

2 0 3

Prove that A3 – 6A2 + 7A + 2I = 0

Answer:

We have A2 = A * A

=> A2 = 1 0 2 1 0 2

0 2 1 0 2 1

2 0 3 2 0 3

= 1 + 0 + 4 0 + 0 + 0 2 + 0 + 6

0 + 0 + 2 0 + 4 + 0 0 + 2 + 3

2 + 0 + 6 0 + 0 + 0 4 + 0 + 9

= 5 0 8

2 4 5

8 0 13

Again A3 = A2 * A

=> A2 = 5 0 8 1 0 2

2 4 5 0 2 1

8 0 13 2 0 3

= 5 + 0 + 16 0 + 0 + 0 10 + 0 + 24

2 + 0 + 10 0 + 8 + 0 4 + 4 + 15

8 + 0 + 26 0 + 0 + 0 15 + 0 + 39



= 21 0 34

12 8 23

34 0 55

Now, A3 – 6A2 + 7A + 2I

= 21 0 34 - 6 5 0 8 + 7 1 0 2 +2 1 0 0

12 8 23 2 4 5 0 2 1 0 1 0

34 0 55 8 0 13 2 0 3 0 0 1

= 21 0 34 - 30 0 48 + 7 0 14 + 2 0 0

12 8 23 12 24 30 0 14 7 0 2 0

34 0 55 48 0 78 14 0 21 0 0 2

= 21 – 30 + 7 + 2 0 – 0 + 0 + 0 34 – 48 + 14 + 0 12 – 12 + 0 + 0 8 – 24 + 14 + 2 23 – 30 + 7 + 0 34 – 48 + 14 + 0 0 – 0 + 0 + 0 55 – 78 + 21 + 2

= O

So, A3 – 6A2 + 7A + 2I = 0

Some Exercises:

Question 1. Fill in the blanks in each of the following: (Exemplar)

1. _________ matrix is both symmetric and skew symmetric matrix.

2. Sum of two skew symmetric matrices is always _________ matrix.

3. The negative of a matrix is obtained by multiplying it by _________.

4. The product of any matrix by the scalar _________ is the null matrix.

5. A matrix which is not a square matrix is called a _________ matrix.

6. Matrix multiplication is _________ over addition.

7. If A is a symmetric matrix, then A3 is a _________ matrix.

8. If A is a skew symmetric matrix, then A2 is a _________.



9. If A and B are square matrices of the same order, then

(i) (AB)′ = _________.

(ii) (kA)′ = _________. (k is any scalar)

(iii) [k (A – B)]′ = _________.

10. If A is skew symmetric, then kA is a _________. (k is any scalar)

11. If A and B are symmetric matrices, then

(i) AB – BA is a _________.

(ii) BA – 2AB is a _________.

12. If A is symmetric matrix, then B′AB is _________.

13. If A and B are symmetric matrices of same order, then AB is symmetric if and only if

_________.

14. In applying one or more row operations while finding A–1 by elementary row

operations, we obtain all zeros in one or more, then A–1 _________.

Question 2: State which of the following statements are True or False (Exemplar)

1. A matrix denotes a number.

2. Matrices of any order can be added.

3. Two matrices are equal if they have same number of rows and same number of

columns.

4. Matrices of different order can not be subtracted.

5. Matrix addition is associative as well as commutative.

6. Matrix multiplication is commutative.

7. A square matrix where every element is unity is called an identity matrix.

8. If A and B are two square matrices of the same order, then A + B = B + A.

9. If A and B are two matrices of the same order, then A – B = B – A.

10. If matrix AB = O, then A = O or B = O or both A and B are null matrices.

11. Transpose of a column matrix is a column matrix.

12. If A and B are two square matrices of the same order, then AB = BA.



13. If each of the three matrices of the same order are symmetric, then their sum is a

symmetric matrix.

14. If A and B are any two matrices of the same order, then (AB)′ = A′B′.

15. . If (AB)′ = B′ A′, where A and B are not square matrices, then number of rows in A is

equal to number of columns in B and number of columns in A is equal to number of rows

in B.

16. If A, B and C are square matrices of same order, then AB = AC always implies that B

= C.

17. AA′ is always a symmetric matrix for any matrix A.

18. If A = 2 3 1

1 4 2

and B = 2 3

4 5

2 1

, then AB and BA are defined and equal.

19. If A is skew symmetric matrix, then A2

is a symmetric matrix.

20. (AB)–1 = A

–1. B

–1, where A and B are invertible matrices satisfying commutative

property with respect to multiplication.

----------------------------

After solving the above questions you can test your understanding by taking the following

quiz

Click here to take Quiz on Matrices

https://docs.google.com/forms/d/e/1FAIpQLSeMFOvYHO2ChzqkISZLK1izWoE4UO_To23CR46wo1lZS9Wueg/viewform?usp=sf_link

matrices - kar

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