Class XII Chapter 3 – Matrices Mathscbseocean.weebly.com/uploads/2/8/1/5/28152469/cmn34.pdf · Class XII Chapter 3 – Matrices Maths Page 7 of 83 Website: Email: [email protected]

Class XII Chapter 3 – Matrices Maths

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Exercise 3.1

Question 1:

In the matrix , write:

(i) The order of the matrix (ii) The number of elements,

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

Answer

(i) In the given matrix, the number of rows is 3 and the number of columns is 4.

Therefore, the order of the matrix is 3 × 4.

(ii) Since the order of the matrix is 3 × 4, there are 3 × 4 = 12 elements in it.

(iii) a13 = 19, a21 = 35, a33 = −5, a24 = 12, a23 =

Question 2:

If a matrix has 24 elements, what are the possible order it can have? What, if it has 13

elements?

Answer

We know that if a matrix is of the order m × n, it has mn elements. Thus, to find all the

possible orders of a matrix having 24 elements, we have to find all the ordered pairs of

natural numbers whose product is 24.

The ordered pairs are: (1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6), and

(6, 4)

Hence, the possible orders of a matrix having 24 elements are:

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

(1, 13) and (13, 1) are the ordered pairs of natural numbers whose product is 13.

Hence, the possible orders of a matrix having 13 elements are 1 × 13 and 13 × 1.

Question 3:


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If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5

elements?

Answer

We know that if a matrix is of the order m × n, it has mn elements. Thus, to find all the

possible orders of a matrix having 18 elements, we have to find all the ordered pairs of

natural numbers whose product is 18.

The ordered pairs are: (1, 18), (18, 1), (2, 9), (9, 2), (3, 6,), and (6, 3)

Hence, the possible orders of a matrix having 18 elements are:

1 × 18, 18 × 1, 2 × 9, 9 × 2, 3 × 6, and 6 × 3

(1, 5) and (5, 1) are the ordered pairs of natural numbers whose product is 5.

Hence, the possible orders of a matrix having 5 elements are 1 × 5 and 5 × 1.

Question 5:

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

(i) (ii)

Answer

In general, a 3 × 4 matrix is given by

(i)


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Therefore, the required matrix is

(ii)


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Therefore, the required matrix is

Question 6:

Find the value of x, y, and z from the following equation:

(i) (ii)

(iii)

Answer

(i)

As the given matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements, we get:


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x = 1, y = 4, and z = 3

(ii)

As the given matrices are equal, their corresponding elements are also equal.


x + y = 6, xy = 8, 5 + z = 5

Now, 5 + z = 5 ⇒ z = 0

We know that:

(x − y)2 = (x + y)2 − 4xy

⇒ (x − y)2 = 36 − 32 = 4

⇒ x − y = ±2

Now, when x − y = 2 and x + y = 6, we get x = 4 and y = 2

When x − y = − 2 and x + y = 6, we get x = 2 and y = 4

∴x = 4, y = 2, and z = 0 or x = 2, y = 4, and z = 0

(iii)

As the two matrices are equal, their corresponding elements are also equal.


x + y + z = 9 … (1)

x + z = 5 … (2)

y + z = 7 … (3)

From (1) and (2), we have:

y + 5 = 9

⇒ y = 4

Then, from (3), we have:

4 + z = 7

⇒ z = 3

∴ x + z = 5

⇒ x = 2

∴ x = 2, y = 4, and z = 3


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Question 7:

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

Answer

As the two matrices are equal, their corresponding elements are also equal.


a − b = −1 … (1)

2a − b = 0 … (2)

2a + c = 5 … (3)

3c + d = 13 … (4)

From (2), we have:

b = 2a

Then, from (1), we have:

a − 2a = −1

⇒ a = 1

⇒ b = 2

Now, from (3), we have:

2 ×1 + c = 5

⇒ c = 3

From (4) we have:

3 ×3 + d = 13

⇒ 9 + d = 13 ⇒ d = 4

∴a = 1, b = 2, c = 3, and d = 4

Question 8:

is a square matrix, if

(A) m < n

(B) m > n


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(C) m = n

(D) None of these

Answer

The correct answer is C.

It is known that a given matrix is said to be a square matrix if the number of rows is

equal to the number of columns.

Therefore, is a square matrix, if m = n.

Question 9:

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

(A)

(B) Not possible to find

(C)

(D)

Answer

The correct answer is B.

It is given that

Equating the corresponding elements, we get:


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We find that on comparing the corresponding elements of the two matrices, we get two

different values of x, which is not possible.

Hence, it is not possible to find the values of x and y for which the given matrices are

equal.

Question 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

Answer

The correct answer is D.

The given matrix of the order 3 × 3 has 9 elements and each of these elements can be

either 0 or 1.

Now, each of the 9 elements can be filled in two possible ways.

Therefore, by the multiplication principle, the required number of possible matrices is 29

= 512


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Exercise 3.2

Question 1:

Let

Find each of the following

(i) (ii) (iii)

(iv) (v)

Answer

(i)

(ii)

(iii)

(iv) Matrix A has 2 columns. This number is equal to the number of rows in matrix B.

Therefore, AB is defined as:


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(v) Matrix B has 2 columns. This number is equal to the number of rows in matrix A.

Therefore, BA is defined as:

Question 2:

Compute the following:

(i) (ii)

(iii)

(v)

Answer

(i)

(ii)


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(iii)

(iv)

Question 3:

Compute the indicated products

(i)

(ii)


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(iii)

(iv)

(v)

(vi)

Answer

(i)

(ii)

(iii)


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(iv)

(v)

(vi)


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Question 4:

If , and , then

compute and . Also, verify that

Answer


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Question 5:

If and then compute .

Answer


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Question 6:

Simplify

Answer

Question 7:

Find X and Y, if

(i) and

(ii) and

Answer

(i)


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Adding equations (1) and (2), we get:

(ii)

Multiplying equation (3) with (2), we get:


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Multiplying equation (4) with (3), we get:


Now,


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Question 8:

Find X, if and

Answer


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Question 9:

Find x and y, if

Answer

Comparing the corresponding elements of these two matrices, we have:

∴x = 3 and y = 3

Question 10:


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Solve the equation for x, y, z and t if

Answer

Comparing the corresponding elements of these two matrices, we get:

Question 11:

If , find values of x and y.

Answer


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2x − y = 10 and 3x + y = 5

Adding these two equations, we have:

5x = 15

⇒ x = 3

Now, 3x + y = 5

⇒ y = 5 − 3x

⇒ y = 5 − 9 = −4

∴x = 3 and y = −4

Question 12:

Given , find the values of x, y, z and

w.

Answer



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Question 13:

If , show that .

Answer


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Question 14:

Show that

(i)

(ii)

Answer


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(i)

(ii)


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Question 15:

Find if

Answer

We have A2 = A × A


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Question 16:

If , prove that

Answer


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Question 17:

If and , find k so that

Answer


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Comparing the corresponding elements, we have:

Thus, the value of k is 1.

Question 18:

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

Answer


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Question 19:

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:


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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:

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.

Question 20:


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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:

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

Question 21:

Assume X, Y, Z, W and P are matrices of order , and

respectively. The restriction on n, k and p so that 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.


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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.

Question 22:

Assume X, Y, Z, W and P are matrices of order , and

respectively. If n = p, then the order of the matrix is

A p × 2 B 2 × n C n × 3 D p × n

Answer


Matrix X is of the order 2 × n.

Therefore, matrix 7X is also of the same order.

Matrix Z is of the order 2 × p, i.e., 2 × n [Since n = p]

Therefore, matrix 5Z is also of the same order.

Now, both the matrices 7X and 5Z are of the order 2 × n.

Thus, matrix 7X − 5Z is well-defined and is of the order 2 × n.


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Exercise 3.3

Question 1:

Find the transpose of each of the following matrices:

(i) (ii) (iii)

Answer

(i)

(ii)

(iii)

Question 2:

If and , then verify that

(i)

(ii)

Answer

We have:


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(i)

(ii)


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Question 3:

If and , then verify that

(i)

(ii)

Answer

(i) It is known that

Therefore, we have:


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(ii)

Question 4:

If and , then find

Answer

We know that

Question 5:

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


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(i)

(ii)

Answer

(i)

(ii)


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Question 6:

If (i) , then verify that

(ii) , then verify that

Answer

(i)


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(ii)


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Question 7:

(i) Show that the matrix is a symmetric matrix

(ii) Show that the matrix is a skew symmetric matrix

Answer

(i) We have:

Hence, A is a symmetric matrix.

(ii) We have:


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Hence, A is a skew-symmetric matrix.

Question 8:

For the matrix , verify that

(i) is a symmetric matrix

(ii) is a skew symmetric matrix

Answer

(i)

Hence, is a symmetric matrix.

(ii)

Hence, is a skew-symmetric matrix.

Question 9:


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Find and , when

Answer

The given matrix is

Question 10:

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

(i)


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(ii)

(iii)

(iv)

Answer

(i)

Thus, is a symmetric matrix.


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Thus, is a skew-symmetric matrix.

Representing A as the sum of P and Q:

(ii)



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(iii)


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(iv)





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Question 11:

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

Answer

The correct answer is A.

A and B are symmetric matrices, therefore, we have:

Thus, (AB − BA) is a skew-symmetric matrix.

Question 12:

If , then , if the value of α is

A. B.

C. π D.

Answer



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Comparing the corresponding elements of the two matrices, we have:

Exercise 3.4

Question 1:

Find the inverse of each of the matrices, if it exists.

Answer

We know that A = IA


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Question 2:


Answer

We know that A = IA


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Question 3:


Answer

We know that A = IA

Question 4:


Answer

We know that A = IA


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Question 5:


Answer

We know that A = IA


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Question 6:


Answer

We know that A = IA


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Question 7:


Answer

We know that A = AI


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Question 8:


Answer

We know that A = IA


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Question 9:


Answer

We know that A = IA

Question 10:


Answer

We know that A = AI


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Question 11:


Answer

We know that A = AI


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Question 12:


Answer

We know that A = IA

Now, in the above equation, we can see all the zeros in the second row of the matrix on

the L.H.S.

Therefore, A−1 does not exist.

Question 13:


Answer


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We know that A = IA

Question 14:


Answer

We know that A = IA

Applying , we have:


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Now, in the above equation, we can see all the zeros in the first row of the matrix on the

L.H.S.

Therefore, A−1 does not exist.

Question 16:


Answer

We know that A = IA

Applying R2 → R2 + 3R1 and R3 → R3 − 2R1, we have:


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Question 17:



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Answer

We know that A = IA

Applying , we have:


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Question 18:

Matrices A and B will be inverse of each other only if

A. AB = BA

C. AB = 0, BA = I

B. AB = BA = 0

D. AB = BA = I

Answer


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Answer: D

We know that if A is a square matrix of order m, and if there exists another square

matrix B of the same order m, such that AB = BA = I, then B is said to be the inverse of

A. In this case, it is clear that A is the inverse of B.

Thus, matrices A and B will be inverses of each other only if AB = BA = I.


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Miscellaneous Solutions

Question 1:

Let , show that , where I is the identity matrix of

order 2 and n ∈ N

Answer

It is given that

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

For n = 1, we have:

Therefore, the result is true for n = 1.

Let the result be true for n = k.

That is,

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

Consider

From (1), we have:


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

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

Question 2:

If , prove that

Answer

It is given that


For n = 1, we have:



That is


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Thus by the principle of mathematical induction, we have:

Question 3:

If , then prove where n is any positive integer

Answer

It is given that


For n = 1, we have:


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That is,



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

Question 4:

If A and B are symmetric matrices, prove that AB − BA is a skew symmetric matrix.

Answer

It is given that A and B are symmetric matrices. Therefore, we have:


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Thus, (AB − BA) is a skew-symmetric matrix.

Question 5:

Show that the matrix is symmetric or skew symmetric according as A is symmetric

or skew symmetric.

Answer

We suppose that A is a symmetric matrix, then … (1)

Consider

Thus, if A is a symmetric matrix, then is a symmetric matrix.

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

Then,


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

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

symmetric matrix accordingly.

Question 6:

Solve system of linear equations, using matrix method.

Answer

The given system of equations can be written in the form of AX = B, where

Thus, A is non-singular. Therefore, its inverse exists.


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Question 7:

For what values of ?

Answer

We have:

∴4 + 4x = 0

⇒ x = −1

Thus, the required value of x is −1.


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Question 8:

If , show that

Answer

Question 9:


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Find x, if

Answer

We have:

Question 10:

A manufacturer produces three products x, y, z which he sells in two markets.

Annual sales are indicated below:

Market Products

I 10000 2000 18000

II 6000 20000 8000

(a) If unit sale prices of x, y and z are Rs 2.50, Rs 1.50 and Rs 1.00, respectively, find

the total revenue in each market with the help of matrix algebra.


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(b) If the unit costs of the above three commodities are Rs 2.00, Rs 1.00 and 50 paise

respectively. Find the gross profit.

Answer

(a) The unit sale prices of x, y, and z are respectively given as Rs 2.50, Rs 1.50, and Rs

1.00.

Consequently, the total revenue in market I can be represented in the form of a matrix

as:

The total revenue in market II can be represented in the form of a matrix as:

Therefore, the total revenue in market I isRs 46000 and the same in market II isRs

53000.

(b) The unit cost prices of x, y, and z are respectively given as Rs 2.00, Rs 1.00, and 50

paise.

Consequently, the total cost prices of all the products in market I can be represented in

the form of a matrix as:


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Since the total revenue in market I isRs 46000, the gross profit in this marketis (Rs

46000 − Rs 31000) Rs 15000.

The total cost prices of all the products in market II can be represented in the form of a

matrix as:

Since the total revenue in market II isRs 53000, the gross profit in this market is (Rs

53000 − Rs 36000) Rs 17000.

Question 11:

Find the matrix X so that

Answer

It is given that:

The matrix given on the R.H.S. of the equation is a 2 × 3 matrix and the one given on

the L.H.S. of the equation is a 2 × 3 matrix. Therefore, X has to be a 2 × 2 matrix.

Now, let

Therefore, we have:

Equating the corresponding elements of the two matrices, we have:


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Thus, a = 1, b = 2, c = −2, d = 0

Hence, the required matrix X is

Question 12:

If A and B are square matrices of the same order such that AB = BA, then prove by

induction that . Further, prove that for all n ∈ N

Answer

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

For n = 1, we have:


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Thus, by the principle of mathematical induction, we have

Now, we prove that for all n ∈ N

For n = 1, we have:






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Thus, by the principle of mathematical induction, we have , for all natural

numbers.

Question 13:

Choose the correct answer in the following questions:

If is such that then

A.

B.

C.

D.

Answer

Answer: C

On comparing the corresponding elements, we have:


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Question 14:

If the matrix A is both symmetric and skew symmetric, then

A. A is a diagonal matrix

B. A is a zero matrix

C. A is a square matrix

D. None of these

Answer

Answer: B

If A is both symmetric and skew-symmetric matrix, then we should have

Therefore, A is a zero matrix.

Question 15:

If A is square matrix such that then is equal to

A. A B. I − A C. I D. 3A

Answer

Answer: C


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Class XII Chapter 4 – Determinants Maths

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Head Office: 1/3-H-A-2, Street # 6, East Azad Nagar, Delhi-110051

(One Km from ‘Welcome’ Metro Station)

Exercise 4.1

Question 1:

Evaluate the determinants in Exercises 1 and 2.

Answer

= 2(−1) − 4(−5) = − 2 + 20 = 18

Question 2:

Evaluate the determinants in Exercises 1 and 2.

(i) (ii)

Answer

(i) = (cos θ)(cos θ) − (−sin θ)(sin θ) = cos2 θ+ sin2 θ = 1

(ii)

= (x2 − x + 1)(x + 1) − (x − 1)(x + 1)

= x3 − x2 + x + x2 − x + 1 − (x2 − 1)

= x3 + 1 − x2 + 1

= x3 − x2 + 2

Question 3:

If , then show that

Answer

The given matrix is .


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Question 4:

If , then show that

Answer

The given matrix is .

It can be observed that in the first column, two entries are zero. Thus, we expand along

the first column (C1) for easier calculation.

From equations (i) and (ii), we have:


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Hence, the given result is proved.

Question 5:

Evaluate the determinants

(i) (iii)

(ii) (iv)

Answer

(i) Let .

It can be observed that in the second row, two entries are zero. Thus, we expand along

the second row for easier calculation.

(ii) Let .

By expanding along the first row, we have:


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(iii) Let


(iv) Let

By expanding along the first column, we have:

Question 6:

If , find .

Answer

Let



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Question 7:

Find values of x, if

(i) (ii)

Answer

(i)

(ii)

Question 8:

If , then x is equal to


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(A) 6 (B) ±6 (C) −6 (D) 0

Answer

Answer: B

Hence, the correct answer is B.


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Exercise 4.2

Question 1:

Using the property of determinants and without expanding, prove that:

Answer

Question 2:


Answer

Here, the two rows R1 and R3 are identical.

∆ = 0.


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Question 3:


Answer

Question 4:


Answer


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By applying C3 → C3 + C2, we have:

Here, two columns C1 and C3 are proportional.

∆ = 0.

Question 5:


Answer

Applying R2 → R2 − R3, we have:


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Applying R1 ↔R3 and R2 ↔R3, we have:

Applying R1 → R1 − R3, we have:

Applying R1 ↔R2 and R2 ↔R3, we have:

From (1), (2), and (3), we have:


Question 6:

By using properties of determinants, show that:


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Answer

We have,

Here, the two rows R1 and R3 are identical.

∴∆ = 0.

Question 7:


Answer


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Applying R2 → R2 + R1 and R3 → R3 + R1, we have:

Question 8:


(i)

(ii)

Answer

(i)


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Applying R1 → R1 − R3 and R2 → R2 − R3, we have:

Applying R1 → R1 + R2, we have:

Expanding along C1, we have:


(ii) Let .

Applying C1 → C1 − C3 and C2 → C2 − C3, we have:


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Applying C1 → C1 + C2, we have:



Question 9:



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Answer


Applying R3 → R3 + R2, we have:

Expanding along R3, we have:


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Question 10:


(i)

(ii)

Answer

(i)

Applying R1 → R1 + R2 + R3, we have:

Applying C2 → C2 − C1, C3 → C3 − C1, we have:


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(ii)





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Question 11:


(i)

(ii)

Answer

(i)


Applying C2 → C2 − C1, C3 → C3 − C1, we have:


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(ii)

Applying C1 → C1 + C2 + C3, we have:




Question 12:



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Answer






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Question 13:


Answer

Applying R1 → R1 + bR3 and R2 → R2 − aR3, we have:


Question 14:



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Answer

Taking out common factors a, b, and c from R1, R2, and R3 respectively, we have:


Applying C1 → aC1, C2 → bC2, and C3 → cC3, we have:



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Question 15:

Choose the correct answer.

Let A be a square matrix of order 3 × 3, then is equal to

A. B. C. D.

Answer

Answer: C

A is a square matrix of order 3 × 3.

Hence, the correct answer is C.

Question 16:

Which of the following is correct?


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A. Determinant is a square matrix.

B. Determinant is a number associated to a matrix.

C. Determinant is a number associated to a square matrix.

D. None of these

Answer

Answer: C

We know that to every square matrix, of order n. We can associate a number

called the determinant of square matrix A, where element of A.

Thus, the determinant is a number associated to a square matrix.

Hence, the correct answer is C.


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Exercise 4.3

Question 1:

Find area of the triangle with vertices at the point given in each of the following:

(i) (1, 0), (6, 0), (4, 3) (ii) (2, 7), (1, 1), (10, 8)

(iii) (−2, −3), (3, 2), (−1, −8)

Answer

(i) The area of the triangle with vertices (1, 0), (6, 0), (4, 3) is given by the relation,

(ii) The area of the triangle with vertices (2, 7), (1, 1), (10, 8) is given by the relation,

(iii) The area of the triangle with vertices (−2, −3), (3, 2), (−1, −8)

is given by the relation,


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Hence, the area of the triangle is .

Question 2:

Show that points

are collinear

Answer

Area of ∆ABC is given by the relation,

Thus, the area of the triangle formed by points A, B, and C is zero.


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Hence, the points A, B, and C are collinear.

Question 3:

Find values of k if area of triangle is 4 square units and vertices are

(i) (k, 0), (4, 0), (0, 2) (ii) (−2, 0), (0, 4), (0, k)

Answer

We know that the area of a triangle whose vertices are (x1, y1), (x2, y2), and

(x3, y3) is the absolute value of the determinant (∆), where

It is given that the area of triangle is 4 square units.

∴∆ = ± 4.

(i) The area of the triangle with vertices (k, 0), (4, 0), (0, 2) is given by the relation,

∆ =

∴∴∴∴−k + 4 = ± 4

When −k + 4 = − 4, k = 8.

When −k + 4 = 4, k = 0.

Hence, k = 0, 8.


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(ii) The area of the triangle with vertices (−2, 0), (0, 4), (0, k) is given by the relation,

∆ =

∴k − 4 = ± 4

When k − 4 = − 4, k = 0.

When k − 4 = 4, k = 8.

Hence, k = 0, 8.

Question 4:

(i) Find equation of line joining (1, 2) and (3, 6) using determinants

(ii) Find equation of line joining (3, 1) and (9, 3) using determinants

Answer

(i) Let P (x, y) be any point on the line joining points A (1, 2) and B (3, 6). Then, the

points A, B, and P are collinear. Therefore, the area of triangle ABP will be zero.

Hence, the equation of the line joining the given points is y = 2x.

(ii) Let P (x, y) be any point on the line joining points A (3, 1) and


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B (9, 3). Then, the points A, B, and P are collinear. Therefore, the area of triangle ABP

will be zero.

Hence, the equation of the line joining the given points is x − 3y = 0.

Question 5:

If area of triangle is 35 square units with vertices (2, −6), (5, 4), and (k, 4). Then k is

A. 12 B. −2 C. −12, −2 D. 12, −2

Answer

Answer: D

The area of the triangle with vertices (2, −6), (5, 4), and (k, 4) is given by the relation,

It is given that the area of the triangle is ±35.

Therefore, we have:


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When 5 − k = −7, k = 5 + 7 = 12.

When 5 − k = 7, k = 5 − 7 = −2.

Hence, k = 12, −2.



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Exercise 4.4

Question 1:

Write Minors and Cofactors of the elements of following determinants:

(i) (ii)

Answer

(i) The given determinant is .

Minor of element aij is Mij.

∴M11 = minor of element a11 = 3

M12 = minor of element a12 = 0

M21 = minor of element a21 = −4

M22 = minor of element a22 = 2

Cofactor of aij is Aij = (−1)i + j Mij.

∴A11 = (−1)1+1 M11 = (−1)2 (3) = 3

A12 = (−1)1+2 M12 = (−1)3 (0) = 0

A21 = (−1)2+1 M21 = (−1)3 (−4) = 4

A22 = (−1)2+2 M22 = (−1)4 (2) = 2

(ii) The given determinant is .

Minor of element aij is Mij.


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∴M11 = minor of element a11 = d

M12 = minor of element a12 = b

M21 = minor of element a21 = c

M22 = minor of element a22 = a

Cofactor of aij is Aij = (−1)i + j Mij.

∴A11 = (−1)1+1 M11 = (−1)2 (d) = d

A12 = (−1)1+2 M12 = (−1)3 (b) = −b

A21 = (−1)2+1 M21 = (−1)3 (c) = −c

A22 = (−1)2+2 M22 = (−1)4 (a) = a

Question 2:

(i) (ii)

Answer

(i) The given determinant is .

By the definition of minors and cofactors, we have:

M11 = minor of a11=

M12 = minor of a12=


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M13 = minor of a13 =




M31 = minor of a31=



A11 = cofactor of a11= (−1)1+1 M11 = 1

A12 = cofactor of a12 = (−1)1+2 M12 = 0

A13 = cofactor of a13 = (−1)1+3 M13 = 0

A21 = cofactor of a21 = (−1)2+1 M21 = 0

A22 = cofactor of a22 = (−1)2+2 M22 = 1

A23 = cofactor of a23 = (−1)2+3 M23 = 0

A31 = cofactor of a31 = (−1)3+1 M31 = 0

A32 = cofactor of a32 = (−1)3+2 M32 = 0

A33 = cofactor of a33 = (−1)3+3 M33 = 1

(ii) The given determinant is .

By definition of minors and cofactors, we have:

M11 = minor of a11=


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M12 = minor of a12=





M31 = minor of a31=



A11 = cofactor of a11= (−1)1+1 M11 = 11

A12 = cofactor of a12 = (−1)1+2 M12 = −6

A13 = cofactor of a13 = (−1)1+3 M13 = 3

A21 = cofactor of a21 = (−1)2+1 M21 = 4

A22 = cofactor of a22 = (−1)2+2 M22 = 2

A23 = cofactor of a23 = (−1)2+3 M23 = −1

A31 = cofactor of a31 = (−1)3+1 M31 = −20

A32 = cofactor of a32 = (−1)3+2 M32 = 13

A33 = cofactor of a33 = (−1)3+3 M33 = 5

Question 3:


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Using Cofactors of elements of second row, evaluate .

Answer

The given determinant is .

We have:

M21 =

∴A21 = cofactor of a21 = (−1)2+1 M21 = 7

M22 =

∴A22 = cofactor of a22 = (−1)2+2 M22 = 7

M23 =

∴A23 = cofactor of a23 = (−1)2+3 M23 = −7

We know that ∆ is equal to the sum of the product of the elements of the second row

with their corresponding cofactors.


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∴∆ = a21A21 + a22A22 + a23A23 = 2(7) + 0(7) + 1(−7) = 14 − 7 = 7

Question 4:

Using Cofactors of elements of third column, evaluate

Answer

The given determinant is .

We have:

M13 =

M23 =

M33 =

∴A13 = cofactor of a13 = (−1)1+3 M13 = (z − y)

A23 = cofactor of a23 = (−1)2+3 M23 = − (z − x) = (x − z)

A33 = cofactor of a33 = (−1)3+3 M33 = (y − x)

We know that ∆ is equal to the sum of the product of the elements of the second row

with their corresponding cofactors.


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Hence,

Question 5:

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

(i)

(ii)

Answer

(i)


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(ii)


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Exercise 4.5

Question 1:

Find adjoint of each of the matrices.

Answer

Question 2:

Find adjoint of each of the matrices.

Answer


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Question 3:

Verify A (adj A) = (adj A) A = I .

Answer


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Question 4:

Verify A (adj A) = (adj A) A = I .

Answer


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Question 6:

Find the inverse of each of the matrices (if it exists).

Answer


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Question 7:


Answer


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Question 8:


Answer

Question 9:



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Answer

Question 10:


.

Answer


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Question 11:


Answer


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Question 12:

Let and . Verify that

Answer


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(AB)−1 = B−1A−1


Question 13:

If , show that . Hence find .

Answer


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Question 14:

For the matrix , find the numbers a and b such that A2 + aA + bI = O.

Answer


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We have:

Comparing the corresponding elements of the two matrices, we have:

Hence, −4 and 1 are the required values of a and b respectively.


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Question 15:

For the matrix show that A3 − 6A2 + 5A + 11 I = O. Hence, find

A−1.

Answer


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From equation (1), we have:

Question 16:

If verify that A3 − 6A2 + 9A − 4I = O and hence find A−1

Answer


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From equation (1), we have:


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Question 17:

Let A be a nonsingular square matrix of order 3 × 3. Then is equal to

A. B. C. D.

Answer B

We know that,


Question 18:

If A is an invertible matrix of order 2, then det (A−1) is equal to

A. det (A) B. C. 1 D. 0

Answer

Since A is an invertible matrix,


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Exercise 4.6

Question 1:

Examine the consistency of the system of equations.

x + 2y = 2

2x + 3y = 3

Answer

The given system of equations is:

x + 2y = 2

2x + 3y = 3


∴ A is non-singular.

Therefore, A−1 exists.

Hence, the given system of equations is consistent.

Question 2:


2x − y = 5

x + y = 4

Answer


2x − y = 5

x + y = 4



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Question 3:


x + 3y = 5

2x + 6y = 8

Answer


x + 3y = 5

2x + 6y = 8


∴ A is a singular matrix.


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Thus, the solution of the given system of equations does not exist. Hence, the system of

equations is inconsistent.

Question 4:


x + y + z = 1

2x + 3y + 2z = 2

ax + ay + 2az = 4

Answer


x + y + z = 1

2x + 3y + 2z = 2

ax + ay + 2az = 4

This system of equations can be written in the form AX = B, where





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Question 5:


3x − y − 2z = 2

2y − z = −1

3x − 5y = 3

Answer


3x − y − 2z = 2

2y − z = −1

3x − 5y = 3

This system of equations can be written in the form of AX = B, where

∴ A is a singular matrix.

Thus, the solution of the given system of equations does not exist. Hence, the system of

equations is inconsistent.


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Question 6:


5x − y + 4z = 5

2x + 3y + 5z = 2

5x − 2y + 6z = −1

Answer


5x − y + 4z = 5

2x + 3y + 5z = 2

5x − 2y + 6z = −1





Question 7:


Answer


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Question 8:


Answer




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Question 9:


Answer




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Question 10:


5x + 2y = 3

3x + 2y = 5

Answer



Question 11:


Answer



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Question 12:


x − y + z = 4

2x + y − 3z = 0


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x + y + z = 2

Answer



Question 13:



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2x + 3y + 3z = 5

x − 2y + z = −4

3x − y − 2z = 3

Answer

The given system of equations can be written in the form AX = B, where



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Question 14:


x − y + 2z = 7

3x + 4y − 5z = −5

2x − y + 3z = 12

Answer



Question 15:


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If , find A−1. Using A−1 solve the system of equations

Answer

Now, the given system of equations can be written in the form of AX = B, where


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Question 16:

The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60. The cost of 2 kg onion, 4 kg

wheat and 6 kg rice is Rs 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70.

Find cost of each item per kg by matrix method.

Answer

Let the cost of onions, wheat, and rice per kg be Rs x, Rs y,and Rs z respectively.

Then, the given situation can be represented by a system of equations as:



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Now,

X = A−1 B

Hence, the cost of onions is Rs 5 per kg, the cost of wheat is Rs 8 per kg, and the cost of

rice is Rs 8 per kg.


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Miscellaneous Solutions

Question 1:

Prove that the determinant is independent of θ.

Answer

Hence, ∆ is independent of θ.

Question 2:

Without expanding the determinant, prove that

Answer


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Question 3:

Evaluate

Answer



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Question 4:

If a, b and c are real numbers, and ,

Show that either a + b + c = 0 or a = b = c.

Answer



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Hence, if ∆ = 0, then either a + b + c = 0 or a = b = c.

Question 5:

Solve the equations

Answer


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Question 6:

Prove that

Answer


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Question 8:

Let verify that

(i)

(ii)

Answer

(i)

We have,


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(ii)


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Question 9:

Evaluate

Answer


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Question 10:

Evaluate

Answer


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Question 11:

Using properties of determinants, prove that:

Answer


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Question 12:


Answer


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Question 13:


Answer


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Question 14:


Answer


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Question 15:


Answer


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Question 16:

Solve the system of the following equations

Answer

Let

Then the given system of equations is as follows:

This system can be written in the form of AX = B, where


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A


Now,

A11 = 75, A12 = 110, A13 = 72

A21 = 150, A22 = −100, A23 = 0

A31 = 75, A32 = 30, A33 = − 24


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Question 17:


If a, b, c, are in A.P., then the determinant

A. 0 B. 1 C. x D. 2x

Answer

Answer: A

Here, all the elements of the first row (R1) are zero.

Hence, we have ∆ = 0.



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Question 18:


If x, y, z are nonzero real numbers, then the inverse of matrix is

A. B.

C. D.

Answer

Answer: A


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Question 19:


Let , where 0 ≤ θ≤ 2π, then

A. Det (A) = 0

B. Det (A) ∈ (2, ∞)

C. Det (A) ∈ (2, 4)


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D. Det (A)∈ [2, 4]

Answer

sAnswer: D


12

MATRICES & DETERMINANTS

SCHEMATIC DIAGRAM

INTRODUCTION:

MATRIX: If mn elements can be arranged in the form of m row and n column in a

rectangular array then this arrangement is called a matrix.

Order of a matrix: A matrix having m row and n column is called a matrix of 𝑚 × 𝑛 order.

Addition and subtraction of matrices: Two matrices A and B can be added or subtracted if

they are of the same order i.e. if A and B are two matrices of order 𝑚 × 𝑛 then A ± B is also

a matrix of order m×n.

Multiplication of matrices: The product of two matrices A and B can be defined if the

number of rows of B is equal to the number of columns of A i.e. if A be an 𝑚 × 𝑛 matrix and

B be an 𝑛 × 𝑝 matrix then the product of matrices A and B is another matrix of order 𝑚 × 𝑝.

Transpose of a Matrix: If A = [aij] be an m × n matrix, then the matrix obtained by

interchanging.

the rows and columns of A is called the transpose of A. Transpose of the matrix A is

denoted by 𝐴/ 𝑜𝑟 𝐴𝑇.

Properties of transpose of the Matrices: For any matrices A and B of suitable orders, we

have

(i) AATT (𝑖𝑖)(𝐾𝐴)𝑇 = 𝐾𝐴𝑇(𝑖𝑖𝑖)(𝐴 + 𝐵)𝑇 = 𝐴𝑇 + 𝐵𝑇(𝑖𝑣)(𝐴𝐵)𝑇 = 𝐵𝑇𝐴𝑇

Symmetric Matrix: A square matrix M is said to be symmetric if 𝐴𝑇 = 𝐴

e.g.[𝑎 𝑏𝑏 𝑐

] , [𝑥 𝑦 𝑧𝑦 𝑢 𝑣𝑧 𝑣 𝑤

]

Note: there will be symmetry about the principal diagonal in Symmetric Matrix.

Skew symmetric Matrix: A square matrix M is said to be skew symmetric if 𝐴𝑇 = −𝐴

e.g.[

0 𝑒 𝑓−𝑒 0 𝑔−𝑓 −𝑔 0

]

Note: All the principal diagonal element of a skew symmetric Matrix are zero.

Determinant: For every Square Matrix we can associate a number which is called the

Determinant of the square Matrix.

Determinant of a matrix of order one

Let A = [a ] be the matrix of order 1, then determinant of A is defined to be equal to a.

Determinant of a matrix of order two

Let A= [𝑎 𝑏𝑥 𝑦

] be a Square Matrix of order2× 2 then the determinant of A is denoted by |𝐴|

and defined by |𝐴| = |𝑎 𝑏𝑥 𝑦

|= ay-bx

13

Determinant of a matrix of order 3× 𝟑: Let us consider the determinant of a square matrix

of order

3× 3, |𝐴|= |𝑎 𝑏 𝑐𝑝 𝑞 𝑟𝑥 𝑦 𝑧

|

Expansion along first row |𝐴| = 𝑎(𝑞𝑧 − 𝑦𝑟) − 𝑏(𝑝𝑧 − 𝑥𝑟) + 𝑐(𝑝𝑦 − 𝑞𝑥)

We can expand the determinant with respect to any row or any column.

Minors and cofactors:

Minor of an element𝑎𝑖𝑗of a determinant is the determinant obtained by

deleting its ith row and jth column in which element𝑎𝑖𝑗 lies. Minor of an elemen𝑎𝑖𝑗is

denoted by 𝑀𝑖𝑗.

Cofactors: cofactors of an element𝑎𝑖𝑗 𝑑𝑒𝑛𝑜𝑡𝑒𝑑 by 𝐴𝑖𝑗 and is defined by 𝐴𝑖𝑗 = (−1)𝑖+𝑗𝑀𝑖𝑗

where 𝑀𝑖𝑗 is the minor of 𝑎𝑖𝑗.

Adjoint of a Matrix: Let A = [𝛼 𝛽𝛾 𝛿

] be a Matrix of order 2 × 2

Then adj(A) = [𝜹 −𝜷

−𝜸 𝜶]

Again letA = [𝑥 𝑦 𝑧𝑝 𝑞 𝑟𝑎 𝑏 𝑐

] be a Matrix of order 3 × 3

Then adj(A) =

[ |

𝑞 𝑟𝑏 𝑐

| − |𝑝 𝑟𝑎 𝑐

| |𝑝 𝑞𝑎 𝑏

|

− |𝑦 𝑧𝑏 𝑐

| |𝑥 𝑧𝑎 𝑐

| − |𝑥 𝑦𝑎 𝑏

|

|𝑦 𝑧𝑞 𝑟| − |

𝑥 𝑧𝑝 𝑟| |

𝑥 𝑦𝑝 𝑞| ]

𝑇

=

[

𝑞𝑐 − 𝑏𝑟 −(𝑝𝑐 − 𝑎𝑟) 𝑝𝑏 − 𝑎𝑞−(𝑦𝑐 − 𝑏𝑧) 𝑥𝑐 − 𝑎𝑧 −(𝑏𝑥 − 𝑎𝑦)

𝑦𝑟 − 𝑞𝑧 −(𝑥𝑟 − 𝑝𝑧) 𝑥𝑞 − 𝑝𝑦]

𝑇

= [𝑞𝑐 − 𝑏𝑟 𝑏𝑧 − 𝑎𝑦 𝑦𝑟 − 𝑞𝑧𝑎𝑟 − 𝑝𝑐 𝑥𝑐 − 𝑎𝑧 𝑝𝑧 − 𝑥𝑟𝑝𝑏 − 𝑎𝑞 𝑎𝑦 − 𝑏𝑥 𝑥𝑞 − 𝑝𝑦

]

Inverse of a Matrix: Inverse of a Square Matrix A is defined as𝑨−𝟏 =𝒂𝒅𝒋(𝑨)

|𝑨|

Note: If A be a given Square Matrix of order n then

(i) A(adj(A) = adj(A)A=|𝑨|𝑰where I is the Identity Matrix of order n.

(ii) A square Matrix A is said to be singular and non-singular according as |𝑨| =

𝟎 𝒂𝒏𝒅 |𝑨| ≠ 𝟎

(iii) |𝒂𝒅𝒋(𝑨)|= |𝑨|𝒏−𝟏 (𝑭𝒐𝒓 𝒂 𝒔𝒒𝒖𝒂𝒓𝒆 𝑴𝒂𝒕𝒓𝒊𝒙 𝒐𝒇 𝒐𝒓𝒅𝒆𝒓 𝟑 × 𝟑 |𝒂𝒅𝒋(𝑨)| = |𝑨|𝟐)

IMPORTANT SOLVED PROBLEMS

Q1. If A=[−245

] , 𝐵 = [1 3 −6] , Verify that (AB)1=B1A1

Solution: - We have

If A=[−245

] , 𝐵 = [1 3 −6]

14

Then AB =[−245

] [1 3 −6] = [−2 −6 124 12 −245 15 −30

]

Now A1 = [−2 4 5] , B1= [13

−6]

B1A1 = [13

−6] [−2 4 5] = [

−2 4 5−6 12 1512 −24 −30

] = (AB)1

Clearly (AB)1=B1A1

Q2. If 𝑥 [23] + 𝑦 [

−11

] = [105

] then find the value of x and y.

Sol. Given 𝑥 [23] + 𝑦 [

−11

] = [105

]

[2𝑥3𝑥

] + [−𝑦𝑦 ] = [

105

]or[2𝑥 − 𝑦3𝑥 + 𝑦

] = [105

]

So 2x – y = 10 and 3x + y = 5

On solving we get x = 3 and y = -4

Q3. If F(x) = [𝑐𝑜𝑠𝑥 −𝑠𝑖𝑛𝑥 0𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥 00 0 1

] prove that F(x) F(y) = F(x+y)

Sol. Given F(x) = [𝑐𝑜𝑠𝑥 −𝑠𝑖𝑛𝑥 0𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥 00 0 1

] so F(y) = [𝑐𝑜𝑠𝑦 −𝑠𝑖𝑛𝑦 0𝑠𝑖𝑛𝑦 𝑐𝑜𝑠𝑦 00 0 1

]

Hence F(x) .F(y) = [𝑐𝑜𝑠𝑥 −𝑠𝑖𝑛𝑥 0𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥 00 0 1

] [𝑐𝑜𝑠𝑦 −𝑠𝑖𝑛𝑦 0𝑠𝑖𝑛𝑦 𝑐𝑜𝑠𝑦 00 0 1

] =

[𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝑦 − 𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦 −𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑦 − 𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑦 0𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑦 + 𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑦 −𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦 + 𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝑦 0

0 0 1]

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

0 0 1

]

Hence F(x) F(y) = F(x+y)

Q4. Express the given Matrix as the sum of a symmetric and skew symmetric matrix

A= [6 −2 2

−2 3 −12 −1 3

]

Sol. Here 𝐴𝑇 = [6 −2 2

−2 3 −12 −1 3

]

P = 1

2(𝐴 + 𝐴𝑇) =

1

2[12 −4 4−4 6 −24 −2 6

] = [6 −2 2

−2 3 −12 −1 3

]

Now 𝑃𝑇 = 𝑃 so P = 1

2(𝐴 + 𝐴𝑇) is a symmetric Matrix.

Also let Q = = 1

2(𝐴 − 𝐴𝑇) =

1

2[0 0 00 0 00 0 0

] = [0 0 00 0 00 0 0

]

15

𝑄𝑇 = −𝑄 Hence Q is an Skew Symmetric Matrix.

Now P + Q= [6 −2 2

−2 3 −12 −1 3

] + [0 0 00 0 00 0 0

] = [6 −2 2

−2 3 −12 −1 3

] = 𝐴

Thus A is represented as the sum of a symmetric and skew symmetric matrix.

Q5. Using the property of determinant prove that|𝑎 − 𝑏 − 𝑐 2𝑎 2𝑎

2𝑏 𝑏 − 𝑐 − 𝑎 2𝑏2𝑐 2𝑐 𝑐 − 𝑎 − 𝑏

| =

(𝑎 + 𝑏 + 𝑐)3

Sol.Applying𝑅1 → 𝑅1 + 𝑅2 + 𝑅3 𝑤𝑒 𝑔𝑒𝑡

L.H.S = |𝑎 + 𝑏 + 𝑐 𝑎 + 𝑏 + 𝑐 𝑎 + 𝑏 + 𝑐

2𝑏 𝑏 − 𝑐 − 𝑎 2𝑏2𝑐 2𝑐 𝑐 − 𝑎 − 𝑏

|

Taking common a + b + c from first Row we get

L.H.S =(a + b+ c) |1 1 12𝑏 𝑏 − 𝑐 − 𝑎 2𝑏2𝑐 2𝑐 𝑐 − 𝑎 − 𝑏

|

Now applying 𝐶2 → 𝐶2 − 𝐶1 , 𝐶3 → 𝐶3 − 𝐶1 𝑤𝑒 𝑔𝑒𝑡

L.H.S =(a + b+ c) |

1 0 02𝑏 −(𝑎 + 𝑏 + 𝑐) 02𝑐 0 −(𝑎 + 𝑏 + 𝑐)

|

Expanding along first Row L.H.S = (a+b+c)[(𝑎 + 𝑏 + 𝑐)2 − 0] = (𝑎 + 𝑏 + 𝑐)3 = R.H.S

Hence proved

Q6. Solve the system of equations x + 2y – 3z = - 4, 2x + 3y + 2z = 2, 3x – 3y – 4z = 11

Sol. The given system of equation can be written as A X = B where

A = [1 2 −32 3 23 −3 −4

]𝑋 = [𝑥𝑦𝑧] 𝐵 = [

−4211

]

Now |𝐴| = |1 2 −32 3 23 −3 −4

| = −6 + 28 + 45 = 67

adj(A) = [−6 14 −1517 5 913 −8 −1

]

𝑇

= [−6 17 1314 5 −8

−15 9 −1]𝑵𝒐𝒘 𝑨−𝟏 =

𝒂𝒅𝒋(𝑨)

|𝑨| =

𝟏

𝟔𝟕[−6 17 1314 5 −8

−15 9 −1]

Hence 𝑋 = 𝑨−𝟏𝑩 = 𝟏

𝟔𝟕[−6 17 1314 5 −8

−15 9 −1] [

−4211

] =𝟏

𝟔𝟕[24 + 34 + 143−56 + 10 − 8860 + 18 − 11

]

So[𝑥𝑦𝑧] =

𝟏

𝟔𝟕[

201−13467

] = [3

−21

]

Hence x = 3 , y = -2 , z = 1

Flow chart:

16

Step 1.Write the given system of equation in the form of A X = B

Step2. Find |𝑨|

Step3. Find adj(A)

Step4. Find 𝑨−𝟏 =𝒂𝒅𝒋(𝑨)

|𝑨|

Step5. Find 𝑿 = 𝑨−𝟏𝑩

Step6 Find the value of x , y and z

ASSIGNMENTS

(i). Order, Addition, Multiplication and transpose of matrices:

LEVEL I

1. If a matrix has 6 elements, what are the possible orders it can have?

2. Construct a 3 × 2 matrix whose elements are given by aij = |i – 3j |

3. If A = , B = , then find A –2 B.

4. If A = and B = , write the order of AB and BA.

LEVEL II

1. For the following matrices A and B, verify (AB)T = BTAT, where A = , B = .

2. Give example of matrices A & B such that AB = O, but BA ≠ O, where O is a zero matrix and

A, B are both non zero matrices.

3. If B is skew symmetric matrix, write whether the matrix (ABAT) is symmetric or skew symmetric.

4. If A = and I = , find a and b so that A2 + aI = bA

LEVEL III

1. If A = , then find the value of A2 –3A + 2I

2. Express the matrix A as the sum of a symmetric and a skew symmetric matrix, where:

A =

3. If A = [𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃−𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃

] 𝑡ℎ𝑒𝑛 𝑝𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 = 𝐴𝑛 = [𝑐𝑜𝑠𝑛𝜃 𝑠𝑖𝑛𝑛𝜃−𝑠𝑖𝑛𝑛𝜃 𝑐𝑜𝑠𝑛𝜃

] , 𝑛𝜖𝑁

(ii) Cofactors &Adjoint of a matrix

LEVEL I

1. Find the co-factor of a12 in A =

2. Find the adjoint of the matrix A =

LEVEL II

Verify A(adjA) = (adjA) A = I if

17

1. A =

2. A =

(iii)Inverse of a Matrix & Applications

LEVEL I

1. If A = , write A-1 in terms of A

2. If A is square matrix satisfying A2 = I, then what is the inverse of A ?

3. For what value of k , the matrix A = is not invertible ?

LEVEL II

1. If A = , show that A2 –5A – 14I = 0. Hence find A-1

2. If A, B, C are three non-zero square matrices of same order, find the condition

on A such that AB = AC B = C.

3. Find the number of all possible matrices A of order 3 × 3 with each entry 0 or 1 and

for which

0

0

1

z

y

x

A has exactly two distinct solutions.

LEVEL III

1. If A = , find A-1 and hence solve the following system of

equations: 2x – 3y + 5z = 11, 3x + 2y – 4z = - 5, x + y – 2z = - 3.

2. Using matrices, solve the following system of equations:

(i) x + 2y - 3z = - 4 , 2x + 3y + 2z = 2 , 3x - 3y – 4z = 11

(ii) 4x + 3y + 2z = 60 , x + 2y + 3z = 45 , 6x + 2y + 3z = 70

3. Find the product AB, where A = , B = and use it to

Solve the equations x – y = 3, 2x + 3y + 4z = 17, y + 2z = 7

4. Using matrices solve the following system of equations:

2111

0312

4111

zyx

zyx

zyx

5. A trust caring for indicate children gets rupees 30,000/- every month from its donors. The

trust spends half of the funds received for medical and educational care of the children and

for that it charges 2% of the spent amount from them and deposits the balance note in a

private bank to get the money multiplied so that in future the trust goes on functioning

regularly. What % of interest should the trust get from the bank to get a total of rupees 1800/-

18

every month? Use matrix method to find the rate of interest? Do u think people should donate

to such trusts?

6. Using elementary transformations, find the inverse of the matrix

(iv)To Find The Difference Between

LEVEL I

1. Evaluate 7575

1515

CosSin

SinCos

2. What is the value of , where I is identity matrix of order 3?

3. If A is non singular matrix of order 3 and = 3, then find

4. For what valve of a, is a singular matrix?

LEVEL II

1. If A is a square matrix of order 3 such that = 64, find

2. If A is a nonsingular matrix of order 3 and = 7, then find

LEVEL III

1. If A = and 3 = 125, then find a.

2. A square matrix A, of order 3, has = 5, find

(v).Properties of Determinants

LEVEL I

1. Find positive valve of x if =

2. Evaluate

LEVEL II

Using properties of determinants, prove the following :

1.

2.

3. = (1 + pxyz)(x - y)(y - z) (z - x)

LEVEL III

1. Using properties of determinants, solve the following for x :

abc4

bacc

bacb

aacb

322

22

22

22

ba1

ba1a2b2

a2ba1ab2

b2ab2ba1

19

a. = 0

b. = 0

c. = 0

2. If a, b, c, are positive and unequal, show that the following determinant is negative:

=

3.

4.

5.

6.

7. = 2abc( a + b + c)3

8. If a, b, c are real numbers, and

Show that either a + b +c = 0 or a = b = c.

ANSWERS

1. Order, Addition, Multiplication and transpose of matrices:

LEVEL I

1. 1 6, 6 1 , 2 3 , 3 2 2. 3. 4. 2 2, 3 3

LEVEL II

3.skew symmetric 4. a = 8, b = 8

222

2

2

2

cba1

1ccbca

bc1bab

acab1a

abc3cba

baaccb

accbba

cba333

0

baabba

accaac

cbbccb

22

22

22

3

22

22

22

)cabcab(

ababbaba

accacaca

bccbcbbc

0

accbba

cbbaac

baaccb

20

LEVEL III.

1. 2. +

(ii). Cofactors &Adjoint of a matrix

LEVEL I

1. 46 2.

(iii)Inverse of a Matrix & Applications

LEVEL I

1. A-1 = -A 2. A-1 = A 3. k = 17

LEVEL II

1. 3. 512

LEVEL III

1.x = 1, y = 2, z = 3. 2. (i) x = 3, ,y = -2, z = 1. (ii) x=7,y=4,z=10 3. AB = 6I, x =

, y = - 1, z =

4.x = ½, y = -1, z = 1. 6.

(iv). To Find The Difference Between

LEVEL I

1. 0 2. 27 3.24 4.

LEVEL II

1. 8 2. 49

LEVEL III

1.a = 3 2. 125

(v). Properties of Determinants

LEVEL I

1. x = 4 2. + + +

LEVEL II

2. [Hint: Apply C1 C1–bC3 and C2 C2+aC3]

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