Plus Two Math's Solution Miscellaneous Chapter 3 Matrices

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Board	SCERT, Kerala
Text Book	NCERT Based
Class	Plus Two
Subject	Math's Textbook Solution
Chapter	Chapter 3
Exercise	Miscellaneous
Chapter Name	Matrices
Category	Plus Two Kerala

Kerala Syllabus Plus Two Math's Textbook Solution Chapter  3 Matrices miscellaneous Exercises 

Chapter 3 : Matrices Solution

• Exercises 3.1
• Exercises 3.2
• Exercises 3.3
• Exercises 3.4
• Miscellaneous Exercise Chapter 3

Kerala plus two maths NCERT textbooks, we provide complete solutions for the exercise and answers provided at the end of each chapter. We also cover the entire syllabus given by the Board of secondary education, Kerala state.

Chapter 3: Matrices miscellaneous Exercise   

Question 1:

Find the matrix X so that 

Solution 1:

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 matrices. Therefore, X has to be a 2 × 2 matrices.

Now, let

Therefore, we have:

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

Thus, a = 1, b = 2, c = −2, d = 0

Hence, the required matrix X is 

Question 2:

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

Solution 2:

We can prove it by mathematical induction

Question 3:

Choose the correct answer in the following questions:

If is such that then

A. 

B. 

C. 

D. 

Solution 3:

C

Question 4:

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

Solution 4:

A is a zero matrix.

Question 5:

For what values of

Solution 5:

We have:

∴4 + 4x = 0

⇒ x = −1

Thus, the required value of x is −1.

Question 6:

If, show that 

Solution 6:

It is given that 

Question 7:

Let, show that, where I is the identity matrix of order 2 and n ∈ N 

Solution 7:

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:

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

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

Question 8:

If, prove that 

Solution 8:

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.

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

Thus by the principle of mathematical induction, we have:

Question 9:

If, then prove where n is any positive integer

Solution 9:

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.

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

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

 

Question 10:

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

Solution 10:

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

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

Question 11:

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

Solution 11:

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

Consider

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

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

Then, 

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

Hence, if A is a symmetric or skew-symmetric matrix, thenis a symmetric or skew-symmetric matrix accordingly.

Question 12:

Find the values of x, y, z if the matrix satisfy the equation 

Solution 12:

Now, 

On comparing the corresponding elements, we have:

Question 13:

Find x, if 

Solution 13:

We have:

 Question 14:

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

Annual sales are indicated below: 

(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.

(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.

Solution 14:

(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:

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 15:

Find the matrix X so that 

Solution 15:

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:

Thus, a = 1, b = 2, c = −2, d = 0

Hence, the required matrix X is 

Question 16:

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

Solution 16:

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

For n = 1, we have:

Therefore, the result is true for n = 1.

Let the result be true for n = k.

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

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

Thus, by the principle of mathematical induction, we have

Now, we prove that for all n ∈ N

For n = 1, we have:

Therefore, the result is true for n = 1.

Let the result be true for n = k.

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

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

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

Question 17:

Choose the correct answer in the following questions:

If is such that then

A. 

B. 

C. 

D. 

Solution 17:

Answer: C

On comparing the corresponding elements, we have:

Question 18:

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

Solution 18:

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

Therefore, A is a zero matrix.

Question 19:

If A is square matrix such that then is equal to

A. A B. I − A C. I D. 3A

Solution 19:

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            Plus Two Math's Part I

• Chapter 1: Relations and Functions
• Chapter 2: Inverse Trigonometric Functions
• Chapter 3: Matrices
• Chapter 4: Determinants
• Chapter 5: Continuity and Differentiability
• Chapter 6: Application of Derivatives



             Plus Two Math's Part II

• Chapter  7: Application of Integrals
• Chapter  8: Integrals
• Chapter  9: Differential Equations
• Chapter 10: Vector Algebra
• Chapter 11: Three Dimensional Geometry
• Chapter 12: Linear Programming
• Chapter 13: Probability

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