**Topics and Sub Topics in Class 11 Maths Chapter 3 Matrices:**

Section Name | Topic Name |

3 | Matrices |

3.1 | Introduction |

3.2 | Matrix |

3.3 | Types of Matrices |

3.4 | Operations on Matrices |

3.5 | Transpose of a Matrix |

3.6 | Symmetric and Skew Symmetric Matrices |

3.7 | Elementary Operation (Transformation) of a Matrix |

3.8 | Invertible Matrices |

**Contents**show

## NCERT Solutions for Class 12th Chapter 3 Maths Chapter 3 Matrices

#### Page No 64:

#### Question 1:

In the matrix, write:

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

(iii) Write the elements *a*_{13}, *a*_{21}, *a*_{33}, *a*_{24}, *a*_{23}

#### 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)** *a*_{13} = 19, *a*_{21} = 35, *a*_{33} = −5, *a*_{24} = 12, *a*_{23} =

#### 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:

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 4:

Construct a 2 × 2 matrix,, whose elements are given by:

(i)

(ii)

(iii)

#### Answer:

In general, a 2 × 2 matrix is given by

**(i)**

Therefore, the required matrix is

**(ii)**

Therefore, the required matrix is

**(iii)**

Therefore, the required matrix is

#### 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)**

Therefore, the required matrix is

**(ii)**

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:

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

**(ii)**

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

Comparing the corresponding elements, we get:

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

Now, 5 + *z* = 5 ⇒ *z* = 0

We know that:

(*x* − *y*)^{2} = (*x* + *y*)^{2} − 4*xy*

⇒ (*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.

Comparing the corresponding elements, we get:

*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

#### 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.

Comparing the corresponding elements, we get:

*a* − *b* = −1 … (1)

2*a* − *b* = 0 … (2)

2*a* + *c* = 5 … (3)

3*c* + *d* = 13 … (4)

From (2), we have:

*b* = 2*a*

Then, from (1), we have:

*a *− 2*a* = −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

#### Page No 65:

#### Question 8:

is a square matrix, if

**(A)** *m* < *n*

**(B)** *m* > *n*

**(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:

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 2^{9} = 512

#### Page No 80:

#### 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:

**(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)**

**(iii)**

**(iv)**

#### Question 3:

Compute the indicated products

(i)

(ii)

(iii)

(iv)

(v)

(vi)

#### Answer:

**(i)**

**(ii)**

**(iii)**

**(iv)**

**(v)**

**(****vi)**

#### Page No 81:

#### Question 4:

If, and, then compute and. Also, verify that

#### Answer:

#### Question 5:

If andthen compute.

#### Answer:

#### Question 6:

Simplify

#### Answer:

#### Question 7:

Find *X *and *Y*, if

(i) and

(ii) and

#### Answer:

**(i)**

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

**(ii)**

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

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

From (5) and (6), we have:

Now,

#### Question 8:

Find *X*, if and

#### Answer:

#### 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:

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:

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

2*x* − *y* = 10 and 3*x* + *y* = 5

Adding these two equations, we have:

5*x* = 15

⇒ *x* = 3

Now, 3*x* + *y* = 5

⇒ *y* = 5 − 3*x*

⇒ *y* = 5 − 9 = −4

∴*x* = 3 and *y* = −4

#### Question 12:

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

*w*.

#### Answer:

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

#### Page No 82:

#### Question 13:

If, show that.

#### Answer:

#### Question 14:

Show that

(i)

(ii)

#### Answer:

**(i)**

**(ii)**

#### Question 15:

Find if

#### Answer:

We have *A*^{2} = *A* × *A*

#### Question 16:

If, prove that

#### Answer:

#### Question 17:

If and, find *k* so that

#### Answer:

Comparing the corresponding elements, we have:

Thus, the value of *k* is 1.

#### Question 18:

Ifand *I* is the identity matrix of order 2, show that

#### Answer:

#### 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:

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.

#### Page No 83:

#### Question 20:

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.

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:

The correct answer is B.

Matrix *X* is of the order 2 × *n*.

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

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

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

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

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

#### Page No 88:

#### 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:

**(i)**

**(ii)**

#### Question 3:

If

and, then verify that

(i)

(ii)

#### Answer:

**(i)** It is known that

Therefore, we have:

**(ii)**

#### Question 4:

If

and, then find

#### Answer:

We know that

#### Question 5:

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

(i)

(ii)

#### Answer:

**(i)**

**(ii)**

#### Page No 89:

#### Question 6:

If (i) , then verify that

(ii) , then verify that

#### Answer:

**(i)**

**(ii)**

#### 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:

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:

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)

(ii)

(iii)

(iv)

#### Answer:

**(i)**

Thus, is a symmetric matrix.

Thus, is a skew-symmetric matrix.

Representing *A* as the sum of *P* and *Q*:

**(ii)**

Thus, is a symmetric matrix.

Thus, is a skew-symmetric matrix.

Representing *A* as the sum of *P* and *Q*:

**(iii)**

Thus, is a symmetric matrix.

Thus, is a skew-symmetric matrix.

Representing *A* as the sum of *P* and *Q*:

**(iv)**

Thus, is a symmetric matrix.

Thus, is a skew-symmetric matrix.

Representing *A* as the sum of *P* and *Q*:

#### Page No 90:

#### 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:

The correct answer is B.

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

#### Page No 97:

#### Question 1:

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

#### Answer:

We know that *A* = *IA*

#### Question 2:

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

#### Answer:

We know that *A* = *IA*

#### Question 3:

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

#### Answer:

We know that *A* = *IA*

#### Question 4:

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

#### Answer:

We know that *A* = *IA*

#### Question 5:

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

#### Answer:

We know that *A* = *IA*

#### Question 6:

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

#### Answer:

We know that *A* = *IA*

#### Question 7:

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

#### Answer:

We know that *A* = *AI*

#### Question 8:

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

#### Answer:

We know that *A* = *IA*

#### Question 9:

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

#### Answer:

We know that *A* = *IA*

#### Question 10:

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

#### Answer:

We know that *A* = *AI*

#### Question 11:

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

#### Answer:

We know that *A* = *AI*

#### Question 12:

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

#### 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:

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

#### Answer:

We know that *A* = *IA*

#### Question 14:

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

#### Answer:

We know that *A* = *IA*

Applying, we have:

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 15:

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

#### Answer:

We know that *A* = *IA*

#### Question 16:

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

#### Answer:

We know that *A* = *IA*

Applying R_{2} → R_{2} + 3R_{1} and R_{3} → R_{3} − 2R_{1}, we have:

#### Question 17:

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

#### Answer:

We know that *A* = *IA*

Applying, we have:

#### 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:

**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*.

#### Page No 100:

#### 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:

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

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 3:

If, then prove where *n* is any positive integer

#### 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.

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

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:

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, 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 6:

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

#### Answer:

Now,

On comparing the corresponding elements, we have:

#### Question 7:

For what values of?

#### Answer:

We have:

∴4 + 4*x* = 0

⇒ *x* = −1

Thus, the required value of *x* is −1.

#### Question 8:

If, show that

#### Answer:

#### Question 9:

Find *x*, if

#### Answer:

We have:

#### Page No 101:

#### 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.

(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:

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:

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:

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 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:

#### 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. **3*A*

#### Answer:

**Answer: C**

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