NCERT Solutions For Class 12 Maths Chapter 3

NCERT Solutions Class 12 Maths Matrices Symmetric Matrices

Ncert Solutions For Class 12 Maths Chapter 3 PDF Download

NCERT solutions for class 12 maths chapter 3 – matrices is one of the most important topic for the students to practice. For solving questions, it is crucial to have good knowledge of the topic. Students can study various concepts of maths by visiting our site BYJU’S. To ease the fear of maths for the students, we at BYJU’S provide NCERT Solution for Class 12 Maths Chapter 3 Matrices. Student can download the NCERT Solution for class 12 Maths Chapter 3 pdf or can view it online by following the link. These solutions are provided in a detailed manner, where one can find step-by-step solution to all the questions of class 12 maths chapter 3.

Class 12 maths chapter 3 NCERT solutions are prepared by our subject experts under the guidelines of NCERT to assists students in their examination. The chapter 3 in class 12 mathematics is easy to remember and practice from. The topic of these chapters is Matrices and Symmetry of Matrices which are easy to learn and prepare from. You can find the entire list of class 12 maths exercises below containing all the topics in the respective chapter. The topics in mathematics for this chapter contains information and solutions to the majority of the problems asked. The symmetry of matrices are important to remember as a student can score the maximum amount of marks.

NCERT Solutions Class 12 Maths Chapter 3 Exercises

EXERCISE – 3.1

Q-1: In a matrix given below:

A = \(\begin{bmatrix} 3 & 6 & 20 & -8 \\ 36 & -3 & \frac{ 7 }{ 2 } & 13 \\ \sqrt{ 5 } & 2 & -6 & 18 \end{bmatrix}\), write

(a) the order of the matrix.

(b) the number of element.

(c) Write the elements of a13, a21, a23, a24, a33

 

Solution:

(a) In the matrix given in the question, there are 3 rows and 4 columns. Hence, the order of the matrix will be 3 × 4.

(b) As the order of the given matrix is 3 × 4, there are 3 × 4 = 12 elements in the matrix.

(c) a13 = 20, a21 = 35, a23 = \(\frac{ 7 }{ 2 }\), a24 = 13, a33 = -5

 

 

Q-2: Let us consider a matrix A having 24 elements. Then, find the possible order for the matrix. What if there are 15 element of the same matrix?

Solution:

We know it well that if the order of the matrix is m × n, then the number of elements in the matrix will be mn. Then, to get all the possible orders of the matrix which having 24 elements, we need to find all of the possible pairs of the natural numbers having product 24.

Thus,

The possible ordered pairs will be:

( 1, 24 ), ( 2, 12 ) , ( 3, 8 ) , ( 4, 6 ) , ( 6, 4 ) , ( 8, 3 ), ( 12, 2 ), ( 24, 1 )

Therefore, the possible orders for the matrix with 24 elements are:

( 1 × 24 ), ( 2 × 12 ) , ( 3 × 8 ) , ( 4 × 6 ) , ( 6 × 4 ) , ( 8 × 3 ), ( 12 × 2 ), ( 24 × 1 )

If the number of elements in the matrix is 15, then the possible ordered pairs will be:

( 1, 15 ), ( 3, 5 ), ( 5, 3 ), ( 15, 1 )

Therefore, the possible orders for the matrix with 13 elements are:

( 1 × 15 ), ( 3 × 5 ) , ( 5 × 3 ) , ( 15 × 1 )

 

 

Q-3: Let us consider a matrix A having 21 elements. Then, find the possible order for the matrix. What if there are 10 element of the same matrix?

 

Solution:

We know it well that if the order of the matrix is m × n, then the number of elements in the matrix will be mn. Then, to get all the possible orders of the matrix which having 21 elements, we need to find all of the possible pairs of the natural numbers having product 21.

Thus,

The possible ordered pairs will be:

( 1, 21 ), ( 3, 7 ) , ( 7, 3 ), ( 21, 1 )

Therefore, the possible orders for the matrix with 24 elements are:

( 1 × 21 ), ( 3 × 7 ), ( 7 × 3 ), ( 21 × 1 )

If the number of elements in the matrix is 10, then the possible ordered pairs will be:

( 1, 10 ), ( 2, 5 ), ( 5, 2 ), ( 10, 1 )

Therefore, the possible orders for the matrix with 13 elements are:

( 1 × 10 ), ( 2 × 5 ) , ( 5 × 2 ) , ( 10 × 1 )

 

 

Q-4: Construct a 2 × 2 matrix, A = [ mij ], whose elements will be given by

(a) mij = \(\frac{ \left ( i + j \right )^{ 2 } }{ 2 }\)

(b) mij = \(\frac{  i }{ j }\)

(c) mij = \(\frac{ \left ( i + 2j \right )^{ 2 } }{ 2 }\)

 

Solution:

(a) As per the data given in the question:

mij = \(\frac{ \left ( i + j \right )^{ 2 } }{ 2 }\)

A ( 2 × 2 ) matrix is usually given by,

A = \(\begin{bmatrix} a_{ 11 } & a_{ 12 } \\ a_{ 21 } & a_{ 22 } \end{bmatrix}\)

Then,

a11 = \(\frac{ \left ( 1 + 1 \right )^{ 2 } }{ 2 } = \frac{ 4 }{ 2 } = 2\)

a12 = \(\frac{ \left ( 1 + 2 \right )^{ 2 } }{ 2 } = \frac{ 9 }{ 2 }\)

a21 = \(\frac{ \left ( 2 + 1 \right )^{ 2 } }{ 2 } = \frac{ 9 }{ 2 }\)

a22 = \(\frac{ \left ( 2 + 2 \right )^{ 2 } }{ 2 } = \frac{ 16 }{ 2 } = 8\)

Hence, the required matrix is A = \(\begin{bmatrix} 2 & \frac{ 9 }{ 2 } \\ \frac{ 9 }{ 2 } & 8 \end{bmatrix}\).

 

(b) As per the data given in the question:

mij = \(\frac{ i }{ j }\)

A ( 2 × 2 ) matrix is usually given by:

A = \(\begin{bmatrix} a_{ 11 } & a_{ 12 } \\ a_{ 21 } & a_{ 22 } \end{bmatrix}\)

Then,

a11 = \(\frac{ 1 }{ 1 } = 1\)

a12 = \(\frac{ 1 }{ 2 }\)

a21 = \(\frac{ 2 }{ 1 } = \frac{ 2 }{ 1 }\)

a22 = \(\frac{ 2 }{ 2 } = 1\)

Hence, the required matrix is A = \(\begin{bmatrix} 1 & \frac{ 1 }{ 2 } \\ 2 & 1 \end{bmatrix}\)

 

(c) As per the data given in the question:

mij = \(\frac{ \left ( i + 2j \right )^{ 2 } }{ 2 }\)

A ( 2 × 2 ) matrix is usually given by,

A = \(\begin{bmatrix} a_{ 11 } & a_{ 12 } \\ a_{ 21 } & a_{ 22 } \end{bmatrix}\)

Then,

a11 = \(\frac{ \left ( 1 + 2 \times 1 \right )^{ 2 } }{ 2 } = \frac{ 9 }{ 2 }\)

a12 = \(\frac{ \left ( 1 + 2 \times 2 \right )^{ 2 } }{ 2 } = \frac{ 25 }{ 2 }\)

a21 = \(\frac{ \left ( 2 + 2 \times 1 \right )^{ 2 } }{ 2 } = \frac{ 16 }{ 2 } = 8\)

a22 = \(\frac{ \left ( 2 + 2 \times 2 \right )^{ 2 } }{ 2 } = \frac{ 36 }{ 2 } = 18\)

Hence, the required matrix is A = \(\begin{bmatrix} \frac{ 9 }{ 2 } & \frac{ 25 }{ 2 } \\ 8 & 18 \end{bmatrix}\)

 

Q-5: Construct a 3 × 4 matrix, A = [ mij ], whose elements will be given by

(a) mij = \(\frac{ 1 }{ 2 }| -3i + j |\)

(b) mij = \(2i – j\)

 

Solution:

(a) As per the data given in the question:

mij = \(\frac{ 1 }{ 2 }| -3i + j |\)

A ( 3 × 4 ) matrix is usually given by:

A = \(\begin{bmatrix} a_{ 11 } & a_{ 12 } & a_{ 13 } & a_{ 14 } \\ a_{ 21 } & a_{ 22 } & a_{ 23 } & a_{ 24 } \\ a_{ 31 } & a_{ 32 } & a_{ 33 } & a_{ 34 } \end{bmatrix}\)Then,

a11 = \(\frac{ 1 }{ 2 }| -3 \times 1 + 1 | = 1\)

a12 = \( \frac{ 1 }{ 2 }| -3 \times 1 + 2 |= \frac{ 1 }{ 2 }\)

a13 = \( \frac{ 1 }{ 2 }| -3 \times 1 + 3 |= 0\)

a14 = \( \frac{ 1 }{ 2 }| -3 \times 1 + 4 | = \frac{ 1 }{ 2 }\)

a21 = \(\frac{ 1 }{ 2 }| -3 \times 2 + 1 | = \frac{ 5 }{ 2 }\)

a22 = \( \frac{ 1 }{ 2 }| -3 \times 2 + 2 |= 2\)

a23 = \( \frac{ 1 }{ 2 }| -3 \times 2 + 3 |= \frac{ 3 }{ 2 }\)

a24 = \( \frac{ 1 }{ 2 }| -3 \times 2 + 4 | = 1\)

a31 = \(\frac{ 1 }{ 2 }| -3 \times 3 + 1 | = 4\)

a32 = \( \frac{ 1 }{ 2 }| -3 \times 3 + 2 |= \frac{ 7 }{ 2 }\)

a33 = \( \frac{ 1 }{ 2 }| -3 \times 3 + 3 |= 3\)

a34 = \( \frac{ 1 }{ 2 }| -3 \times 3 + 4 | = \frac{ 5 }{ 2 }\)

Hence, the required matrix is A = \(\begin{bmatrix} 1 & \left ( \frac{ 1 }{ 2 } \right ) & 0 & \left ( \frac{ 1 }{ 2 } \right ) \\ \\ \left ( \frac{ 5 }{ 2 } \right ) & 2 & \left ( \frac{3 }{ 2 } \right ) & 1 \\ \\ 4 & \left ( \frac{ 7 }{ 2 } \right ) & 3 & \left ( \frac{ 5 }{ 2 } \right ) \end{bmatrix}\)

 

(b) As per the data given in the question:

mij = \(2i + j\)

A ( 3 × 4 ) matrix is usually given by,

A = \(\begin{bmatrix} a_{ 11 } & a_{ 12 } & a_{ 13 } & a_{ 14 } \\ a_{ 21 } & a_{ 22 } & a_{ 23 } & a_{ 24 } \\ a_{ 31 } & a_{ 32 } & a_{ 33 } & a_{ 34 } \end{bmatrix}\)Then,

a11 = \(2 \times 1 – 1 = 1\)

a12 = \(2 \times 1 – 2 = 0\)

a13 = \( 2 \times 1 – 3 = -1\)

a14 = \( 2 \times 1 – 4 = -2\)

a21 = \( 2 \times 2 – 1 = 3\)

a22 = \( 2 \times 2 – 2 = 2\)

a23 = \( 2 \times 2 – 3 = 1\)

a24 = \( 2 \times 2 – 4 = 0\)

a31 = \( 2 \times 3 – 1 = 5\)

a32 = \( 2 \times 3 – 2 = 4\)

a33 = \( 2 \times 3 – 3 = 3\)

a34 = \( 2 \times 3 – 4 = 2\)

Hence, the required matrix is A = \(\begin{bmatrix} 1 & 0 & -1 & -2 \\ 3 & 2 & 1 & 0 \\ 5 & 4 & 3 & 2 \end{bmatrix}\)

 

 

Q-6: What will be the value of a, b and c in the following equation:

(a) \(\begin{bmatrix} 6 & 3 \\ a & 7 \end{bmatrix} = \begin{bmatrix} b & c \\ 1 & 7 \end{bmatrix}\)

(b) \(\begin{bmatrix} a + b  &  2 \\ 5 + c  &  ab  \end{bmatrix} = \begin{bmatrix} 9 & 2 \\ 6 & 8\end{bmatrix}\)

(c) \(\begin{bmatrix} a + b + c \\ a + c \\ b + c \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \\ 7 \end{bmatrix}\)

 

Solution:

(a) \(\begin{bmatrix} 6 & 3 \\ a & 7 \end{bmatrix} = \begin{bmatrix} b & c \\ 1 & 7 \end{bmatrix}\)

Since, the given matrices are equal, then their corresponding elements will also be equal.

Now,

By comparing the corresponding elements, we will get:

a = 1, b = 6 and c = 3

 

(b) \(\begin{bmatrix} a + b  &  2 \\ 5 + c  &  ab  \end{bmatrix} = \begin{bmatrix} 9 & 2 \\ 6 & 8\end{bmatrix}\)

Since, the given matrices are equal, then their corresponding elements will also be equal.

Now,

By comparing the corresponding elements, we will get

a + b = 9 ………………… (i)

5 + c = 6 ………………… (ii)

And,

ab = 8 ………………… (iii)

From (ii), we will get:

c = 6 – 5 = 1

From (i), we have

a = 9 – b ………………… (iv)

Substituting (iv) in equation (iii), we will get:

( 9 – b )b = 8

⟹ 9b – b2 = 8

⟹ b2 – 9b + 8 = 0

⟹ b2 – 8b – b + 8 = 0

⟹ b( b – 8 ) -1( b – 8 ) = 0

⟹ ( b – 8 )( b – 1 ) = 0

b = 1, 8

For b = 1, a = 9 – 1 = 8

For b = 8, a = 9 – 8 = 1

Hence, a = 1, b = 8 and c = 1 or, a = 8, b = 1 and c = 1

 

(c) \(\begin{bmatrix} a + b + c \\ a + c \\ b + c \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \\ 7 \end{bmatrix}\)

Since, the given matrices are equal, then their corresponding elements will also be equal.

Now,

By comparing the corresponding elements, we will get

a + b + c = 9 ………………(i)

a + c = 5 .………………(ii)

b + c = 7.………………(iii)

From equation (i), we have:

a + ( b + c ) = 9 ………………..(iv)

From equation (iii) and (iv), we will get:

a + 7 = 9

a = 2

Putting the value of an in equation (ii), we will get:

2 + c = 5

c = 3

Putting the value of c in equation (iii), we will get:

b + 3 = 7

b = 4

Hence, a = 2, b = 4 and c= 3.

 

 

Q-7: What will be the value of p, q, r and s in the following equation:

\(\begin{bmatrix} p – q & 2p – r \\ 2p – q & 3r + s \end{bmatrix} = \begin{bmatrix} -2 & 6 \\ 0 & 14 \end{bmatrix}\)

 

Solution:

\(\begin{bmatrix} p – q & 2p – r \\ 2p – q & 3r + s \end{bmatrix} = \begin{bmatrix} -2 & 6 \\ 0 & 14 \end{bmatrix}\)

Since, the given matrices are equal, then their corresponding elements will also be equal.

Now,

By comparing the corresponding elements, we will get:

p – q = -2 …………(i)

2p – r = 0 …………(ii)

2p + r = 6 …………(iii)

3r + s = 14 …………(iv)

From equation (ii), we have:

r = 2p

Putting it in equation (iii), we will get:

2p + 2p = 6

4p = 6

⟹ p = \(\frac{ 6 }{ 4 }\)

⟹ p = \(\frac{ 3 }{ 2 }\)

Putting the value of p in equation (ii), we will get:

r = 2 × \(\frac{ 3 }{ 2 }\) = 3

Putting the value of r in equation (iv), we will get:

3 × 3 + s = 14

s = 5

Putting the value of p in equation (i), we will get:

\(\frac{ 3 }{ 2 }\) – q = -2

⟹ q = \(\frac{ 3 }{ 2 }\) + 2

⟹ q = \(\frac{ 3 + 2 \times 2 }{ 2 } = \frac{ 7 }{ 2 }\)

Hence, p = \(\frac{ 3 }{ 2 }\), q = \(\frac{ 7 }{ 2 }\), r = 3 and s = 5.

 

 

Q-8: Let us consider a matrix A = [aij]m × n. The matrix will be a square matrix, if

(a)  m < n                   (b) m > n                        (c) m = n                           (d) None of these

 

Solution:

We know that a matrix can be a square matrix if and only if the number of rows and number of the columns of the matrix.

i.e., m = n

Hence, A = [aij]m × n will be a square matrix when m = n

Therefore, (c) is the correct answer.

 

 

Q-9. What will be the value of a and b in the following equation:

 \(\begin{bmatrix} 3a + 7 & 5 \\ b + 1 & 2 – 3a \end{bmatrix} = \begin{bmatrix} 0 & b – 2 \\ 8 & 4 \end{bmatrix}\)

(a) a = \(\frac{ -1 }{ 3 }\), b = 7              

(b) Not possible to find

(c) y = 7, x \(\frac{ -2 }{ 3 }\)

(d) x = \(\frac{ -1 }{ 3 }, y = \frac{ -2 }{ 3 }\)

 

Solution:

From the matrix, we have:

3a + 7 = 0

a = \(\frac{ -7 }{ 3 }\)

b + 1 = 8

b = 7

Also,

5 = b – 2

⟹ b = 3

2 – 3a = 4

⟹ 3a = -2

⟹ a = \(\frac{ -2 }{ 3 }\)

Hence, the correct answer is (b).

 

 

Q-10. The number of all the possible matrix of order 3 × 3 having each entry as 0 and 1 is-

(a) 512                                (b) 81                                     (c) 18                                    (d) 27

 

Solution:

As the given matrix has order 3 × 3, i.e., the matrix have 9 elements and each of the element of this matrix will be either 0 or 1.

As we have only two digits and there are 9 elements in the matrix, so, the matrix will be filled in two possible ways only.

Hence, by the principle of multiplication, required number for the possible matrices will be 29 = 512

Therefore, the correct answer is (a).

 

 

EXERCISE – 3.2

 

 

Q-1: A = \(\begin{bmatrix} 3 & 5 \\ 4 & 3 \end{bmatrix}\), B = \(\begin{bmatrix} 2 & 4 \\ -3 & 6 \end{bmatrix}\),  C = \(\begin{bmatrix} -3 & 6 \\ 4 & 5 \end{bmatrix}\)

Find the value of each of the following from the matrices given:

(a) A + B   (b) A – B   (c) 3A – C   (d) AB   (e) BA

 

Solution:

(a) A + B = \(\begin{bmatrix} 3 & 5 \\ 4 & 3 \end{bmatrix}\) + \(\begin{bmatrix} 2 & 4 \\ -3 & 6 \end{bmatrix}\)

= \(\begin{bmatrix} 3 + 2 & 5 + 4 \\ 4 – 3 & 3 + 6 \end{bmatrix} \)

= \( \begin{bmatrix} 5 & 9 \\ 1 & 9 \end{bmatrix}\)

 

(b) A – B = \(\begin{bmatrix} 3 & 5 \\ 4 & 3 \end{bmatrix}\) – \(\begin{bmatrix} 2 & 4 \\ -3 & 6 \end{bmatrix}\)

= \(\begin{bmatrix} 3 – 2 & 5 – 4 \\ 4 + 3 & 3 – 6 \end{bmatrix} \)

= \( \begin{bmatrix} 1 & 1 \\ 7 & -3 \end{bmatrix}\)

 

(c) 3A – C = \(3\begin{bmatrix} 3 & 5 \\ 4 & 3 \end{bmatrix} – \begin{bmatrix} -3 & 6 \\ 4 & 5 \end{bmatrix}\)

= \(\begin{bmatrix} 3 \times 3 & 5 \times 3 \\ 4 \times 3 & 3 \times 3 \end{bmatrix} – \begin{bmatrix} -3 & 6 \\ 4 & 5 \end{bmatrix}\)

= \(\begin{bmatrix} 9 & 15 \\ 12 & 9 \end{bmatrix} – \begin{bmatrix} -3 & 6 \\ 4 & 5 \end{bmatrix}\)

= \(\begin{bmatrix} 9 + 3 & 15 – 6 \\ 12 – 4 & 9 – 5 \end{bmatrix}\)

= \(\begin{bmatrix} 12 & 9 \\ 8 & 4 \end{bmatrix}\)

 

(d) Since, the matrix A and B, both are a square matrix as the number of rows and the number of columns is the same in both of the matrices, then

Multiplication of the two matrices A and B will be defined as:

AB = \(\begin{bmatrix} 3 & 5 \\ 4 & 3 \end{bmatrix}\)\(\begin{bmatrix} 2 & 4 \\ -3 & 6 \end{bmatrix}\)

= \(\begin{bmatrix} 3( 2 ) + 5( -3 ) & 3( 4 ) + 5( 6 ) \\ 4( 2 ) + 3( -3 ) & 4( 4 ) + 3( 6 ) \end{bmatrix}\)

= \(\begin{bmatrix} 6 – 15 & 12 + 30 \\ 8 – 9 & 16 + 18 \end{bmatrix}\)

= \(\begin{bmatrix} -9 & 42 \\ -1 & 34 \end{bmatrix}\)

 

(e) Since, the matrix A and B, both are a square matrix as the number of rows and the number of columns is the same in both of the matrices, then

Multiplication of the two matrices B and A will be defined as:

BA = \(\begin{bmatrix} 2 & 4 \\ -3 & 6 \end{bmatrix}\) \(\begin{bmatrix} 3 & 5 \\ 4 & 3 \end{bmatrix}\)

= \(\begin{bmatrix} 2( 3 ) + 3( 4 ) & 2( 5 ) + 4( 3 ) \\ -3( 3 ) + 6( 4 ) & -3( 5 ) + 6( 3 ) \end{bmatrix}\)

= \(\begin{bmatrix} 6 + 12 & 10 + 12 \\ – 9 + 24 & -15 + 18 \end{bmatrix}\)

= \(\begin{bmatrix} 18 & 22 \\ 15 & 3 \end{bmatrix}\)

 

Q-2: Solve the following:

(i) \(\begin{bmatrix} x & y \\ -y & x \end{bmatrix} + \begin{bmatrix} x & y \\ y & x \end{bmatrix}\)

(ii) \(\begin{bmatrix} x^{ 2 } + y^{ 2 } & y^{ 2 } + z^{ 2 } \\ x^{ 2 } + z^{ 2 } & x^{ 2 } + y^{ 2 } \end{bmatrix} + \begin{bmatrix} 2xy & 2yz \\ -2xz & -2xy \end{bmatrix}\)

(iii) \(\begin{bmatrix} -2 & 5 & -7 \\ 9 & 6 & 17 \\ 3 & 9 & 6 \end{bmatrix} + \begin{bmatrix} 13 & 8 & 7\\ 9 & 1 & 6 \\ 4 & 3 & 5 \end{bmatrix}\)

(iv) \(\begin{bmatrix} cos^{ 2 }a & sin^{ 2 }a \\ sin^{ 2 }a & cos^{ 2 }a \end{bmatrix} + \begin{bmatrix} sin^{ 2 }a & cos^{ 2 }a \\ cos^{ 2 }a & sin^{ 2 }a \end{bmatrix}\)

 

Solution:

(i) \(\begin{bmatrix} x & y \\ -y & x \end{bmatrix} + \begin{bmatrix} x & y \\ y & x \end{bmatrix}\)

= \(\begin{bmatrix} x + x & y + y \\ -y + y & x + x \end{bmatrix}\)

= \(\begin{bmatrix} 2x & 2y \\ 0 & 2x \end{bmatrix}\)

 

(ii) \(\begin{bmatrix} x^{ 2 } + y^{ 2 } & y^{ 2 } + z^{ 2 } \\ x^{ 2 } + z^{ 2 } & x^{ 2 } + y^{ 2 } \end{bmatrix} + \begin{bmatrix} 2xy & 2yz \\ -2xz & -2xy \end{bmatrix}\)

= \(\begin{bmatrix} x^{ 2 } + y^{ 2 } + 2xy  & y^{ 2 } + z^{ 2 } + 2yz \\ x^{ 2 } + z^{ 2 } – 2xz  & x^{ 2 } + y^{ 2 } – 2ab \end{bmatrix}\)

= \(\begin{bmatrix} \left ( x + y \right )^{ 2 } & \left ( y + z \right )^{ 2 } \\ \left ( x – y \right )^{ 2 } & \left ( x – y \right )^{ 2 } \end{bmatrix}\)

 

(iii) \(\begin{bmatrix} -2 & 5 & -7 \\ 9 & 6 & 17 \\ 3 & 9 & 6 \end{bmatrix} + \begin{bmatrix} 13 & 8 & 7\\ 9 & 1 & 6 \\ 4 & 3 & 5 \end{bmatrix}\)

= \(\begin{bmatrix} -2 + 13 & 5 + 8 & -7 + 7 \\ 9 + 9 & 6 + 1 & 17 + 6 \\ 3 + 4 & 9 + 3 & 6 + 5 \end{bmatrix}\)

= \(\begin{bmatrix} 11 & 13 & 0 \\ 18 & 6 & 23 \\ 7 & 12 & 11 \end{bmatrix}\)

 

(iv) \(\begin{bmatrix} cos^{ 2 }a & sin^{ 2 }a \\ sin^{ 2 }a & cos^{ 2 }a \end{bmatrix} + \begin{bmatrix} sin^{ 2 }a & cos^{ 2 }a \\ cos^{ 2 }a & sin^{ 2 }a \end{bmatrix}\)

= \(\begin{bmatrix} cos^{ 2 }a + sin^{ 2 }a & sin^{ 2 }a + cos^{ 2 }a \\ sin^{ 2 }a + cos^{ 2 }a & cos^{ 2 }a + sin^{ 2 }a \end{bmatrix}\)

= \(\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \;\;\;\;\;\;\;\;\left ( as sin^{ 2 }a + cos^{ 2 }a = 1 \right )\)

 

 

Q-3: Compute the following:

(i) \(\begin{bmatrix} x & y \\ -y & x \end{bmatrix}\begin{bmatrix} x & -y \\ y & a \end{bmatrix}\)

(ii) \(\begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \end{bmatrix}\)

(iii) \(\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 2 \end{bmatrix}\)

(iv) \(\begin{bmatrix} 3 & 4 & 5 \\ 4 & 5 & 6 \\ 5 & 6 & 7 \end{bmatrix}\begin{bmatrix} 2 & -4 & 6 \\ 0 & 3 & 4 \\ 3 & 0 & 5 \end{bmatrix}\)

(v) \(\begin{bmatrix} 3 & 2 \\ 4 & 3 \\ -2 & 2 \end{bmatrix}\begin{bmatrix} 1 & 0 & 1 \\ -1 & 3 & 2 \end{bmatrix}\)

(vi) \(\begin{bmatrix} 4 & -2 & 4 \\ -2 & 0 & 3 \end{bmatrix} \begin{bmatrix} 3 & -4 \\ 2 & 1 \\ 4 & 2 \end{bmatrix}\)

 

Solution:

(i) \(\begin{bmatrix} x & y \\ -y & x \end{bmatrix}\begin{bmatrix} x & -y \\ y & a \end{bmatrix}\)

= \(\begin{bmatrix} x( x )+ y( y ) & x( -y ) + y( x ) \\ -y( x ) + x( y ) & -y( -y ) + x( x ) \end{bmatrix}\)

= \(\begin{bmatrix} x^{ 2 } + y^{ 2 } & -xy + xy \\ -yx + xy & y^{ 2 } + x^{ 2 } \end{bmatrix}\)

= \(\begin{bmatrix} x^{ 2 } + y^{ 2 } & 0  \\ 0 & y^{ 2 } + x^{ 2 } \end{bmatrix}\)

 

(ii) \(\begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \end{bmatrix}\)

= \(\begin{bmatrix} 2( 1 ) & 2( 2 ) & 2( 3 )\\ 3( 1 ) & 3( 2 ) & 3( 3 )\\ 4( 1 ) & 4( 2 ) & 4( 3 ) \end{bmatrix}\)

= \(\begin{bmatrix} 2  & 4  & 6 \\ 3  & 6  & 9 \\ 4  & 8  & 12 \end{bmatrix}\)

 

(iii) \(\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 2 \end{bmatrix}\)

= \(\begin{bmatrix} 2( 2 ) – 3( 3 ) & 2( 3 ) – 3( 4 ) & 2( 4 ) – 3( 2 )\\ 3( 2 ) + 4( 3 ) & 3( 3 ) + 4( 4 ) & 3( 4 ) + 4( 2 ) \end{bmatrix}\)

= \(\begin{bmatrix} 4 – 9 & 6 – 12 & 8 – 6 \\ 6 + 12 & 9 + 16 & 12 + 8 \end{bmatrix}\)

= \(\begin{bmatrix} -5 & -6 & -2 \\ 18 & 25 & 20 \end{bmatrix}\)

 

(iv) \(\begin{bmatrix} 3 & 4 & 5 \\ 4 & 5 & 6 \\ 5 & 6 & 7 \end{bmatrix}\begin{bmatrix} 2 & -4 & 6 \\ 0 & 3 & 4 \\ 3 & 0 & 5 \end{bmatrix}\)

= \(\begin{bmatrix} 3( 2 ) + 4( 0 ) + 5( 3 ) & 3( -4 ) + 4( 3 ) + 5( 0 ) & 3( 6 ) + 4( 4 ) + 5( 5 ) \\ 4( 2 ) + 5( 0 ) + 6( 3 ) & 4( -4 ) + 5( 3 ) + 6( 0 ) & 4( 6 ) + 5( 4 ) + 6( 5 ) \\ 5( 2 ) + 6( 0 ) + 7( 3 ) & 5( -4 ) + 6( 3 ) + 7( 0 ) & 5( 6 ) + 6( 4 ) + 7( 5 ) \end{bmatrix}\)

= \(\begin{bmatrix} 6 + 0 + 15 & -12 + 12 + 0 & 18 + 16 + 25 \\ 8 + 0 + 18 & -16 + 15 + 0 & 24 + 20 + 30 \\ 10 + 0 + 21 & -20 + 18 + 0 & 30 + 24 + 35 \end{bmatrix}\)

= \(\begin{bmatrix} 21 & 0 & 59 \\ 26 & -1 & 74 \\ 31 & -2 & 89 \end{bmatrix}\)

 

(v) \(\begin{bmatrix} 3 & 2 \\ 4 & 3 \\ -2 & 2 \end{bmatrix}\begin{bmatrix} 1 & 0 & 1 \\ -1 & 3 & 2 \end{bmatrix}\)

= \(\begin{bmatrix} 3( 1 ) + 2( -1 ) & 3( 0 ) + 2( 3 ) & 3( 1 ) + 2( 2 ) \\ 4( 1 ) + 3( -1 ) & 4( 0 ) + 3( 3 ) & 4( 1 ) + 3( 2 ) \\ -2( 1 ) + 2( -1 ) & -2( 0 ) + 2( 3 ) & -2( 1 ) + 2( 2 ) \end{bmatrix}\)

= \(\begin{bmatrix} 3 – 2 & 0 + 6 & 3 + 4 \\ 4 – 3 & 0 + 9 & 4 + 6 \\ -2 – 2 & 0 + 6 & -2 + 4 \end{bmatrix}\)

= \(\begin{bmatrix} 1 & 6 & 7 \\ 1 & 9 & 10 \\ -4 & 6 & 2 \end{bmatrix}\)

 

(vi) \(\begin{bmatrix} 4 & -2 & 4 \\ -2 & 0 & 3 \end{bmatrix} \begin{bmatrix} 3 & -4 \\ 2 & 1 \\ 4 & 2 \end{bmatrix}\)

= \(\begin{bmatrix} 4( 3 ) – 2( 2 ) + 4( 4 ) & 4( -4 ) – 2( 1 ) + 4( 2 )\\ -2( 3 ) + 0( 2 ) + 3( 4 ) & -2( -4 ) + 0( 1 ) + 4( 2 ) \end{bmatrix}\)

= \(\begin{bmatrix} 12 – 4 + 16 & -16 – 2 + 8 \\ -6 + 0 + 12 & 8 + 0 + 6 \end{bmatrix}\)

= \(\begin{bmatrix} 24 & -10 \\ 6 & 14 \end{bmatrix}\)

 

Q-4: If, A = \(\begin{bmatrix} 2 & 3 & -4 \\ 5 & 0 & 3 \\ 2 & -2 & 2 \end{bmatrix}\), B = \(\begin{bmatrix} 4 & -2 & 3 \\ 5 & 3 & 6 \\ 3 & 1 & 4 \end{bmatrix}\) and  C = \(\begin{bmatrix} 5 & 2 & 3 \\ 1 & 4 & 3 \\ 2 & -3 & 4 \end{bmatrix}\), then

Find ( A + B ) and ( B – C ). Also, check whether A + ( B – C ) = ( A + B ) – C

 

Solution:

A + B = \(\begin{bmatrix} 2 & 3 & -4 \\ 5 & 0 & 3 \\ 2 & -2 & 2 \end{bmatrix}\) + \(\begin{bmatrix} 4 & -2 & 3 \\ 5 & 3 & 6 \\ 3 & 1 & 4 \end{bmatrix}\)

= \(\begin{bmatrix} 2 + 4  & 3 – 2  & -4 + 3  \\ 5 + 5  & 0 + 3 & 3 + 6  \\ 2 + 3 & -2 + 0  & 2 + 4  \end{bmatrix}\)

= \(\begin{bmatrix} 6  & 1  & -1  \\ 10  & 3  & 9  \\ 5 & -2 & 6 \end{bmatrix}\)

 

B – C = \(\begin{bmatrix} 4 & -2 & 3 \\ 5 & 3 & 6 \\ 3 & 1 & 4 \end{bmatrix}\) – \(\begin{bmatrix} 5 & 2 & 3 \\ 1 & 4 & 3 \\ 2 & -3 & 4 \end{bmatrix}\)

= \(\begin{bmatrix} 4 – 5 & -2 – 2 & 3 – 3 \\ 5 – 1 & 3 – 4 & 6 – 3 \\ 3 – 2 & 1 – ( -3 ) & 4 – 4 \end{bmatrix}\)

= \(\begin{bmatrix} -1 & -4 & 0 \\ 4 & -1 & 3 \\ 1 & 4 & 0 \end{bmatrix}\)

 

A + ( B – C ) = \(\begin{bmatrix} 2 & 3 & -4 \\ 5 & 0 & 3 \\ 2 & -2 & 2 \end{bmatrix}\) + \(\begin{bmatrix} -1 & -4 & 0 \\ 4 & -1 & 3 \\ 1 & 4 & 0 \end{bmatrix}\)

= \(\begin{bmatrix} 2 – 1  & 3 – 4  & -4 + 0  \\ 5 + 4  & 0 – 1  & 3 + 3  \\ 2 + 1 & -2 + 4  & 2 + 0  \end{bmatrix}\)

= \(\begin{bmatrix} 1  & -1  & -4  \\ 9  & -1  & 6  \\ 3  & 2  & 2 \end{bmatrix}\)

 

( A + B ) – C = \(\begin{bmatrix} 6  & 1  & -1  \\ 10  & 3  & 9  \\ 5 & -2 & 6 \end{bmatrix}\) – \(\begin{bmatrix} 5 & 2 & 3 \\ 1 & 4 & 3 \\ 2 & -3 & 4 \end{bmatrix}\)

= \(\begin{bmatrix} 6 – 5 & 1 – 2 & -1 – 3 \\ 10 – 1 & 3 – 4 & 9 – 3 \\ 5 – 2 & -2 + 3 & 6 – 4 \end{bmatrix}\)

= \(\begin{bmatrix} 1 & -1 & -4 \\ 9 & -1 & 6 \\ 3 & 1 & 2 \end{bmatrix}\)

Therefore, A + ( B – C ) = ( A + B ) – C = \(\begin{bmatrix} 1 & -1 & -4 \\ 9 & -1 & 6 \\ 3 & 1 & 2 \end{bmatrix}\)

Hence, proved.

 

 

Q-5: If, A = \(\begin{bmatrix} \frac{ 2 }{ 3 } & 2 & \frac{ 5 }{ 3 } \\ \\ \frac{ 1 }{ 3 } & \frac{ 2 }{ 3 } & \frac{ 4 }{ 3 } \\ \\ \frac{ 7 }{ 3 } & 3 & \frac{ 2 }{ 3 } \end{bmatrix}\) and B = \(\begin{bmatrix} \frac{ 2 }{ 5 } & \frac{ 3 }{ 5 } & 2 \\ \\ \frac{ 1 }{ 5 } & \frac{ 2 }{ 5 } & \frac{ 4 }{ 5 } \\ \\ \frac{ 7 }{ 5 } & \frac{ 6 }{ 5 } & \frac{ 2 }{ 5 } \end{bmatrix}\), then

Find the value of 3A – 5B.

 

Solution:

3A – 5B = 3 \(\begin{bmatrix} \frac{ 2 }{ 3 } & 2 & \frac{ 5 }{ 3 } \\ \\ \frac{ 1 }{ 3 } & \frac{ 2 }{ 3 } & \frac{ 4 }{ 3 } \\ \\ \frac{ 7 }{ 3 } & 3 & \frac{ 2 }{ 3 } \end{bmatrix}\) – 5 \(\begin{bmatrix} \frac{ 2 }{ 5 } & \frac{ 3 }{ 5 } & 2 \\ \\ \frac{ 1 }{ 5 } & \frac{ 2 }{ 5 } & \frac{ 4 }{ 5 } \\ \\ \frac{ 7 }{ 5 } & \frac{ 6 }{ 5 } & \frac{ 2 }{ 5 } \end{bmatrix}\)

= \(\begin{bmatrix} 2 & 6 & 5 \\ \\ 1  & 2 & 4 \\ \\ 7 & 9 & 2 \end{bmatrix}\) – \(\begin{bmatrix} 2 & 3 & 10 \\ \\ 1 & 2 & 4 \\ \\ 7 & 6 & 2 \end{bmatrix}\)

= \(\begin{bmatrix} 0 & 3 & -5 \\ \\ 0  & 0 & 0 \\ \\ 0 & 3 & 0 \end{bmatrix}\)

 

Q-6: Find the value of:

\(cos\Theta \begin{bmatrix} cos\Theta & sin\Theta \\ -sin\Theta & cos\Theta \end{bmatrix}\) + \(sin\Theta \begin{bmatrix} sin\Theta & -cos\Theta \\ cos\Theta & sin\Theta \end{bmatrix}\).

 

Solution:

\(cos\Theta \begin{bmatrix} cos\Theta & sin\Theta \\ -sin\Theta & cos\Theta \end{bmatrix}\) + \(sin\Theta \begin{bmatrix} sin\Theta & -cos\Theta \\ cos\Theta & sin\Theta \end{bmatrix}\)

= \(\begin{bmatrix} cos^{ 2 }\Theta & cos\Theta sin\Theta \\ -sin\Theta cos\Theta & cos^{ 2 }\Theta \end{bmatrix}\)  + \(\begin{bmatrix} sin^{ 2 }\Theta & -sin\Theta cos\Theta \\ sin\Theta cos\Theta & sin^{ 2 }\Theta \end{bmatrix}\)

= \(\begin{bmatrix} cos^{ 2 }\Theta + sin^{ 2 }\Theta & cos\Theta sin\Theta – cos\Theta sin\Theta \\ -sin\Theta cos\Theta + cos\Theta sin\Theta & cos^{ 2 }\Theta + sin^{ 2 }\Theta \end{bmatrix}\)

= \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)                        

 

 

Q-7: What will be the value of A and B, if

(i) A + B = \(\begin{bmatrix} 8 & 0 \\ 4 & 6 \end{bmatrix}\) and A – B = \(\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}\)

(ii) 2A + 3B = \(\begin{bmatrix} 3 & 4  \\  5  & 0 \end{bmatrix}\) and 3A + 2B = \(\begin{bmatrix} 3 & -3 \\ -2 & 6 \end{bmatrix}\)

 

Solution:

(i) A + B = \(\begin{bmatrix} 8 & 0 \\ 4 & 6 \end{bmatrix}\)    …………………..(a)

A – B = \(\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}\)         …………………..(b)

By adding equation (a) and (b), we will get:

2A = \(\begin{bmatrix} 8 & 0 \\ 4 & 6 \end{bmatrix}\) + \(\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}\)

2A = \(\begin{bmatrix} 8 + 4 & 0 + 0 \\ 0 + 4 & 6 + 4 \end{bmatrix}\)

2A = \(\begin{bmatrix} 12 & 0 \\ 4 & 10 \end{bmatrix}\)

Thus,

A = \(\frac{ 1 }{ 2 } \) \(\begin{bmatrix} 12 & 0 \\ 4 & 10 \end{bmatrix}\)

A = \(\begin{bmatrix} 6 & 0 \\ 2 & 5 \end{bmatrix}\)

A + B = \(\begin{bmatrix} 8 & 0 \\ 4 & 6 \end{bmatrix}\)

\(\begin{bmatrix} 6 & 0 \\ 2 & 5 \end{bmatrix}\) + B = \(\begin{bmatrix} 8 & 0 \\ 4 & 6 \end{bmatrix}\)

B = \(\begin{bmatrix} 8 & 0 \\ 4 & 6 \end{bmatrix}\) – \(\begin{bmatrix} 6 & 0 \\ 2 & 5 \end{bmatrix}\)

B = \(\begin{bmatrix} 8 – 6 & 0 – 0 \\ 4 – 2  & 6 – 5 \end{bmatrix}\)

B = \(\begin{bmatrix} 2  & 0 \\ 2   & 1  \end{bmatrix}\)

Hence, A = \(\begin{bmatrix} 6 & 0 \\ 2 & 5 \end{bmatrix}\) and, B = \(\begin{bmatrix} 2  & 0 \\ 2   & 1  \end{bmatrix}\)

 

(ii) 2A + 3B = \(\begin{bmatrix} 3 & 4  \\  5  & 0 \end{bmatrix}\)    …………..(a)

3A + 2B = \(\begin{bmatrix} 3 & -3 \\ -2 & 6 \end{bmatrix}\)          ……………(b)

By multiplying equation (a) by 2, we will get:

2( 2A + 3B ) = 2 \(\begin{bmatrix} 3 & 4  \\  5  & 0 \end{bmatrix}\)

4A + 6B = \(\begin{bmatrix} 6 & 8  \\ 10  & 0 \end{bmatrix}\)   ……………(c)

By multiplying equation (b) by 3, we will get:

3( 3A + 2B ) = 3 \(\begin{bmatrix} 3 & -3 \\ -2 & 6 \end{bmatrix}\)

9A + 6B = \(\begin{bmatrix} 9 & -9 \\ -6 & 12 \end{bmatrix}\)   …………….(d)

By subtracting equation (c) and (d), we will get:

( 4A + 6B ) – ( 9A + 6B ) = \(\begin{bmatrix} 6 – 9 & 8 + 9  \\ 10 + 6  & 0 – 12 \end{bmatrix}\)

-5A = \(\begin{bmatrix} -3 & 17  \\ 16  & -12 \end{bmatrix}\)

A = – \(\frac{ 1 }{ 5 }\) \(\begin{bmatrix} -3 & 17  \\ 16  & -12 \end{bmatrix}\)

  A = \(\begin{bmatrix} \frac{ 3 }{ 5 }& \frac{ -17 }{ 5 }  \\ \frac{ -16 }{ 5 }  & \frac{ 12 }{ 5 } \end{bmatrix}\)

Now,

2A + 3B  = \(\begin{bmatrix} 3 & 4  \\  5  & 0 \end{bmatrix}\)

2 \(\begin{bmatrix} \frac{ 3 }{ 5 }& \frac{ -17 }{ 5 }  \\ \frac{ -16 }{ 5 }  & \frac{ 12 }{ 5 } \end{bmatrix}\) + 3B = \(\begin{bmatrix} 3 & 4  \\  5  & 0 \end{bmatrix}\)

3B = \(\begin{bmatrix} 3 & 4  \\  5  & 0 \end{bmatrix}\) – \(\begin{bmatrix} \frac{ 6 }{ 5 } & \frac{ -34 }{ 5 }  \\ \frac{ -32 }{ 5 }  & \frac{ 24 }{ 5 } \end{bmatrix}\)

3B = \(\begin{bmatrix} 3 – \frac{ 6 }{ 5 } & 4 – \frac{ -34 }{ 5 }  \\ 5 – \frac{ -32 }{ 5 }  &  0 – \frac{ 24 }{ 5 } \end{bmatrix}\)

3B = \(\begin{bmatrix} \frac{ 9 }{ 5 } &  \frac{ 54 }{ 5 }  \\ \frac{ 57 }{ 5 }  &  \frac{ 24 }{ 5 } \end{bmatrix}\)

B = \(\frac{ 1 }{ 3 }\) \(\begin{bmatrix} \frac{ 9 }{ 5 } &  \frac{ 54 }{ 5 }  \\ \frac{ 57 }{ 5 }  &  \frac{ 24 }{ 5 } \end{bmatrix}\)

B = \(\begin{bmatrix} \frac{ 3 }{ 5 } &  \frac{ 18 }{ 5 }  \\ \frac{ 19 }{ 5 }  &  \frac{ 8 }{ 5 } \end{bmatrix}\)

Hence, A = \(\begin{bmatrix} \frac{ 3 }{ 5 }& \frac{ -17 }{ 5 }  \\ \frac{ -16 }{ 5 }  & \frac{ 12 }{ 5 } \end{bmatrix}\) and B = \(\begin{bmatrix} \frac{ 3 }{ 5 } &  \frac{ 18 }{ 5 }  \\ \frac{ 19 }{ 5 }  &  \frac{ 8 }{ 5 } \end{bmatrix}\)

 

Q-8: What will be the value of A if,

B = \(\begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix}\) and 2A + B = \(\begin{bmatrix} 2 & 1 \\ -4 & 3 \end{bmatrix}\)

 

Solution:

2A + B = \(\begin{bmatrix} 2 & 1 \\ -4 & 3 \end{bmatrix}\)

2A = \(\begin{bmatrix} 2 & 1 \\ -4 & 3 \end{bmatrix}\) – \(\begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix}\)

2A = \(\begin{bmatrix} 2 – 4 & 1 – 3 \\ -4 – 2 & 3 – 5  \end{bmatrix}\)

2A = \(\begin{bmatrix} -2 & -2 \\ -6 & -2 \end{bmatrix}\)

A = \(\frac{ 1 }{ 2 }\) \(\begin{bmatrix} -2 & -2 \\ -6 & -2 \end{bmatrix}\)

A = \(\begin{bmatrix} -1 & -1 \\ -3 & -1 \end{bmatrix}\)

Hence, A = \(\begin{bmatrix} -1 & -1 \\ -3 & -1 \end{bmatrix}\).

 

 

Q-9: Calculate the value of a and b, if

\(2 \begin{bmatrix} 2 & 4 \\ 1 & a \end{bmatrix} + \begin{bmatrix} b & 1 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 7 \\ 2 & 9 \end{bmatrix}\).

 

Solution:

\(2 \begin{bmatrix} 2 & 4 \\ 1 & a \end{bmatrix} + \begin{bmatrix} b & 1 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 7 \\ 2 & 9 \end{bmatrix}\)

⟹ \( \begin{bmatrix} 4 & 8 \\ 2 & 2a \end{bmatrix} + \begin{bmatrix} b & 1 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 7 \\ 2 & 9 \end{bmatrix}\)

⟹ \(\begin{bmatrix} 4 + b  & 8 + 1  \\ 2 + 2  & 2a + 2 \end{bmatrix} = \begin{bmatrix} 6 & 7 \\ 2 & 9 \end{bmatrix}\)

⟹ \(\begin{bmatrix} 4 + b  & 9  \\  4  & 2a + 2 \end{bmatrix} = \begin{bmatrix} 6 & 7 \\ 2 & 9 \end{bmatrix}\)

By comparing the corresponding elements of the above two matrices, we will get:

4 + b = 6

b = 2

2a + 2 = 9

2a = 7

a = \(\frac{ 7 }{ 2 }\)

Hence, a = \(\frac{ 7 }{ 2 }\) and b = 2.

 

Q-10: Calculate the value of a, b, c and d, if

\(2 \begin{bmatrix} a & b \\ c & d \end{bmatrix} + 3 \begin{bmatrix} 2 & -2 \\ 0 & 3 \end{bmatrix} =  3 \begin{bmatrix} 4 & 6 \\ 5 & 7 \end{bmatrix}\).

 

Solution:

\(2\begin{bmatrix} a & b \\ c & d \end{bmatrix} + 3 \begin{bmatrix} 2 & -2 \\ 0 & 3 \end{bmatrix} =  3 \begin{bmatrix} 4 & 6 \\ 5 & 7 \end{bmatrix}\)

\(\begin{bmatrix} 2a & 2b \\ 2c & 2d \end{bmatrix} + \begin{bmatrix} 6 & -6 \\ 0 & 9 \end{bmatrix} = \begin{bmatrix} 12 & 18 \\ 15 & 21 \end{bmatrix}\)

\(\begin{bmatrix} 2a + 6  & 2b – 6 \\ 2c + 0 & 2d + 9 \end{bmatrix} = \begin{bmatrix} 12 & 18 \\ 15 & 21 \end{bmatrix}\)

By comparing the corresponding elements of the above two matrices, we will get:

2a + 6 = 12

⟹ 2a = 6

a = 3

2b – 6 = 18

⟹ 2b = 24

b = 12

2c + 0 = 15

⟹ 2c = 15

c = \(\frac{ 15 }{ 2 }\)

2d + 9 = 21

⟹ 2d = 12

d = 6

Hence, a = 3, b = 12, c = \(\frac{ 15 }{ 2 }\) and d = 6.

 

 

Q-11: What will be the value of x and y in the matrices given below:

 \(x \begin{bmatrix} 3 \\ 4 \end{bmatrix} + y \begin{bmatrix} -2 \\ 2 \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}\)?

 

Solution:

\(x \begin{bmatrix} 3 \\ 4 \end{bmatrix} + y \begin{bmatrix} -2 \\ 2 \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}\\\)

⟹ \(\\\begin{bmatrix} 3x \\ 4x \end{bmatrix} + \begin{bmatrix} -2y \\ 2y  \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}\\\)

⟹ \(\\\begin{bmatrix} 3x – 2y \\ 4x + 2y \end{bmatrix}  = \begin{bmatrix} 10 \\ 5 \end{bmatrix}\)

By comparing the corresponding elements of the above two matrices, we will get:

3x – 2y = 10………….. (i)

4x + 2y = 5 ……………(ii)

Adding equation (i) and (ii), we will get:

7x = 15

x = \(\frac{ 15 }{ 7 }\)

Putting the value of x in equation (i), we will get:

3 × \(\frac{ 15 }{ 7 }\) – 2y = 10

⟹ \(\frac{ 45 }{ 7 }\) – 2y = 10

⟹ 2y = \(\frac{ 45 }{ 7 }\) – 10

⟹ 2y = \(\frac{ 45 – 70 }{ 7 }\)

⟹ 2y = \(\frac{ -25 }{ 7 }\)

⟹ y = \(\frac{ -25 }{ 14 }\)

Hence, x = \(\frac{ 15 }{ 7 }\) and, y = \(\frac{ -25 }{ 14 }\)

 

 

Q-12: What will be the value of a, b, c and d in the following matrix

\(3\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a & 7 \\ -2 & 2d \end{bmatrix} + \begin{bmatrix} 5 & a + b \\ c + d & 4 \end{bmatrix}\)?

 

Solution:

\(3\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a & 7 \\ -2 & 2d \end{bmatrix} + \begin{bmatrix} 5 & a + b \\ c + d & 4 \end{bmatrix}\)

⟹ \( \begin{bmatrix} 3a & 3b \\ 3c & 3d \end{bmatrix} = \begin{bmatrix} a + 5  & 7 + a + b \\ -2 + c + d & 2d + 4 \end{bmatrix}\\\\\)

⟹ \( \begin{bmatrix} 3a & 3b \\ 3c & 3d \end{bmatrix} = \begin{bmatrix} a + 5  & a + b + 7  \\  c + d – 2  & 2d + 4 \end{bmatrix}\)

By comparing the corresponding elements of the above two matrices, we will get:

3a = a + 5

2a = 5

a = \(\frac{ 5 }{ 2 }\)

3b = a + b + 7

⟹ 3b = \(\frac{ 5 }{ 2 }\) + b + 7

⟹ 2b = \(\frac{ 5 }{ 2 }\) + 7

⟹ 2b = \(\frac{ 5 + 14 }{ 2 }\)

⟹ 2b = \(\frac{ 19 }{ 2 }\)

b = \(\frac{ 19 }{ 4 }\)

3d = 2d + 4

d = 4

3c = c + d – 2

⟹ 2c = 4 – 2

⟹ 2c = 2

c = 1

Hence, a = \(\frac{ 5 }{ 2 }\), b = \(\frac{ 19 }{ 4 }\), c = 1 and d = 4.

 

 

Q-13:If F(a) = \(\begin{bmatrix} cos a & -sin a & 0 \\ sin a & cos a & 0 \\ 0 & 0 & 1 \end{bmatrix}\), prove that F(a) F(b) = F( a + b ).

 

Solution:

F(a) = \(\begin{bmatrix} cos a & -sin a & 0 \\ sin a & cos a & 0 \\ 0 & 0 & 1 \end{bmatrix}\), F(b) = \(\begin{bmatrix} cos b & -sin b & 0 \\ sin b & cos b & 0 \\ 0 & 0 & 1 \end{bmatrix}\)

F(a + b) = \(\begin{bmatrix} cos \left ( a + b \right ) & -sin \left ( a + b \right ) & 0 \\ sin \left ( a + b \right ) & cos \left ( a + b \right ) & 0 \\ 0 & 0 & 1 \end{bmatrix}\)

F(a) F(b) = \(\begin{bmatrix} cos a & -sin a & 0 \\ sin a & cos a & 0 \\ 0 & 0 & 1 \end{bmatrix}\)\(\begin{bmatrix} cos b & -sin b & 0 \\ sin b & cos b & 0 \\ 0 & 0 & 1 \end{bmatrix}\)

= \(\begin{bmatrix} \left ( cos a \;cos b – sin a \;sin b + 0 \right ) & \left ( -cos a\; sin b – sin a \;cos b + 0 \right ) & 0 \\ \left ( sin a \;cos b + cos a \;sin b + 0 \right ) & -sin a \;sin b + cos a\; cos b + 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}\)

= \(\begin{bmatrix} cos \left ( a + b \right ) & -sin \left ( a + b \right ) & 0 \\ sin \left ( a + b \right ) & cos \left ( a + b \right ) & 0 \\ 0 & 0 & 1 \end{bmatrix}\) = F(a + b)

Hence, proved.

 

Q-14: Prove that:

(i) \(\begin{bmatrix} 6 & -2 \\ 7 & 8 \end{bmatrix}\begin{bmatrix} 3 & 2 \\ 4 & 5 \end{bmatrix}\neq \begin{bmatrix} 3 & 2 \\ 4 & 5 \end{bmatrix}\begin{bmatrix} 6 & -2 \\ 7 & 8 \end{bmatrix}\).

(ii) \(\begin{bmatrix} 2 & 3 & 4 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}\begin{bmatrix} -2 & 1 & 0 \\ 0 & -2 & 1 \\ 3 & 4 & 5 \end{bmatrix}\neq \begin{bmatrix} -2 & 1 & 0 \\ 0 & -2 & 1 \\ 3 & 4 & 5 \end{bmatrix}\begin{bmatrix} 2 & 3 & 4 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}\)

 

Solution:

(i) \(\begin{bmatrix} 6 & -2 \\ 7 & 8 \end{bmatrix}\begin{bmatrix} 3 & 2 \\ 4 & 5 \end{bmatrix}\)

=  \(\\\begin{bmatrix} 6( 3 ) – 2( 4 ) & 6( 2 ) – 2( 5 ) \\ 7( 3 ) – 8( 4 ) & 7( 2 ) – 8( 5 ) \end{bmatrix}\)

= \(\\\begin{bmatrix} 18 – 8 & 12 – 10 \\ 21 – 24  & 14 – 40  \end{bmatrix}\)

= \(\\\begin{bmatrix} 10  & 2  \\ -3  & -26  \end{bmatrix}\)

\(\\\begin{bmatrix} 3 & 2 \\ 4 & 5 \end{bmatrix}\begin{bmatrix}  6 & -2 \\ 7 & 8 \end{bmatrix}\)

=  \(\\\begin{bmatrix} 3( 6 ) + 2( 7 ) &  3( -2 ) + 2( 8 ) \\ 4( 6 ) + 5( 7 )  &  4( -2 ) + 5( 8 ) \end{bmatrix}\)

= \(\\\begin{bmatrix} 18 + 14 & -6 + 16 \\ 24 + 35  & -8 + 40  \end{bmatrix}\)

= \(\\\begin{bmatrix} 32  &  10  \\ 59  & 32  \end{bmatrix}\)

Therefore, \(\begin{bmatrix} 6 & -2 \\ 7 & 8 \end{bmatrix}\begin{bmatrix} 3 & 2 \\ 4 & 5 \end{bmatrix}\) ≠ \(\begin{bmatrix} 3 & 2 \\ 4 & 5 \end{bmatrix}\begin{bmatrix}  6 & -2 \\ 7 & 8 \end{bmatrix}\)

Hence, proved.

 

(ii) \(\\\begin{bmatrix} 2 & 3 & 4 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}\begin{bmatrix} -2 & 1 & 0 \\ 0 & -2 & 1 \\ 3 & 4 & 5 \end{bmatrix}\)

= \(\\\begin{bmatrix} 2( -2 ) + 3( 0 ) + 4( 3 ) & 2( 1 ) + 3( -2 ) + 4( 4 ) & 2( 0 ) + 3( 1 ) + 4( 5 ) \\ 0( -2 ) + 1( 0 ) + 0( 3 ) & 0( 1 ) + 1( -2 ) + 0( 4 ) & 0( 0 ) + 1( 1 ) + 0( 5 ) \\ 1( -2 ) + 1( 0 ) + 0( 3 ) & 1( 1 ) + 1( -2 ) + 0( 4 ) & 1( 0 ) + 1( 1 ) + 0( 5 ) \end{bmatrix}\)

= \(\\\begin{bmatrix} -4 + 0 + 12 & 2 – 6 + 16 & 0 + 3 + 20 \\ 0 + 0 + 0 & 0  – 2  + 0 & 0 + 1 + 0 \\ -2 + 0 + 0 & 1 – 2 + 0 & 0 + 1 + 0 \end{bmatrix}\)

= \(\\\begin{bmatrix} 8  & 12  & 23  \\ 0  &   – 2  & 1  \\  -2  &  -1  &  1  \end{bmatrix}\)

\(\\\begin{bmatrix} -2 & 1 & 0 \\ 0 & -2 & 1 \\ 3 & 4 & 5 \end{bmatrix}\begin{bmatrix} 2 & 3 & 4 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}\)

= \(\\\begin{bmatrix} -2( 2 ) + 1( 0 ) + 0( 1 ) & -2( 3 ) + 1( 1 ) + 0( 1 ) & -2( 4 ) + 1( 0 ) + 0( 0 ) \\ 0( 2 ) – 2( 0 ) + 1( 1 ) & 0( 3 ) – 2( 1 ) + 1( 1 ) & 0( 4 ) – 2( 0 ) + 1( 0 ) \\ 3( 2 ) + 4( 0 ) + 5( 1 ) & 3( 3 ) + 4( 1 ) + 5( 1 ) & 3( 4 ) + 4( 0 ) + 5( 0 ) \end{bmatrix}\)

= \(\\\begin{bmatrix} -4 + 0 + 0 & -6 + 1 + 0 & -8 + 0 + 0  \\ 0 + 0  + 1 & 0 – 2 + 1 & 0 – 0 + 0 \\ 6 + 0 + 5 & 9 + 4 + 5 & 12 + 0 + 0 \end{bmatrix}\)

= \(\\\begin{bmatrix} -4  & -5  & -8  \\  1 &  – 1 &  0 \\ 11  & 18  & 12  \end{bmatrix}\)

Therefore, \(\begin{bmatrix} 2 & 3 & 4 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}\begin{bmatrix} -2 & 1 & 0 \\ 0 & -2 & 1 \\ 3 & 4 & 5 \end{bmatrix}\) ≠ \(\begin{bmatrix} -2 & 1 & 0 \\ 0 & -2 & 1 \\ 3 & 4 & 5 \end{bmatrix}\begin{bmatrix} 2 & 3 & 4 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}\)

Hence, proved.

 

Q-15: What will be the value of A2 – 5A + 6I, if  A = \(\begin{bmatrix} 3 & 0 & 1 \\ 3 & 2 & 4 \\ 1 & -1 & 0 \end{bmatrix}\)?

Solution:

A2 = A × A

A2 = \(\\\begin{bmatrix} 3 & 0 & 1 \\ 3 & 2 & 4 \\ 1 & -1 & 0 \end{bmatrix}\) × \(\begin{bmatrix} 3 & 0 & 1 \\ 3 & 2 & 4 \\ 1 & -1 & 0 \end{bmatrix}\)

= \(\\\begin{bmatrix} 3( 3 ) + 3( 3 ) + 1( 1 ) & 3( 0 ) + 3( 2 ) + 1( -1 ) & 3( 1 ) + 3( 4 ) + 1( 0 ) \\ 3( 3 ) + 2( 3 ) + 4( 1 ) & 3( 0 ) + 2( 2 ) + 4( -1 ) & 3( 1 ) + 2( 4 ) + 4( 0 ) \\ 1( 3 ) – 1( 3 ) + 0( 1 ) & 1( 0 ) – 1( 2 ) + 0( -1 ) & 1( 1 ) – 1( 4 ) + 0( 0 ) \end{bmatrix}\)

= \(\\\begin{bmatrix} 9 + 9 + 1  & 0 + 6 – 1 & 3 + 12 + 0  \\ 9 + 6 + 4  & 0 + 4 – 4  & 3 + 8 + 0 \\ 3 – 3 + 0 & 0 – 2 + 0 & 1 – 4 + 0 \end{bmatrix}\)

= \(\\\begin{bmatrix} 19  & 5  & 15   \\ 19  & 0  & 11 \\  0 &  – 2  & -3 \end{bmatrix}\)

 

5A = 5\(\begin{bmatrix} 3 & 0 & 1 \\ 3 & 2 & 4 \\ 1 & -1 & 0 \end{bmatrix}\)

= \(\\\begin{bmatrix} 15 & 0 & 5 \\ 15  & 10  & 20  \\ 5  & -5  & 0 \end{bmatrix}\)

 

6I = 6 \(\begin{bmatrix} 1  & 0  & 0 \\ 0  & 1  & 0 \\ 0  & 0  & 1 \end{bmatrix}\)

= \(\\\begin{bmatrix} 6  & 0  & 0 \\ 0  & 6  & 0 \\ 0  & 0  & 6 \end{bmatrix}\)

 

A2 – 5A + 6I = \(\begin{bmatrix} 19  & 5  & 15   \\ 19  & 0  & 11 \\  0 &  – 2  & -3 \end{bmatrix}\) – \(\begin{bmatrix} 15 & 0 & 5 \\ 15  & 10  & 20  \\ 5  & -5  & 0 \end{bmatrix}\) + \(\begin{bmatrix} 6  & 0  & 0 \\ 0  & 6  & 0 \\ 0  & 0  & 6 \end{bmatrix}\)

= \(\\\begin{bmatrix} 19 + 15 + 6  & 5 + 0 + 0   & 15 + 5 + 0  \\ 19 + 15 + 0  & 0 + 10 + 6  & 11 + 20 + 0 \\ 0 + 5 + 0  & -2 -5 + 0  & -3 + 0 + 6 \end{bmatrix}\)

= \(\\\begin{bmatrix} 40  & 5  & 20  \\ 34  & 16  & 31 \\ 5  & -7  & 3 \end{bmatrix}\)

Hence,

A2 – 5A + 6I = \(\begin{bmatrix} 40  & 5  & 20  \\ 34  & 16  & 31 \\ 5  & -7  & 3 \end{bmatrix}\)

 

 

Q-16: If A = \(\begin{bmatrix} 0 & -tan\frac{ \beta  }{ 2 } \\ tan\frac{ \beta  }{ 2 } & 0 \end{bmatrix}\) and the identity matrix of order 2 be represented by I. Prove that:

I + A = ( I – A ) \(\begin{bmatrix} cos \beta & -sin \beta \\ sin \beta & cos \beta \end{bmatrix}\)

 

Solution:

LHS = I + A = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) + \(\begin{bmatrix} 0 & -tan\frac{ \beta }{ 2 } \\ tan\frac{ \beta }{ 2 } & 0 \end{bmatrix}\)

= \(\\\begin{bmatrix} 1 & -tan\frac{ \alpha }{ 2 } \\ tan\frac{ \alpha }{ 2 } & 1 \end{bmatrix}\) ……… (i)

 

RHS = ( I – A ) \(\begin{bmatrix} cos \beta & -sin \beta \\ sin \beta & cos \beta \end{bmatrix}\)

= \(\\\left ( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} – \begin{bmatrix} 0 & -tan\frac{ \beta }{ 2 } \\ tan\frac{ \beta }{ 2 } & 0 \end{bmatrix} \right ) \begin{bmatrix} cos \beta & – sin\beta \\ sin\beta & cos\beta \end{bmatrix}\)

= \(\\\begin{bmatrix} 1 & tan\frac{ \beta }{ 2 } \\ -tan\frac{ \beta }{ 2 } & 1 \end{bmatrix} \;\; \begin{bmatrix} cos \beta & – sin\beta \\ sin\beta & cos\beta \end{bmatrix}\)

= \(\\\begin{bmatrix} cos \beta + sin \beta tan\frac{ \beta }{ 2 } & -sin \beta + cos \beta tan\frac{ \beta }{ 2 } \\ -cos \beta tan\frac{ \beta }{ 2 } + sin \beta tan\frac{ \beta }{ 2 } & sin \beta + cos \beta \end{bmatrix}\) ………..(ii)

= \(\\\begin{bmatrix} 1 – 2sin^{ 2 }\frac{ \beta }{ 2 } + 2sin\frac{ \beta }{ 2 }cos\frac{ \beta }{ 2 }tan\frac{ \beta }{ 2 } & -2sin\frac{ \beta }{ 2 }cos\frac{ \beta }{ 2 } + \left ( 2cos^{ 2 }\frac{ \beta }{ 2 } – 1 \right )tan\frac{ \beta }{ 2 } \\ -\left ( 2cos^{ 2 }\frac{ \beta }{ 2 } – 1 \right )tan\frac{ \beta }{ 2 } + 2sin\frac{ \beta }{ 2 }cos\frac{ \beta }{ 2 } & 2sin\frac{ \beta }{ 2 }cos\frac{ \beta }{ 2 }tan\frac{ \beta }{ 2 } + 1 – 2sin^{ 2 }\frac{ \beta }{ 2 } \end{bmatrix}\)

= \(\\\begin{bmatrix} 1 – 2sin^{ 2 }\frac{ \beta }{ 2 } + 2sin^{ 2 }\frac{ \beta }{ 2 } & -2sin\frac{ \beta }{ 2 }cos\frac{ \beta }{ 2 } + 2sin\frac{ \beta }{ 2 }cos\frac{ \beta } – tan\frac{ \beta }{ 2 } \\ -2sin\frac{ \beta }{ 2 }cos\frac{ \beta }{ 2 } + 2sin\frac{ \beta }{ 2 }cos\frac{ \beta } + tan\frac{ \beta }{ 2 } & 2sin^{ 2 }\frac{ \beta }{ 2 } + 1 – 2sin^{ 2 }\frac{ \beta }{ 2 } \end{bmatrix}\)

= \(\\\begin{bmatrix} 1 & -tan\frac{ \alpha }{ 2 } \\ tan\frac{ \alpha }{ 2 } & 1 \end{bmatrix}\)

= LHS

Hence, proved.

 

 

Q-17: There is the fund of Rs 30,000 from a trust which should be invested in any of the two different types of bonds. The first bond will pay 5% interest every year, and the second bond will pay 7% interest every year. Using the principle of matrix multiplication, determine how Rs 30,000 will be divided between the two different types of bonds, if the trust fund will obtain an annual total interest of:

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

 

Solution:

(a) Let, in the first bond, Rs. q is invested.

The sum of the total money thus invested in the second bond is Rs. ( 30000 – q ).

As per the data given in the question, the first bond needs to pay 5% interest every year and the second bond needs to pay 7% interest every year.

Hence, in order to have an annual interest of Rs. 2000, we have

S.I for 1 year = \(\frac{ Principle \times Rate \times 1 }{ 100 }\)

⟹ \(\begin{bmatrix} q & \left ( 30000 – q \right ) \end{bmatrix}\begin{bmatrix} \frac{ 5 }{ 100 } \\ \frac{ 7 }{ 100 } \end{bmatrix} = 2000\)

⟹ \(\frac{ 5q }{ 100 } + \frac{ 7 \left ( 30000 – q \right ) }{ 100 } = 2000\)

⟹ 5q + 7( 30000 – q ) = 200000

⟹ 5q +  210000 – 7q  = 200000

⟹  210000 – 200000 =   7q  –  5q

⟹ 2q = 10000

q = 5000

In order to have an annual interest of Rs. 2000, the trusty providing fund must invest Rs. 5000 in the first bond and remaining Rs. 25000 in the second bond.

 

(b) Let, in the first bond, Rs. q is invested.

The sum of the total money thus invested in the second bond is Rs. ( 30000 – q ).

As per the data given in the question, the first bond needs to pay 5% interest every year and the second bond needs to pay 7% interest every year.

Hence, in order to have an annual interest of Rs. 1800, we have

S.I for 1 year = \(\frac{ Principle \times Rate \times 1 }{ 100 }\)

⟹ \(\begin{bmatrix} q & \left ( 30000 – q \right ) \end{bmatrix}\begin{bmatrix} \frac{ 5 }{ 100 } \\ \frac{ 7 }{ 100 } \end{bmatrix} = 1800\)

⟹ \(\frac{ 5q }{ 100 } + \frac{ 7 \left ( 30000 – q \right ) }{ 100 } = 1800\)

⟹ 5q + 7( 30000 – q ) = 180000

⟹ 5q +  210000 – 7q  = 180000

⟹  210000 – 180000 =   7q  –  5q

⟹ 2q = 30000

q = 15000

In order to have an annual interest of Rs. 1800, the trusty providing fund must invest Rs. 15000 in the first bond and remaining Rs. 15000 in the second bond.

 

 

Q-18: A particular school has the bookshop of 8 dozen of the physics books, 10 dozen of the economics books, 10 dozen of the chemistry books. Their selling prices are Rs 60, Rs 40 and Rs 80 each, respectively. What will be the total amount of the bookshop which will be received by selling all of the books using the matrix algebra?

 

Solution:

The bookshop of 8 dozen of the physics books, 10 dozen of the economics books, 10 dozen of the chemistry books.

The selling prices for the physics, chemistry and the economics book are Rs. 60, Rs. 40 and Rs. 80, respectively.

The total amount of the bookshop received by selling all of the books by using the matrix algebra will be represented in the form given below:

\(12\begin{bmatrix} 8 & 10 & 10 \end{bmatrix}\begin{bmatrix} 60 \\ 40 \\ 80 \end{bmatrix}\)

= 12 [8 × 60 + 10 × 40 + 10 × 80]

= 12 [480 + 400 + 800]

= 12 [1680]

= 20160

Hence, the bookshop will get Rs. 20160 by the selling of all of the books.

 

 

Q-19: Let us consider P, Q, R, S and T be the matrices with order 2 × m, 3 × n, 2 × o, m× 3 and o× n, respectively. Then, the restriction on m, n and o such that, TQ + SQ is defined as:

  1. n = 3, o = m
  2. n is arbitrary, o = 2
  3. o is arbitrary, n = 3
  4. n = 2, o = 3

 

Solution:

Matrices T and Q are having the orders o × n and 3 × n, respectively.

Thus, hence formed matrix TQ will be defined if n = 3. Consequently, TQ matrix will be have the order o × n. Matrices S and Q are having the orders of m × 3 and 3 × n, respectively.

As the number of columns in the matrix S is equal to the number of rows in the matrix Q, so, the matrix SQ hence is well-defined and will be of the order m × n. Matrices PY and WY will be added at the situation,  when their orders will be the same.

Since, the matrix TQ has the order o × n and the matrix SQ has the order m × n. Hence, we will have o = m, surely.

Therefore, n = 3 and o = m are the restrictions on m, n and o, such that TQ + SQ will be defined.

 

 

Q-20: Let us consider P, Q, R, S and T be the matrices with order 2 × m, 3 × n, 2 × o, m× 3 and o× n, respectively. If m = o, then find the order for the matrix 7P – 5R.

(a). o × 2

(b). 2 × m

(c). m × 3                

(d). o × n

 

Solution:

The matrix P has the order 2 × m.

Thus, the matrix 7P will also have the same order. The matrix R is of the order 2 × o, i.e., 2 × m [As m = o]

Thus, the matrix 5R will also have the same order.

Now, both of the matrices 7P and 5R are of the order 2 × m.

Thus, the matrix 7P – 5R is well-defined and will have the order 2 × m.

 

 

EXERCISE – 3.3

 

 

Q-1: What is the transpose of each of the following matrices?

(i) \(\begin{bmatrix} 6 \\ \frac{ 1 }{ 2 }\\ -2 \end{bmatrix}\)

(ii) \(\begin{bmatrix} 2 & -2 \\ 3 & 4 \end{bmatrix}\)

 

Solution:

(i) Let M = \(\begin{bmatrix} 6 \\ \frac{ 1 }{ 2 }\\ -2 \end{bmatrix}\)

Then,

MT = \(\begin{bmatrix} 6 & \frac{ 1 }{ 2 } & -2 \end{bmatrix}\)

(ii) Let M = \(\begin{bmatrix} 2 & -2 \\ 3 & 4 \end{bmatrix}\)

Then,

MT = \(\begin{bmatrix} 2  & 3 \\ -2  & 4 \end{bmatrix}\)

 

Q-2: If A = \(\begin{bmatrix} -2 & 3 & 4 \\ 6 & 7 & 10 \\ -3 & 2 & 2 \end{bmatrix}\) and B = \(\begin{bmatrix} -5 & 2 & -6 \\ 2 & 3 & 1 \\ 2 & 4 & 2 \end{bmatrix}\), then verify that

(A + B) = A + B

 

Solution:

A = \(\begin{bmatrix} -2 & 3 & 4 \\ 6 & 7 & 10 \\ -3 & 2 & 2 \end{bmatrix}\), B = \(\begin{bmatrix} -5 & 2 & -6 \\ 2 & 3 & 1 \\ 2 & 4 & 2 \end{bmatrix}\)

A + B = \(\begin{bmatrix} -2 & 3 & 4 \\ 6 & 7 & 10 \\ -3 & 2 & 2 \end{bmatrix}\) + \(\begin{bmatrix} -5 & 2 & -6 \\ 2 & 3 & 1 \\ 2 & 4 & 2 \end{bmatrix}\)

= \(\begin{bmatrix} -2 – 5  & 3 + 2  & 4 – 6  \\ 6 + 2  & 7 + 3  & 10 + 1  \\ -3 + 2  & 2 + 4  & 2 + 2  \end{bmatrix}\)

= \(\begin{bmatrix} -7  & 5  & -2 \\ 8  & 10  & 10  \\ -1 & 6 & 4 \end{bmatrix}\)

Thus,

(A + B) = \(\begin{bmatrix} -7  & 8  & -1 \\ 5  & 10  & 6  \\ -2  & 10  & 4 \end{bmatrix}\)

A + B = \(\begin{bmatrix} -2 & 6 & -3 \\ 3  & 8  & 2 \\  4  & 10  & 2 \end{bmatrix}\) + \(\begin{bmatrix} -5 & 2  & 2  \\ 2 & 3 & 4  \\ -6  & 1  & 2 \end{bmatrix}\)

= \(\begin{bmatrix} -2 – 5  & 6 + 2  & -3 + 2  \\ 3 + 2  & 8 + 3  & 2 + 4  \\  4 – 6  & 9 + 1  & 2 + 2 \end{bmatrix}\)

= \(\begin{bmatrix} -7  & 8  & -1 \\ 5  & 11  & 6 \\  -2  & 10  & 4 \end{bmatrix}\)

Thus, (A + B) = A + B

Hence, proved.

 

 

Q-3: If A= \(\begin{bmatrix} -3 & 4 \\ 2 & 3 \end{bmatrix}\) and B = \(\begin{bmatrix} -2 & 1 \\ 2 & 3 \end{bmatrix}\), now find (A + 2B)’.

 

Solution:

We know that:

A = \({ \left ( {A}’ \right ) }'\)

Thus,

\({ A }’ \) = \(\begin{bmatrix} -3 & 4 \\ 2 & 3 \end{bmatrix}\)

A = \(\begin{bmatrix} -3  & 2  \\ 4  & 3 \end{bmatrix}\)

Now,

A + 2B = \(\begin{bmatrix} -3  & 2  \\ 4  & 3 \end{bmatrix}\) + 2\(\begin{bmatrix} -2  & 1  \\ 2  & 3 \end{bmatrix}\)

= \(\begin{bmatrix} -3  & 2  \\ 4  & 3 \end{bmatrix}\) + \(\begin{bmatrix} -4  & 2  \\ 4  & 6 \end{bmatrix}\)

= \(\begin{bmatrix} -3 – 4  & 2 + 2  \\ 4 + 4  & 3 + 6 \end{bmatrix}\)

= \(\begin{bmatrix} -7  & 4  \\ 8  & 9 \end{bmatrix}\)

Therefore,\({ \left ( A + 2B \right ) }'\) = \(\begin{bmatrix} -7  & 8  \\ 4  & 9 \end{bmatrix}\)

 

 

Q-4: Prove that \({ \left ( AB \right ) }’ = { B }'{ A }'\) for the matrices A and B where,

A = \( \begin{bmatrix} 2 \\ -5 \\ 4 \end{bmatrix}\), B = \( \begin{bmatrix} -2 & 3 & 2 \end{bmatrix}\)

 

Solution:

AB = \( \begin{bmatrix} 2 \\ -5 \\ 4 \end{bmatrix}\)\( \begin{bmatrix} -2 & 3 & 2 \end{bmatrix}\)

= \(\begin{bmatrix} -4  & 6  & 4 \\ 10  & -15  & -10 \\ -8  & 12  & 8 \end{bmatrix}\)

Thus,

\({ \left ( AB \right ) }'\) = \(\begin{bmatrix} -4  & 10  & -8 \\ 6  & -15  & 12 \\ 4  & -10  & 8 \end{bmatrix}\)

Now,

\({ A }'\) = \(\begin{bmatrix} 2 & -5 & 4 \end{bmatrix}\)

\({ B }'\) = \( \begin{bmatrix} -2 \\ 3  \\ 2 \end{bmatrix}\)

Thus,

{ B }’ { A }’= \(\begin{bmatrix} -4  & 10  & -8 \\ 6  & -15  & 12 \\ 4  & -10  & 8 \end{bmatrix}\)

= \({ \left ( AB \right ) }'\)

Hence, proved.

 

 

Q-5: If M = \(\begin{bmatrix} cos \beta & sin \beta \\ -sin \beta & cos \beta \end{bmatrix}\), then prove that M’M = I.

 

Solution:

M = \(\begin{bmatrix} cos \beta & sin \beta \\ -sin \beta & cos \beta \end{bmatrix}\)

M’ = \(\begin{bmatrix} cos \beta & -sin \beta \\ sin \beta & cos \beta \end{bmatrix}\)

M’M = \(\begin{bmatrix} cos \beta & -sin \beta \\ sin \beta & cos \beta \end{bmatrix}\) \(\begin{bmatrix} cos \beta & sin \beta \\ -sin \beta & cos \beta \end{bmatrix}\)

= \(\begin{bmatrix} \left ( cos \beta \right )\left ( cos \beta \right ) + \left ( -sin \beta \right )\left ( -sin \beta \right ) & \left ( cos \beta \right )\left ( sin \beta \right ) + \left ( -sin \beta \right )\left ( cos \beta \right ) \\ \left ( sin \beta \right )\left ( cos \beta \right ) + \left ( cos \beta \right )\left ( -sin \beta \right ) & \left ( sin \beta \right )\left ( sin \beta \right ) + \left ( cos \beta \right )\left ( cos \beta \right ) \end{bmatrix}\)

= \(\begin{bmatrix} cos^{ 2 } \beta + sin^{ 2 } & cos \beta . sin \beta – sin \beta . cos \beta \\ sin \beta cos \beta – cos \beta . sin \beta & sin^{ 2 } \beta + cos^{ 2 } \beta \end{bmatrix}\)

= \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)

= I

Hence, proved.

 

 

Q-6: Prove that the matrix A = \(\begin{bmatrix} 2 & -2 & 6 \\ -2 & 3 & 2 \\ 6 & 2 & 4 \end{bmatrix}\) is a symmetric matrix.

 

Solution:

Here, we can see that

\({ A }'\) = \(\begin{bmatrix} 2 & -2 & 6 \\ -2 & 3 & 2 \\ 6 & 2 & 4 \end{bmatrix}\) which is equal to A.

Hence, \({ A }'\) = A

Therefore, the given matrix A = \(\begin{bmatrix} 2 & -2 & 6 \\ -2 & 3 & 2 \\ 6 & 2 & 4 \end{bmatrix}\) is a symmetric matrix.

 

 

Q-7: Prove that ( A + \({ A }'\) ) is a symmetric matrix for the matrix A = \(\begin{bmatrix} 2 & 6 \\ 7 & 8 \end{bmatrix}\).

 

Solution:

\({ A }'\) = \(\begin{bmatrix} 2 & 7 \\ 6 & 8 \end{bmatrix}\)

A + \({ A }'\) = \(\begin{bmatrix} 2 & 6 \\ 7 & 8 \end{bmatrix}\) + \(\begin{bmatrix} 2 & 7 \\ 6 & 8 \end{bmatrix}\)

= \(\begin{bmatrix} 4 & 13 \\ 13 & 16 \end{bmatrix}\)

\({ \left ( A + { A }’ \right ) }'\) = \(\begin{bmatrix} 4 & 13 \\ 13 & 16 \end{bmatrix}\) = A + \({ A }'\)

Hence, (A + \({ A }'\) ) is a symmetric matrix.

 

 

Q-8: If M and N are the symmetric matrices with having same order, then MN – NM will be a

(a) Symmetric matrix

(b) Skew symmetric matrix.

(c) Identity matrix 

(d) Zero matrix

 

Solution:

As per the data given in the question, M and N are two symmetric matrices.

Thus, we have

\({ A }’ = A\) and \({ B }’ = B\) ……..(i)

Let us consider,

\({ \left( MN – NM \right ) }’ = { \left ( MN \right ) }’ – { \left ( NM \right ) }’ \;\;\;\;\;\;\;\; \left [ {\left ( M – N \right )}’ = { M }’ – { N }’ \right ]\)

= \({ N }'{ M }’ – { M }'{ N }’ \;\;\;\;\;\;\;\; \left [ {\left ( MN \right )}’ = { N }'{ M }’ \right ]\)

= \(BA – MN \;\;\;\;\;\;\;\;\;\; \left [ by \; (i) \right ]\)

= – (MN – NM)

Thus, (MN – NM )’ = – (MN – NM)

Hence, (MN – NM) is a skew- symmetric matrix.

Therefore, the correct answer is (b).

 

 

Q-9: Consider M = \(\begin{bmatrix} cos \beta & – sin \beta \\ sin \beta & cos \beta \end{bmatrix}\).

Then M + M’ = I, if β is:

(a) \(\frac{ \pi }{ 3 }\)                                       

(b) \(\frac{ \pi }{ 6 }\)

(c) \(\frac{ 3 \pi }{ 2 }\)                                      

(d) \(\frac{ \pi }{ 2 }\)

 

Solution:

M = \(\begin{bmatrix} cos \beta & -sin \beta \\ sin \beta & cos \beta \end{bmatrix}\)

M’ = \(\begin{bmatrix} cos \beta & sin \beta \\ -sin \beta & cos \beta \end{bmatrix}\)

M + M’ = I

⟹ \(\begin{bmatrix} cos \beta  & -sin \beta \\ sin \beta  & cos \beta \end{bmatrix}\)  + \(\begin{bmatrix} cos \beta  & sin \beta \\ -sin \beta  & cos \beta \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)

⟹ \(\begin{bmatrix} 2 cos \beta  & 0 \\ 0  & 2 cos \beta  \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)

Thus,

\( 2 cos \beta = 1 \)

⟹ cos β = \(\frac{ 1 }{ 2 }\)

⟹ cos β = \(cos \frac{ \pi }{ 3 }\)

β = \(\frac{ \pi }{ 3 }\)

Therefore, (a) is the correct answer.

 

 

EXERCISE – 3.4

 

 

Q-1: What will be the inverse of the matrix \(\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}\), if any exists?

 

Solution:

Let, M = \(\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}\)

As we know, M = IM

⟹ \(\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)M

⟹ \(\begin{bmatrix} 1 & -1 \\ 0 & 5 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix}\)M        [ R2 → R2 – 2 R1 ]

⟹ \(\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ \frac{ -2 }{ 5 }  & \frac{ 1 }{ 5 } \end{bmatrix}\)M        [ R2 → \(\frac{ 1 }{ 5 } \) R2]

⟹ \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} \frac{ 3 }{ 5 }  & \frac{ 1 }{ 5 } \\ \frac{ -2 }{ 5 }  & \frac{ 1 }{ 5 } \end{bmatrix}\)M        [ R2 → R1 +  R2 ]

Hence, M-1 = \(\begin{bmatrix} \frac{ 3 }{ 5 }  & \frac{ 1 }{ 5 } \\ \frac{ -2 }{ 5 }  & \frac{ 1 }{ 5 } \end{bmatrix}\)

 

 

Q-2: What will be the inverse of the matrix \(\begin{bmatrix} 2 & -6 \\ 1 & -2 \end{bmatrix}\), if any exists?

 

Solution:

Let, M = \(\begin{bmatrix} 2 & -6 \\ 1 & -3 \end{bmatrix}\)

As we know, M = IM

⟹ \(\begin{bmatrix} 2 & -6 \\ 1 & -3 \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)M

⟹ \(\begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}\)M        [ C2 → C2  +  3 C1 ]

⟹ \(\begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 3 \\ -1  & 1 \end{bmatrix}\)M        [ C1 → C1 – C2]

⟹ \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -1  & 3  \\ \frac{ -1 }{ 2 }  & 1 \end{bmatrix}\)M        [ C1 → \(\frac{ 1 }{ 2 }\) C1]

Hence, M-1 = \(\begin{bmatrix} \frac{ 3 }{ 5 }  & \frac{ 1 }{ 5 } \\ \frac{ -2 }{ 5 }  & \frac{ 1 }{ 5 } \end{bmatrix}\)

 

 

Q-3: What will be the inverse of the matrix \(\begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}\), if any exists?

 

Solution:

Let, M = \(\begin{bmatrix} 6 & -3 \\ -2 & 1  \end{bmatrix}\)

As we know, M = IM

⟹ \(\begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)M

⟹ \(\begin{bmatrix} 1 & \frac{ -1 }{ 2 } \\ -2  & 1 \end{bmatrix} = \begin{bmatrix} \frac{ 1 }{ 6 }  & 0 \\ 0 & 1 \end{bmatrix}\)M        [ R1 → \(\frac{ 1 }{ 6 }\) R1 ]

⟹ \(\begin{bmatrix} 1 & \frac{ -1 }{ 2 } \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} \frac{ 1 }{ 6 } & 0 \\ \frac{ 1 }{ 3 }  & 1 \end{bmatrix}\)M        [ R2 → R2 +2R1]

Here, in the above matrix in LHS side, there is only zero in the second row.

Hence, M-1 does not exist.

 

 

Q-4: What will be the inverse of the matrix \(\begin{bmatrix} 2  & -3 \\ -1  & 2 \end{bmatrix}\), if any exists?

 

Solution:

Let, M = \(\begin{bmatrix} 2 & -3 \\ -1 & 2  \end{bmatrix}\)

As we know, M = IM

⟹ \(\begin{bmatrix} 2  & -3 \\ -1  & 2 \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)M

⟹ \(\begin{bmatrix} 1  & -1  \\ -1  & 2 \end{bmatrix} = \begin{bmatrix} 1  & 1 \\ 0 & 1 \end{bmatrix}\)M        [ R1 → R1 + R2 ]

⟹ \(\begin{bmatrix} 1  & -1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1  & 1 \\ 1  & 2 \end{bmatrix}\)M        [ R2 → R2 + R1]

⟹ \(\begin{bmatrix} 1  & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2  & 3 \\ 1  & 2 \end{bmatrix}\)M        [ R2 → R1 + R2]

Hence, M-1 = \( \begin{bmatrix} 2  & 3 \\ 1  & 2 \end{bmatrix}\)

 

 

Q-5: What will be the inverse of the matrix \(\begin{bmatrix} 2  & 1 \\ 4  & 2 \end{bmatrix}\), if any exists?

 

Solution:

Let, M = \(\begin{bmatrix} 2 & 1 \\ 4 & 2  \end{bmatrix}\)

As we know, M = IM

⟹ \(\begin{bmatrix} 2  & 1 \\ 4  & 2 \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)M

⟹ \(\begin{bmatrix} 0  & 0  \\ 4  & 2 \end{bmatrix} = \begin{bmatrix} 1  & – \frac{ 1 }{ 2 } \\ 0 & 1 \end{bmatrix}\)M        [ R1 → R1 – \(\frac{ 1 }{ 6 }\) R2 ]

Here, in the above matrix in LHS side, there is only zero in the second row.

Hence, M-1 does not exist.

 

 

Q-6: What will be the inverse of the matrix \(\begin{bmatrix} 3 & 10 \\ 2 & 7 \end{bmatrix}\), if any exists?

 

Solution:

Let, M = \(\begin{bmatrix} 3 & 10 \\ 2 & 7 \end{bmatrix}\)

As we know, M = IM

⟹ \(\begin{bmatrix} 3  & 10 \\ 2 & 7 \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)M

⟹ \(\begin{bmatrix} 1  & 3 \\ 2  & 7 \end{bmatrix} = \begin{bmatrix} 1  & -1 \\ 0 & 1 \end{bmatrix}\)M        [ R1 → R1  –  R2 ]

⟹ \(\begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1  & -1 \\ -2  & 3 \end{bmatrix}\)M        [ R2 → R2 – 2R1]

⟹ \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7  & -10  \\ -2  & 3 \end{bmatrix}\)M        [ R1 →  R1 – 3 R2]

Hence, M-1 = \( \begin{bmatrix} 7  & -10  \\ -2  & 3 \end{bmatrix}\)

 

 

Q-7: What will be the inverse of the matrix \(\begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}\), if any exists?

 

Solution:

Let, M = \(\begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}\)

As we know, M = IM

\(\begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\)M

⟹ \(\begin{bmatrix} 1  & 0  & \frac{ -1 }{ 2 }  \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}\) = \(\begin{bmatrix} \frac{ 1 }{ 2 } & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\)M                [ R1 → \(\frac{ 1 }{ 2 }\)R1 ]

⟹ \(\begin{bmatrix} 1  & 0  & \frac{ -1 }{ 2 }  \\ 0 & 1 & \frac{ 5 }{ 2 } \\ 0 & 1 & 3 \end{bmatrix}\) = \(\begin{bmatrix} \frac{ 1 }{ 2 } & 0 & 0 \\ \frac{ -5 }{ 2 } & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\)M                [ R2 → R2 – 5 R1]

⟹ \(\begin{bmatrix} 1  & 0  & \frac{ -1 }{ 2 }  \\ 0 & 1 & \frac{ 5 }{ 2 } \\ 0 & 1 & \frac{ 1 }{ 2 } \end{bmatrix}\) = \(\begin{bmatrix} \frac{ 1 }{ 2 } & 0 & 0 \\ \frac{ -5 }{ 2 } & 1 & 0 \\ \frac{ 5 }{ 2 } & -1 & 1 \end{bmatrix}\)M                [ R3 → R3 – R2]

⟹ \(\begin{bmatrix} 1  & 0  & \frac{ -1 }{ 2 }  \\ 0 & 1 & \frac{ 5 }{ 2 } \\ 0 & 0 & 1 \end{bmatrix}\) = \(\begin{bmatrix} \frac{ 1 }{ 2 } & 0 & 0 \\ \frac{ -5 }{ 2 } & 1 & 0 \\ 5 & -2 & 2 \end{bmatrix}\)M                [ R3 → 2R3]

⟹ \(\begin{bmatrix} 1  & 0  & 0  \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\) = \(\begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}\)M                [ R1 → R1 + \(\frac{ 1 }{ 2 }\) R3, and R2 → R2 – \(\frac{ 5 }{ 2 }\) R3]

Hence, M-1 = \(\begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}\)

 

 

Q-8: The matrices M and N can be inverse of each other if and only if

(a) MN = NM                                              (b) MN = NM = 0

(c) MN = 0, NM = I                                     (d) MN = NM = I

 

Solution:

As we know, if M is a square matrix of order a, and if there exists another square matrix N of the same order b, in such a manner that MN = NM = I, then N will be  said to be the inverse of M. So, in this case, we can see that M is the inverse of N.

Therefore, matrices M and N will be inverses of each other only if MN = NM = I.

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