RD Sharma Class 12 Exercise 4.1 Algebra of Matrices Solutions Maths

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The 4th chapter of the book contains advanced concepts and problems on matrices. For example, there will be sums on addition, subtraction, division, and multiplication of matrices. You will also explore types of Matrices like skew-symmetric matrix, null matrix, symmetric matrix, etc.

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RD Sharma Class 12 Solutions Chapter 4 Algebra of Matrices - Other Exercise

Algebra of Matrices Exercise 4.1

Algebra of Matrices Exercise 4.1 Question 1

Answer: 1st part:

(8 \times 1), (1 \times 8), (4 \times 2), (2 \times 4)

and for the 2nd part:

(5 \times 1), (1 \times 5)

Given: matrix has 8 elements
Here we have to find the possible order of the given matrix
Hint: If the matrix is of the order

m \times n

elements, it has

m n

elements.
Solution: If the matrix has 8 elements, we will find the ordered pairs

m

and

n

m \times n = 8

Then ordered pairs

m

and

n

will be

m \times n

(8 \times 1), (1 \times 8), (4 \times 2), (2 \times 4)

Now, if it has 5 elements then possible orders are .

(5 \times 1), (1 \times 5)

Algebra of Matrices Exercise 4.1 Question 2 (i)

Answer:

a_{22} + b_{21} = 1

Given:

A = [a_{i j}] = [\begin{matrix} 2 & 3 & - 5 \\ 1 & 4 & 9 \\ 0 & 7 & - 2 \end{matrix}]

and

B = [b_{i j}] = [\begin{matrix} 2 & - 1 \\ - 3 & 4 \\ 1 & 2 \end{matrix}]

Here we have to find out the values of $a_{22} + b_{21}$
Hint: Simply we select the elements in the matrix which elements required and simplify
Solution: We know that

A = [a_{i j}] = [\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix}]

\cdot \cdot \cdot (i)

B = [b_{i j}] = [\begin{matrix} b_{11} & b_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{matrix}]

\cdot \cdot \cdot (i i)

Also given that

A = [a_{i j}] = [\begin{matrix} 2 & 3 & - 5 \\ 1 & 4 & 9 \\ 0 & 7 & - 2 \end{matrix}]

and

B = [b_{i j}] = [\begin{matrix} 2 & - 1 \\ - 3 & 4 \\ 1 & 2 \end{matrix}]

Now comparing with eqn (i) and (ii) we have,

a_{22} = 4

b_{21} = - 3

Hence

a_{22} + b_{21} = 4 + (- 3) = 1

This is the required answer.

Algebra of Matrices Exercise 4.1 Question 2 (i)

Answer:

20

Given: Here given that

A = [a_{i j}] = [\begin{matrix} 2 & 3 & - 5 \\ 1 & 4 & 9 \\ 0 & 7 & - 2 \end{matrix}]

and

B = [b_{i j}] = [\begin{matrix} 2 & - 1 \\ - 3 & 4 \\ 1 & 2 \end{matrix}]

Here we have to find out the values of $a_{11} b_{11} + a_{22} b_{22} = 1$
Hint: Simply we select the elements in the matrix which elements required and simplify
Solution: We know that

A = [a_{i j}] = [\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix}]

\cdot \cdot \cdot (i)

B = [b_{i j}] = [\begin{matrix} b_{11} & b_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{matrix}]

\cdot \cdot \cdot (i i)

Also given that

A = [a_{i j}] = [\begin{matrix} 2 & 3 & - 5 \\ 1 & 4 & 9 \\ 0 & 7 & - 2 \end{matrix}]

and

B = [b_{i j}] = [\begin{matrix} 2 & - 1 \\ - 3 & 4 \\ 1 & 2 \end{matrix}]

Now comparing with eqn(i) and (ii) we have,

a_{11} = 2

a_{22} = 4

b_{11} = 2

b_{22} = 4

Hence,

a_{11} b_{11} + a_{22} b_{22} = 2 \times 2 + 4 \times 4

= 4 + 16

a_{11} b_{11} + a_{22} b_{22} = 20

Algebra of Matrices Exercise 4.1 Question 3

Answer: The order of matrix

R_{1} = 1 \times 4

And the order of matrix

C_{2} = 3 \times 1

Given:

A

be a matrix of order

3 \times 4

Here we have to determine the order of matrices

R_{1}

and

C_{2}

Hint: We have to write the order along row and again along column
Solution: Here

A

be a matrix of order

3 \times 4

So,

A = {[a_{i j}]}_{3 \times 4}

R_{1}

= first row of

A = [\begin{matrix} a_{11} & a_{12} & a_{13} & a_{14} \end{matrix}]

So, order of Matrix

R_{1} = 1 \times 4

Again,

C_{2}

= second column of

A = [\begin{matrix} a_{12} \\ a_{22} \\ a_{32} \end{matrix}]

Therefore, order of

C_{2} = 3 \times 1

Algebra of Matrices Exercise 4.1 Question 4 (i)
Answer: $a_{i j}$ $= i \times j$ , $A = [\begin{matrix} 1 & 2 & 3 \\ 2 & 4 & 6 \end{matrix}]$

Given: Here given that matrix of order

2 \times 3

A = {[a_{i j}]}_{2 \times 3}

Here we have to construct

2 \times 3

matrix as

a_{i j}

= i \times j

Hint: We have to construct the matrix according to the question
Solution: Given

a_{i j}

= i \times j

Let

A = {[a_{i j}]}_{2 \times 3}

So, the elements in a

2 \times 3

matrix are

a_{11}, a_{12}, a_{13}, a_{21}, a_{22}, a_{23}

A = [\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{matrix}]

a_{11} = 1 \times 1 = 1 a_{12} = 1 \times 2 = 2 a_{13} = 1 \times 3 = 3

a_{21} = 2 \times 1 = 2 a_{22} = 2 \times 2 = 4 a_{23} = 2 \times 3 = 6

Substituting these values in Matrix

A

, we get

A = [\begin{matrix} 1 & 2 & 3 \\ 2 & 4 & 6 \end{matrix}]

Hence this is the required answer.

Algebra of Matrices Exercise 4.1 Question 4 (ii)

Answer:

A = [\begin{matrix} 1 & 0 & - 1 \\ 3 & 2 & 1 \end{matrix}]

Given: Here given that matrix of order

2 \times 3

A = {[a_{i j}]}_{2 \times 3}

Here we have to construct

2 \times 3

matrix as

a_{i j} = 2 i - j

Hint: We have to construct the matrix according to the question
Solution: Let

A = {[a_{i j}]}_{2 \times 3}

So, the elements in a

2 \times 3

matrix are

a_{11}, a_{12}, a_{13}, a_{21}, a_{22}, a_{23}

A = [\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{matrix}]

a_{11} = 2 \times 1 - 1 = 2 - 1 = 1 a_{12} = 2 \times 1 - 2 = 2 - 2 = 0 a_{13} = 2 \times 1 - 3 = 2 - 3 = - 1

a_{21} = 2 \times 2 - 1 = 4 - 1 = 3 a_{22} = 2 \times 2 - 2 = 4 - 2 = 2 a_{23} = 2 \times 2 - 3 = 4 - 3 = 1

Substituting these values in Matrix

A

, we get

A = [\begin{matrix} 1 & 0 & - 1 \\ 3 & 2 & 1 \end{matrix}]

Hence this is the required answer.

Algebra of Matrices Exercise 4.1 Question 4 (iii)

Answer:

A = [\begin{matrix} 2 & 3 & 4 \\ 3 & 4 & 5 \end{matrix}]

Given: Here given that matrix of order

2 \times 3

A = {[a_{i j}]}_{2 \times 3}

Here we have to construct the matrix according to

a_{i j} = 2 i + j

Hint: First we have to simply adding the row elements with column element as given
question and then construct matrix.
Solution: Let

A = {[a_{i j}]}_{2 \times 3}

So, the elements in a

2 \times 3

matrix are

a_{11}, a_{12}, a_{13}, a_{21}, a_{22}, a_{23}

A = [\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{matrix}]

a_{11} = 1 + 1 = 2 a_{12} = 1 + 2 = 3 a_{13} = 1 + 3 = 4

a_{21} = 2 + 1 = 3 a_{22} = 2 + 2 = 4 a_{23} = 2 + 3 = 5

Substituting these values in Matrix

A

, we get

A = [\begin{matrix} 2 & 3 & 4 \\ 3 & 4 & 5 \end{matrix}]

Hence this is the required answer.

Algebra of Matrices Exercise 4.1 Question 4 (iv)

Answer:

A = [\begin{matrix} 2 & 4.5 & 8 \\ 4.5 & 8 & 12.5 \end{matrix}]

Given: Here given that matrix of order

2 \times 3

A = {[a_{i j}]}_{2 \times 3}

Here we have to construct the matrix according to

a_{i j} = \frac{{(i + j)}^{2}}{2}

Hint: Adding row and column element and squaring then divide by 2
Solution: Let

A = {[a_{i j}]}_{2 \times 3}

and given that

a_{i j} = \frac{{(i + j)}^{2}}{2}

So, the elements in a

2 \times 3

matrix are

a_{11}, a_{12}, a_{13}, a_{21}, a_{22}, a_{23}

A = [\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{matrix}]

a_{11} = \frac{{(1 + 1)}^{2}}{2} = \frac{4}{2} = 2 a_{12} = \frac{{(1 + 2)}^{2}}{2} = \frac{{(3)}^{2}}{2} = \frac{9}{2} = 4.5 a_{13} = \frac{{(1 + 3)}^{2}}{2} = \frac{{(4)}^{2}}{2} = \frac{16}{2} = 8

a_{21} = \frac{{(2 + 1)}^{2}}{2} = \frac{{(3)}^{2}}{2} = \frac{9}{2} = 4.5 a_{22} = \frac{{(2 + 2)}^{2}}{2} = \frac{{(4)}^{2}}{2} = \frac{16}{2} = 8 a_{23} = \frac{{(2 + 3)}^{2}}{2} = \frac{{(5)}^{2}}{2} = \frac{25}{2} = 12.5

Substituting these values in Matrix

A

, we get

A = [\begin{matrix} 2 & 4.5 & 8 \\ 4.5 & 8 & 12.5 \end{matrix}]

Hence this is the required answer.

Algebra of Matrices Exercise 4.1 Question 5 (i)

Answer:

A = [\begin{matrix} 2 & \frac{9}{2} \\ \frac{9}{2} & 8 \end{matrix}]

Given:

\frac{{(i + j)}^{2}}{2}

Here we have to construct

2 \times 2

matrix according to

\frac{{(i + j)}^{2}}{2}

Hint: Substitute required values in the

2 \times 2

matrix
Solution: Let

A = {[a_{i j}]}_{2 \times 2}

So, the elements in a

2 \times 2

are

a_{11}, a_{12}, a_{21}, a_{22}

A = [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}]

a_{11} = \frac{{(1 + 1)}^{2}}{2} = \frac{4}{2} = 2 a_{12} = \frac{{(1 + 2)}^{2}}{2} = \frac{{(3)}^{2}}{2} = \frac{9}{2} = 4.5

a_{21} = \frac{{(2 + 1)}^{2}}{2} = \frac{{(3)}^{2}}{2} = \frac{9}{2} = 4.5 a_{22} = \frac{{(2 + 2)}^{2}}{2} = \frac{{(4)}^{2}}{2} = \frac{16}{2} = 8

Substituting these values in Matrix

A

, we get

A = [\begin{matrix} 2 & \frac{9}{2} \\ \frac{9}{2} & 8 \end{matrix}]

Algebra of Matrices Exercise 4.1 Question 5 (ii)

Answer:

A = [\begin{matrix} 0 & \frac{1}{2} \\ \frac{1}{2} & 0 \end{matrix}]

Given:

\frac{{(i - j)}^{2}}{2}

Here we have to construct

2 \times 2

matrix according to

\frac{{(i - j)}^{2}}{2}

Hint: Substitute required values in the

2 \times 2

matrix
Solution: Let

A = {[a_{i j}]}_{2 \times 2}

So, the elements in a

2 \times 2

are

a_{11}, a_{12}, a_{21}, a_{22}

A = [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}]

a_{11} = \frac{{(1 - 1)}^{2}}{2} = 0 a_{12} = \frac{{(1 - 2)}^{2}}{2} = \frac{{(1)}^{2}}{2} = \frac{1}{2}

a_{21} = \frac{{(2 - 1)}^{2}}{2} = \frac{1}{2} a_{22} = \frac{{(2 - 2)}^{2}}{2} = 0

Substituting these values in Matrix

A

, we get

A = [\begin{matrix} 0 & \frac{1}{2} \\ \frac{1}{2} & 0 \end{matrix}]

Algebra of Matrices Exercise 4.1 Question 5 (iii)

Answer:

A = [\begin{matrix} \frac{1}{2} & \frac{9}{2} \\ \0 & 2 \end{matrix}]

Given:

a_{i j} = \frac{{(i - 2 j)}^{2}}{2}

Here we have to construct

2 \times 2

matrix according to

\frac{{(i - 2 j)}^{2}}{2}

Hint: Substitute required values in the

2 \times 2

matrix
Solution: Let

A = {[a_{i j}]}_{2 \times 2}

So, the elements in a

2 \times 2

are

a_{11}, a_{12}, a_{21}, a_{22}

A = [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}]

a_{11} = \frac{{(1 - 2 \times 1)}^{2}}{2} = \frac{- 1^{2}}{2} = \frac{1}{2} a_{12} = \frac{{(1 - 2 \times 2)}^{2}}{2} = \frac{{(- 3)}^{2}}{2} = \frac{9}{2}

a_{21} = \frac{{(2 - 1 \times 1)}^{2}}{2} = \frac{{(0)}^{2}}{2} = 0 a_{22} = \frac{{(2 - 2 \times 2)}^{2}}{2} = \frac{- 2^{2}}{2} = \frac{4}{2} = 2

Substituting these values in Matrix

A

, we get

A = [\begin{matrix} \frac{1}{2} & \frac{9}{2} \\ \0 & 2 \end{matrix}]

Algebra of Matrices Exercise 4.1 Question 5 (iv)

Answer:

A = [\begin{matrix} \frac{9}{2} & 8 \\ \frac{25}{2} & 18 \end{matrix}]

Given:

a_{i j} = \frac{{(2 i + j)}^{2}}{2}

Here we have to construct

2 \times 2

matrix according to

\frac{{(2 i + j)}^{2}}{2}

Hint: Substitute required values in the

2 \times 2

matrix
Solution: Let

A = {[a_{i j}]}_{2 \times 2}

So, the elements in a

2 \times 2

are

a_{11}, a_{12}, a_{21}, a_{22}

A = [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}]

a_{11} = \frac{{(2 \times 1 + 1)}^{2}}{2} = \frac{3^{2}}{2} = \frac{9}{2} a_{12} = \frac{{(2 \times 1 + 2)}^{2}}{2} = \frac{{(4)}^{2}}{2} = \frac{16}{2} = 8

a_{21} = \frac{{(2 \times 2 + 1)}^{2}}{2} = \frac{{(5)}^{2}}{2} = \frac{25}{2} a_{22} = \frac{{(2 \times 2 + 2)}^{2}}{2} = \frac{6^{2}}{2} = \frac{36}{2} = 18

Substituting these values in Matrix

A

, we get

A = [\begin{matrix} \frac{9}{2} & 8 \\ \frac{25}{2} & 18 \end{matrix}]

Algebra of Matrices Exercise 4.1 Question 5 (v)

Answer:

A = [\begin{matrix} \frac{1}{2} & 2 \\ \frac{1}{2} & 1 \end{matrix}]

Given:

a_{i j} = \frac{| 2 i - 3 j |}{2}

Here we have to construct

2 \times 2

matrix according to

\frac{| 2 i - 3 j |}{2}

Hint: Substitute required values in the

2 \times 2

matrix according

\frac{| 2 i - 3 j |}{2}

Solution: Let

A = {[a_{i j}]}_{2 \times 2}

So, the elements in a

2 \times 2

are

a_{11}, a_{12}, a_{21}, a_{22}

A = [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}]

a_{11} = \frac{| 2 \times 1 - 3 \times 1 |}{2} = \frac{1}{2} a_{12} = \frac{| 2 \times 1 - 3 \times 2 |}{2} = \frac{4}{2} = 2

a_{21} = \frac{| 2 \times 2 - 3 \times 1 |}{2} = \frac{4 - 3}{2} = \frac{1}{2} a_{22} = \frac{| 2 \times 2 - 3 \times 2 |}{2} = \frac{2}{2} = 1

Substituting these values in Matrix

A

, we get

A = [\begin{matrix} \frac{1}{2} & 2 \\ \frac{1}{2} & 1 \end{matrix}]

Algebra of Matrices Exercise 4.1 Question 5 (vi)

Answer:

A = [\begin{matrix} 1 & \frac{1}{2} \\ \frac{5}{2} & 2 \end{matrix}]

Given:

a_{i j} = \frac{| - 3 i + j |}{2}

Here we have to construct

2 \times 2

matrix according to

a_{i j} = \frac{| - 3 i + j |}{2}

Hint: Substitute required values in the

2 \times 2

matrix
Solution: Let

A = {[a_{i j}]}_{2 \times 2}

So, the elements in a

2 \times 2

are

a_{11}, a_{12}, a_{21}, a_{22}

A = [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}]

a_{11} = \frac{| - 3 \times 1 + 1 |}{2} = \frac{2}{2} = 1 a_{12} = \frac{| - 3 \times 1 + 2 |}{2} = \frac{1}{2}

a_{21} = \frac{| - 3 \times 2 + 1 |}{2} = \frac{5}{2} a_{22} = \frac{| - 3 \times 2 + 2 |}{2} = \frac{4}{2} = 2

Substituting these values in Matrix

A

, we get

A = [\begin{matrix} 1 & \frac{1}{2} \\ \frac{5}{2} & 2 \end{matrix}]

Algebra of Matrices Exercise 4.1 Question 5 (vii)

Answer:

A = [\begin{matrix} e^{2 x} \sin x & e^{2 x} \sin 2 x \\ e^{4 x} \sin x & e^{4 x} \sin 2 x \end{matrix}]

Given:

a_{i j} = e^{2 x} \sin x j

Here we have to construct

2 \times 2

matrix according to

e^{2 x} \sin x j

Hint: Putting the value of each row and column element according to the question in matrix
Solution: Let

A = {[a_{i j}]}_{2 \times 2}

= e^{2 x} \sin x j

So, the elements in a

2 \times 2

are

a_{11}, a_{12}, a_{21}, a_{22}

A = [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}]

a_{11} = e^{2 \times 1 x} \sin x \times 1 = e^{2 x} \sin x a_{12} = e^{2 \times 1 x} \sin x \times 2 = e^{2 x} \sin 2 x

a_{21} = e^{2 \times 2 x} \sin x \times 1 = e^{4 x} \sin x a_{22} = e^{2 \times 2 x} \sin x \times 2 = e^{4 x} \sin 2 x

Substituting these values in Matrix

A

, we get

A = [\begin{matrix} e^{2 x} \sin x & e^{2 x} \sin 2 x \\ e^{4 x} \sin x & e^{4 x} \sin 2 x \end{matrix}]

Algebra of Matrices Exercise 4.1 Question 6 (i)

Answer:

A = [\begin{matrix} 2 & 3 & 4 & 5 \\ 3 & 4 & 5 & 6 \\ 4 & 5 & 6 & 7 \end{matrix}]

Given:

a_{i j} = (i + j)

Here we have to construct

3 \times 4

matrix according to

(i + j)

Hint: Find the sum of

i

and

j

for each element.
Solution: Here

a_{i j} = (i + j)

Let

A = {[a_{i j}]}_{3 \times 4}

So,

A = {[\begin{matrix} 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{matrix}]}_{3 \times 4}

a_{11} = 1 + 1 = 2 a_{12} = 1 + 2 = 3 a_{13} = 1 + 4 = 4 a_{14} = 1 + 4 = 5

a_{21} = 2 + 1 = 3 a_{22} = 2 + 2 = 4 a_{23} = 2 + 3 = 5 a_{24} = 2 + 4 = 6

a_{31} = 3 + 1 = 4 a_{32} = 3 + 2 = 5 a_{33} = 3 + 3 = 6 a_{34} = 3 + 4 = 7

Substituting these values in Matrix

A

, we get

A = [\begin{matrix} 2 & 3 & 4 & 5 \\ 3 & 4 & 5 & 6 \\ 4 & 5 & 6 & 7 \end{matrix}]

Algebra of Matrices Exercise 4.1 Question 6 (ii)

Answer:

A = [\begin{matrix} 0 & - 1 & - 2 & - 3 \\ 1 & 0 & - 1 & - 2 \\ 2 & 1 & 0 & - 1 \end{matrix}]

Given:

a_{i j} = (i - j)

Here we have to construct

3 \times 4

matrix according to

(i - j)

Hint: Find the sum of

i

and

j

for each element.
Solution: Here

a_{i j} = (i - j)

Let

A = {[a_{i j}]}_{3 \times 4}

So,

A = {[\begin{matrix} 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{matrix}]}_{3 \times 4}

a_{11} = 1 - 1 = 0 a_{12} = 1 - 2 = - 1 a_{13} = 1 - 3 = - 2 a_{14} = 1 - 4 = - 3

a_{21} = 2 - 1 = 1 a_{22} = 2 - 2 = 0 a_{23} = 2 - 3 = - 1 a_{24} = 2 - 4 = - 2

a_{31} = 3 - 1 = 2 a_{32} = 3 - 2 = 1 a_{33} = 3 - 3 = 0 a_{34} = 3 - 4 = - 1

Substituting these values in Matrix

A

, we get

A = [\begin{matrix} 0 & - 1 & - 2 & - 3 \\ 1 & 0 & - 1 & - 2 \\ 2 & 1 & 0 & - 1 \end{matrix}]

Algebra of Matrices Exercise 4.1 Question 6 (iii)

Answer:

A = [\begin{matrix} 2 & 2 & 2 & 2 \\ 4 & 4 & 4 & 4 \\ 6 & 6 & 6 & 6 \end{matrix}]

Given:

a_{i j} = (2 i)

Here we have to construct

3 \times 4

matrix according to

(2 i)

Hint: Substitute the required value according

(2 i)

Solution: Here

a_{i j} = (2 i)

Let

A = {[a_{i j}]}_{3 \times 4}

So,

A = {[\begin{matrix} 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{matrix}]}_{3 \times 4}

a_{11} = 2 \times 1 = 2 a_{12} = 2 \times 1 = 2 a_{13} = 2 \times 1 = 2 a_{14} = 2 \times 1 = 2

a_{21} = 2 \times 2 = 4 a_{22} = 2 \times 2 = 4 a_{23} = 2 \times 2 = 4 a_{24} = 2 \times 2 = 4

a_{31} = 2 \times 3 = 6 a_{32} = 2 \times 3 = 6 a_{33} = 2 \times 3 = 6 a_{34} = 2 \times 3 = 6

Substituting these values in Matrix

A

, we get

A = [\begin{matrix} 2 & 2 & 2 & 2 \\ 4 & 4 & 4 & 4 \\ 6 & 6 & 6 & 6 \end{matrix}]

Algebra of Matrices Exercise 4.1 Question 6 (v)

Answer:

A = [\begin{matrix} 1 & \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{5}{2} & 2 & \frac{3}{2} & 1 \\ 4 & \frac{7}{2} & 3 & \frac{5}{2} \end{matrix}]

Given:

a_{i j} = \frac{1}{2} | - 3 i + j |

Here we have to construct

3 \times 4

matrix according to

a_{i j} = \frac{1}{2} | - 3 i + j |

Hint: We have to find all the elements of matrix according to

\frac{1}{2} | - 3 i + j |

Solution: Here

a_{i j} = \frac{1}{2} | - 3 i + j |

Let

A = {[a_{i j}]}_{3 \times 4}

So,

A = {[\begin{matrix} 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{matrix}]}_{3 \times 4}

a_{11} = \frac{1}{2} | - 3 \times 1 + 1 | = \frac{1}{2} \times 2 = 1 a_{12} = \frac{1}{2} | - 3 \times 1 + 2 | = \frac{1}{2} \times 1 = \frac{1}{2} a_{13} = \frac{1}{2} | - 3 \times 1 + 3 | = \frac{1}{2} \times 0 = 0 a_{14} = \frac{1}{2} | - 3 \times 1 + 4 | = \frac{1}{2} \times 1 = \frac{1}{2}

a_{21} \frac{1}{2} | - 3 \times 2 + 1 | = \frac{1}{2} \times 5 = \frac{5}{2} a_{22} \frac{1}{2} | - 3 \times 2 + 2 | = \frac{1}{2} \times 4 = 2 a_{23} \frac{1}{2} | - 3 \times 2 + 3 | = \frac{1}{2} \times 3 = \frac{3}{2} a_{24} \frac{1}{2} | - 3 \times 2 + 4 | = \frac{1}{2} \times 2 = 1

a_{31} = \frac{1}{2} | - 3 \times 3 + 1 | = \frac{1}{2} \times 8 = 4 a_{32} = \frac{1}{2} | - 3 \times 3 + 2 | = \frac{1}{2} \times 7 = \frac{7}{2} a_{33} = \frac{1}{2} | - 3 \times 3 + 3 | = \frac{1}{2} \times 6 = 3 a_{34} = \frac{1}{2} | - 3 \times 3 + 4 | = \frac{1}{2} \times 5 = \frac{5}{2}

Substituting these values in Matrix

A

, we get

A = [\begin{matrix} 1 & \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{5}{2} & 2 & \frac{3}{2} & 1 \\ 4 & \frac{7}{2} & 3 & \frac{5}{2} \end{matrix}]

Algebra of Matrices Exercise 4.1 Question 7 (i)

Answer:

A = [\begin{matrix} 3 & \frac{5}{2} & \frac{7}{3} \\ 6 & 5 & \frac{14}{3} \\ 9 & \frac{15}{2} & 7 \\ 12 & 10 & \frac{28}{3} \end{matrix}]

Given:

a_{i j} = 2 i + \frac{i}{j}

Here we have to construct

4 \times 3

matrix according to

a_{i j} = 2 i + \frac{i}{j}

Hint: First we will find all the elements of matrix according to

2 i + \frac{i}{j}

Solution: Here

a_{i j} = 2 i + \frac{i}{j}

Let

A = {[a_{i j}]}_{4 \times 3}

So, The elements in a

4 \times 3

matrix are

a_{11}, a_{21}, a_{31}, a_{41}, a_{12}, a_{22}, a_{32}, a_{42}, a_{13}, a_{23}, a_{33}, a_{43}

A = {[\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ a_{41} & a_{42} & a_{43} \end{matrix}]}_{4 \times 3}

a_{11} = 2 \times 1 + \frac{1}{1} = 2 + 1 = 3 a_{12} = 2 \times 1 + \frac{1}{2} = 2 + \frac{1}{2} = \frac{5}{2} a_{13} = 2 \times 1 + \frac{1}{3} = 2 + \frac{1}{3} = \frac{7}{2}

a_{21} = 2 \times 2 + \frac{2}{1} = 4 + 2 = 6 a_{22} = 2 \times 2 + \frac{2}{2} = 4 + 1 = 5 a_{23} = 2 \times 2 + \frac{2}{3} = 4 + \frac{2}{3} = \frac{14}{3}

a_{31} = 2 \times 3 + \frac{3}{1} = 6 + 3 = 9 a_{32} = 2 \times 3 + \frac{3}{2} = 6 + \frac{3}{2} = \frac{15}{2} a_{33} = 2 \times 3 + \frac{3}{3} = 6 + 1 = 7

a_{41} = 2 \times 4 + \frac{4}{1} = 8 + 4 = 12 a_{42} = 2 \times 4 + \frac{4}{2} = 8 + 2 = 10 a_{43} = 2 \times 4 + \frac{4}{3} = 8 + \frac{4}{3} = \frac{28}{3}

Substituting these values in Matrix

A

, we get

A = [\begin{matrix} 3 & \frac{5}{2} & \frac{7}{3} \\ 6 & 5 & \frac{14}{3} \\ 9 & \frac{15}{2} & 7 \\ 12 & 10 & \frac{28}{3} \end{matrix}]

Algebra of Matrices Exercise 4.1 Question 7 (ii)

Answer:

A = [\begin{matrix} 0 & - \frac{1}{3} & - \frac{1}{2} \\ \frac{1}{3} & 0 & - \frac{1}{5} \\ \frac{1}{2} & \frac{1}{5} & 0 \\ \frac{3}{5} & \frac{1}{3} & \frac{1}{7} \end{matrix}]

Given: Here

a_{i j} = \frac{i - j}{i + j}

Here we have to construct

4 \times 3

matrix according to

\frac{i - j}{i + j}

Hint: First we will find all the elements of matrix according to

\frac{i - j}{i + j}

Solution: Here

a_{i j} = \frac{i - j}{i + j}

Let

A = {[a_{i j}]}_{4 \times 3}

So, the elements in a

4 \times 3

matrix are

a_{11}, a_{21}, a_{31}, a_{41}, a_{12}, a_{22}, a_{32}, a_{42}, a_{13}, a_{23}, a_{33}, a_{43}

A = {[\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ a_{41} & a_{42} & a_{43} \end{matrix}]}_{4 \times 3}

a_{11} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0 a_{12} = \frac{1 - 2}{1 + 2} = - \frac{1}{3} a_{13} = \frac{1 - 3}{1 + 3} = - \frac{2}{4} = \frac{1}{2}

a_{21} = \frac{2 - 1}{2 + 1} = \frac{1}{3} a_{22} = \frac{2 - 2}{2 + 2} = \frac{0}{4} = 0 a_{23} = \frac{2 - 3}{2 + 3} = - \frac{1}{5}

a_{31} = \frac{3 - 1}{3 + 1} = \frac{2}{4} = \frac{1}{2} a_{32} = \frac{3 - 2}{3 + 2} = \frac{1}{5} a_{33} = \frac{3 - 3}{3 + 3} = \frac{0}{6} = 0

a_{41} = \frac{4 - 1}{4 + 1} = \frac{3}{5} a_{42} = \frac{4 - 2}{4 + 2} = \frac{2}{6} = \frac{1}{3} a_{43} = \frac{4 - 3}{4 + 3} = \frac{1}{7}

Substituting these values in Matrix

A

, we get

A = [\begin{matrix} 0 & - \frac{1}{3} & - \frac{1}{2} \\ \frac{1}{3} & 0 & - \frac{1}{5} \\ \frac{1}{2} & \frac{1}{5} & 0 \\ \frac{3}{5} & \frac{1}{3} & \frac{1}{7} \end{matrix}]

Algebra of Matrices Exercise 4.1 Question 7 (iii)

Answer:

A = [\begin{matrix} 1 & 1 & 1 \\ 2 & 2 & 2 \\ 3 & 3 & 3 \\ 4 & 4 & 4 \end{matrix}]

Given:

a_{i j} = i

Here we have to construct

4 \times 3

matrix according to

a_{i j} = i

Hint: First we will find all the elements of matrix according to

a_{i j} = i

Solution: Here

a_{i j} = i

Let

A = {[a_{i j}]}_{4 \times 3}

So, The elements in a

4 \times 3

matrix are

a_{11}, a_{21}, a_{31}, a_{41}, a_{12}, a_{22}, a_{32}, a_{42}, a_{13}, a_{23}, a_{33}, a_{43}

A = {[\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ a_{41} & a_{42} & a_{43} \end{matrix}]}_{4 \times 3}

a_{11} = 1 a_{12} = 1 a_{13} = 1

a_{21} = 2 a_{22} = 2 a_{23} = 2

a_{31} = 3 a_{32} = 3 a_{33} = 3

a_{41} = 4 a_{42} = 4 a_{43} = 4

Substituting these values in Matrix

A

, we get

A = [\begin{matrix} 1 & 1 & 1 \\ 2 & 2 & 2 \\ 3 & 3 & 3 \\ 4 & 4 & 4 \end{matrix}]

Algebra of Matrices Exercise 4.1 Question 8

Answer: $a = 0, b = 5, x = 2$ and $y = - 1$
Given: Here given that

$[\begin{matrix} 3 x + 4 y & 2 & x - 2 y \\ a + b & 2 a - b & - 1 \end{matrix}]$ $= [\begin{matrix} 2 & 2 & 4 \\ 5 & - 5 & - 1 \end{matrix}]$

We have to find the value of $a, b, x$ and $y$
Hint: If two matrices are equal then the elements of each matrix are also equal.
Solution: Given that two matrices are equal
$∴$ By equating them, we get
$3 x + 4 y$ ……(i)
$x - 2 y$ ……(ii)
$a + b = 5$ ……. (iii)
$2 a - b = - 5$ …….. (iv)
Multiplying equation (ii) by 2 and adding to equation (i), we get
$3 x + 4 y + 2 x - 4 y = 2 + 8 \Rightarrow 5 x = 10 \Rightarrow x = 2$
Now substituting the value of $x$ in eqn (i), we get
$3 \times 2 + 4 y = 2 \Rightarrow 6 + 4 y = 2 \Rightarrow 4 y = - 4 \Rightarrow y = - 1$
Now by adding eqn(iii) and eqn (iv)
$a + b + 2 a - b = 5 + (- 5)$
$\Rightarrow 3 a = 5 - 5$
$\Rightarrow a = 0$
Now, again substituting the value of $a$ in eqn(iii), we get
$a + b = 5$
$\Rightarrow b = 5$
Hence, $a = 0, b = 5, x = 2$ and $y = - 1$

Algebra of Matrices Exercise 4.1 Question 10

Answer:

a = 1, b = 2, c = 3, d = 4

Given:

[\begin{matrix} 2 a + b & a - 2 b \\ 5 c - d & 4 c + 3 d \end{matrix}]

= [\begin{matrix} 4 & - 3 \\ 11 & 24 \end{matrix}]

Here we have to find out the values of

a, b, c

and

d

.
Hint: If two matrices are equal then the elements of each matrix are also equal.
Solution:
Given that two matrices are equal

∴

By equating them, we get

2 a + b = 4

….. (i)

a - 2 b = - 3

……(ii)

5 c - d = 11

……(iii)
and

4 c + 3 d = 24

……(iv)
Multiplying eqn(i) by 2 and adding to eqn(ii) we get

4 a + 2 b + a - 2 b = 8 - 3

\Rightarrow 5 a = 5

\Rightarrow a = 1

Now, substituting the value of a in eqn(i)

2 \times 1 + b = 4

\Rightarrow b = 4 - 2

\Rightarrow b = 2

Multiplying eqn(iii) by 3 and adding to eqn(iv) we get

15 c - 3 d + 4 c + 3 d = 33 + 24 \Rightarrow 19 c = 57 \Rightarrow c = 3

Now, substituting the value of c in eqn(iv) we get

4 \times 3 + 3 d = 24 \Rightarrow 12 + 3 d = 24 \Rightarrow 3 d = 24 - 12 \Rightarrow d = \frac{12}{3} \Rightarrow d = 4

Hence

a = 1, b = 2, c = 3, d = 4

Algebra of Matrices Exercise 4.1 Question 11

Answer:

x = 11, y = 9, z = 3

Given:

A = B

[\begin{matrix} x - 2 & 3 & 2 z \\ 18 z & y + 2 & 6 z \end{matrix}]

= [\begin{matrix} y & z & 6 \\ 6 y & x & 2 y \end{matrix}]

Hint: If

A = B \Rightarrow

If two matrices are equal then the elements of each matrix are also equal.
Solution: Here

A = B

[\begin{matrix} x - 2 & 3 & 2 z \\ 18 z & y + 2 & 6 z \end{matrix}]

= [\begin{matrix} y & z & 6 \\ 6 y & x & 2 y \end{matrix}]

Since corresponding entries of equal matrices are equal, So

x - 2 = y

….. (i)

3 = z

….. (ii)

2 z = 6

….. (iii)

18 z = 6 y

….. (iv)

y + 2 = x

….. (v)

6 z = 2 y

….. (vi)
Equation (ii) gives

z = 3

Putting the value of z in eqn(iv) we get

18 \times 3 = 6 y \Rightarrow y = 9

Putting the value of

y

in eqn(v) we get

y + 2 = x \Rightarrow 9 + 2 = x \Rightarrow x = 11

Hence

x = 11, y = 9, z = 3

Algebra of Matrices Exercise 4.1 Question 12

Answer:

x = 3, y = 7, z = - 2, w = 14

Given:

[\begin{matrix} x & 3 x - y \\ 2 x + z & 3 y - w \end{matrix}]

= [\begin{matrix} 3 & 2 \\ 4 & 7 \end{matrix}]

We have to find the value of

x, y, z

and

w

Hint: If two matrices are equal then the elements of each matrix are also equal.
Solution: Let

[\begin{matrix} x & 3 x - y \\ 2 x + z & 3 y - w \end{matrix}]

= [\begin{matrix} 3 & 2 \\ 4 & 7 \end{matrix}]

Since corresponding entries of equal matrices are equal, So

x = 3

….. (i)

3 x - y = 2

….. (ii)

2 x + z = 4

….. (iii)

3 y - w = 7

….. (iv)
Put the value of

x = 3

in eqn(ii) we get

3 \times 3 - y = 2 \Rightarrow y = 9 - 2 \Rightarrow y = 7

Put the value of

y = 7

in eqn(iv) we get

3 y - w = 7 \Rightarrow 3 \times 7 - w = 7 \Rightarrow w = 21 - 7 \Rightarrow w = 14

Put the value of

x = 3

in eqn(iii) we get

2 x + z = 4 \Rightarrow 2 \times 3 + z = 4 \Rightarrow z = 4 - 6 \Rightarrow z = - 2

Hence,

x = 3, y = 7, z = - 2, w = 14

Algebra of Matrices Exercise 4.1 Question 13

Answer:

x = 1, y = 2, z = 4

and

w = 5

Given:

[\begin{matrix} x - y & z \\ 2 x - y & w \end{matrix}]

= [\begin{matrix} - 1 & 4 \\ 0 & 5 \end{matrix}]

We have to find the value of

x, y, z

and

w

Hint: If two matrices are equal then the elements of each matrix are also equal.
Solution: Here

[\begin{matrix} x - y & z \\ 2 x - y & w \end{matrix}]

= [\begin{matrix} - 1 & 4 \\ 0 & 5 \end{matrix}]

Since corresponding entries of equal matrices are equal, So

x - y = - 1

….. (i)

z = 4

….. (ii)

2 x - y = 0

….. (iii)

w = 5

….. (iv)
Solving eqn(i) and (iii), we get

x - y = - 1 2 x - y = 0 - x = - 1

Putting the value of

x = 1

in eqn (i), we get

1 - y = - 1 \Rightarrow y = 1 + 1 \Rightarrow y = 2

Equation (ii) and (iv) gives the values of

z

and

w

respectively. So,

z = 4, w = 5

Hence,

x = 1, y = 2, z = 4

and

w = 5

Algebra of Matrices Exercise 4.1 Question 14

Answer:

x = - 3, y = - 5, z = 2, a = - 2, b = - 7, c = - 1

Given:

[\begin{matrix} x + 3 & z + 4 & 2 y - 7 \\ 4 x + 6 & a - 1 & 0 \\ b - 3 & 3 b & z + 2 c \end{matrix}]

= [\begin{matrix} 0 & 6 & 3 y - 2 \\ 2 x & - 3 & 2 c + 2 \\ 2 b + 4 & - 21 & 0 \end{matrix}]

Here we have to find out all the values of

x, y, z, a, b, c

Hint: By definition of equal matrices is

A = {[a_{i j}]}_{m \times n}

and

B = {[b_{i j}]}_{m \times n}

are equal then

a_{i j} = b_{i j}

for

i = 1, 2, 3. . . . m

and

j = 1, 2, 3. . . . n

Solution: Given that

[\begin{matrix} x + 3 & z + 4 & 2 y - 7 \\ 4 x + 6 & a - 1 & 0 \\ b - 3 & 3 b & z + 2 c \end{matrix}]

= [\begin{matrix} 0 & 6 & 3 y - 2 \\ 2 x & - 3 & 2 c + 2 \\ 2 b + 4 & - 21 & 0 \end{matrix}]

Equating the entries, we get

x + 3 = 0 \Rightarrow x = - 3

;

z + 4 = 6 \Rightarrow z = 2

;

2 y - 7 = 3 y - 2 \Rightarrow 2 y - 3 y = - 2 + 7 \Rightarrow - y = + 5 \Rightarrow y = - 5

Similarly,

a - 1 = - 3

and

z + 2 c = 0

\Rightarrow a = - 3 + 1

and

2 + 2 c = 0

\Rightarrow a = - 2

and

c = - 1

Lastly,

b - 3 = 2 b + 4 \Rightarrow b - 2 b = 4 + 3 \Rightarrow - b = 7 \Rightarrow b = - 7

Hence,

x = - 3, y = - 5, z = 2, a = - 2, b = - 7, c = - 1

Algebra of Matrices Exercise 4.1 Question 15

Answer:

x + y = 7

- 3

Given: Given that

[\begin{matrix} 2 x + 1 & 5 x \\ 0 & y^{2} + 1 \end{matrix}]

= [\begin{matrix} x + 3 & 10 \\ 0 & 26 \end{matrix}]

Here we have to calculate the value of

x + y

Hint: Here we will use equality of matrices.
Solution: Here

[\begin{matrix} 2 x + 1 & 5 x \\ 0 & y^{2} + 1 \end{matrix}]

= [\begin{matrix} x + 3 & 10 \\ 0 & 26 \end{matrix}]

The corresponding entries of equal matrices are equal, So

2 x + 1 = x + 3 \Rightarrow 2 x - x = 3 - 1 \Rightarrow x = 2

y^{2} + 1 = 26 \Rightarrow y^{2} = 26 - 1 \Rightarrow y = \sqrt{25} \Rightarrow y = \pm 5

Case 1: If

x = 2

and

y = + 5

\Rightarrow x + y = 2 + 5 = 7

Case 2: If

x = 2

and

y = - 5

\Rightarrow x + y = 2 + (- 5) = - 3

Hence

x + y = 7

- 3

Algebra of Matrices Exercise 4.1 Question 16

Answer:

x = 2 o r 4 y = 4 o r 2 z = - 6 w = 4

Given:

[\begin{matrix} x y & 4 \\ z + 6 & x + y \end{matrix}] = [\begin{matrix} 8 & w \\ 0 & 6 \end{matrix}]

Here we have to find the value of

x, y, z, w

Hint: Here we will use equality of matrices.
Solution:

[\begin{matrix} x y & 4 \\ z + 6 & x + y \end{matrix}] = [\begin{matrix} 8 & w \\ 0 & 6 \end{matrix}]

The corresponding entries of equal matrices are equal, So

x y = 8

….. (i)

w = 4

….. (ii)

z + 6 = 0

….. (iii)

x + y = 6

….. (iv)
From equation (ii) and (iii) we get

z = - 6

and

w = 4

From eqn (iv) we have,

x + y = 6 \Rightarrow x = 6 - y

Substituting the value of

x

in eqn (i), we get

(6 - y) y = 8 \Rightarrow y^{2} - 6 y + 8 = 0 \Rightarrow (y - 2) (y - 4) = 0 \Rightarrow y = 2 o r 4

Substituting the value of

y

in eqn (i), we get

x = 4, 2

Hence value of

x, y, z, w

are

4, 2; 2, 4; - 6

and

4

respectively

Algebra of Matrices Exercise 4.1 Question 17 (i)

Answer:

{[a]}_{1 \times 1}

Given: Here we have to give an example of row matrix which is also a column matrix
Hint:
Solution: We know that
Order of row matrix =

1 \times n

Order of column matrix =

m \times 1

So, order of a row as column matrix =

1 \times 1

Hence required matrix =

{[a]}_{1 \times 1}

Algebra of Matrices Exercise 4.1 Question 17 (ii)

Answer:

[\begin{matrix} 4 & 0 & 0 \\ 0 & - 3 & 0 \\ 0 & 0 & 2 \end{matrix}]

Here we have to give an example of diagonal matrix which is not scalar
Given:
Hint: A diagonal matrix has only

a_{11} a_{22} a_{33}

for a

3 \times 3

matrix such that

a_{11} a_{22} a_{33}

are equal or different and all other entries zero
Solution: We know that a diagonal matrix has only

a_{11} a_{22} a_{33}

for a

3 \times 3

matrix such that

a_{11} a_{22} a_{33}

are equal or different and all other entries zero. While scalar matrix has

a_{11} = a_{22} = a_{33} = m

(say)
So, a diagonal matrix which is not scalar must have

a_{11} \neq a_{22} \neq a_{33}

and

a_{i j} = 0

for

i \neq j

.
Hence Required matrix =

[\begin{matrix} 4 & 0 & 0 \\ 0 & - 3 & 0 \\ 0 & 0 & 2 \end{matrix}]

Algebra of Matrices Exercise 4.1 Question 17 (iii)

Answer:

[\begin{matrix} 3 & 2 & - 1 \\ 0 & 4 & 3 \\ 0 & 0 & - 6 \end{matrix}]

Here we have to create an example of triangular matrix.
Given:
Hint: A triangular matrix is a square matrix
Solution: We know that a triangular matrix is a square matrix

A = [a_{i j}]

such that

a_{i j} = 0

for

i > j

Hence required matrix

Algebra of Matrices Exercise 4.1 Question 18

Answer:

A = [\begin{matrix} 5 & 3 & 4 \\ 7 & 2 & 3 \end{matrix}]

and

B = [\begin{matrix} 8 & 7 & 6 \\ 10 & 5 & 7 \end{matrix}]

Given: Here the sales figure of two car dealers during January 2013
Here we have to write

2 \times 3

summarizing sales data.
Hint: Simply summarize for dealer A and dealer B
Solution: Given data is
For January 2013:

\begin{matrix} D e l u x & P r e m i u m & S t a d a r d \\ d e a l e r A & 5 & 3 & 4 \\ d e a l e r B & 7 & 2 & 3 \end{matrix}

Hence,

A = [\begin{matrix} D e l u x & P r e m i u m & S t a n d a r d \\ 5 & 3 & 4 \\ 7 & 2 & 3 \end{matrix}]

And For January – February:

\begin{matrix} D e l u x e & P r e m i u m & S t a d a r d \\ d e a l e r A & 8 & 7 & 6 \\ d e a l e r B & 10 & 5 & 7 \end{matrix}

Hence,

B = [\begin{matrix} D e l u x e & P r e m i u m & S t a d a r d \\ d e a l e r A & 8 & 7 & 6 \\ d e a l e r B & 10 & 5 & 7 \end{matrix}]

Algebra of Matrices Exercise 4.1 Question 19

Answer: $A$ and $B$ are not equal for any value of $y$

Given: $A = [\begin{matrix} 2 x + 1 & 2 y \\ 0 & y^{2} + 2 \end{matrix}]$ and

$B = [\begin{matrix} x + 3 & y^{2} + 2 \\ 0 & - 6 \end{matrix}]$

We have to find out the value of $x$ and $y$
Hint: We will use equality of matrices.
Solution: Here $A = B$
Since equal matrix have all corresponding entries equal. So,

$2 x + 1 = x + 3$ ….. (i)

$2 y = y^{2} + 2$ ….. (ii)

$y 2 - 5 y = - 6$ ….. (iii)

Solving equation (i), We get

$2 x + 1 = x + 3 \Rightarrow 2 x - x = 3 - 1 \Rightarrow x = 2$

Solving equation (ii), We get

$2 y = y^{2} + 2 \Rightarrow y^{2} - 2 y + 2 = 0 \Rightarrow D = b^{2} - 4 a c$

$= {(- 2)}^{2} - 4 (1) (2) = 4 - 8 = - 4$

Here $D < 0$
So, there is no real value of $y$ from equation (ii)
Solving equation (iii), We get

$y^{2} 5 y = - 6 \Rightarrow y^{2} - 5 y + 6 = 0 (y - 3) (y - 2) = 0 y = 3 o r 2$

From solution of equation (i) (ii) and (iii)
we can say that $A$ and $B$ cannot equal for any value of $y$

Algebra of Matrices Exercise 4.1 Question 20

Answer:

x = 3, y = 1

Given:

[\begin{matrix} x + 10 & y^{2} + 2 y \\ 0 & - 4 \end{matrix}]

= [\begin{matrix} 3 x + 4 & 3 \\ 0 & y^{2} - 5 y \end{matrix}]

We have to find out the value of

x

and

y

Hint: We will use equality of matrices.
Solution: Here

[\begin{matrix} x + 10 & y^{2} + 2 y \\ 0 & - 4 \end{matrix}]

= [\begin{matrix} 3 x + 4 & 3 \\ 0 & y^{2} - 5 y \end{matrix}]

Since, corresponding entries of equal matrices are equal. So,

x + 10 = 3 x + 4

….. (i)

y^{2} + 2 y = 3

….. (ii)

- 4 = y^{2} - 5 y

….. (iii)
Solving equation (i), We get

x + 10 = 3 x + 4 \Rightarrow 2 x = 6 \Rightarrow x = 3

Solving equation (ii), We get

y^{2} + 2 y - 3 = 0 \Rightarrow y + 3 y - y - 3 = 0 \Rightarrow y (y + 3) - 1 (y + 3) = 0 \Rightarrow y = 1 o r - 3

Solving equation (iii), We get

- 4 = y^{2} - 5 y \Rightarrow y^{2} - 5 y + 4 = 0 \Rightarrow y^{2} - 4 y - y + 4 = 0 \Rightarrow y (y - 1) - 1 ((y - 4) = 0 \Rightarrow (y - 1) (y - 4) = 0 \Rightarrow y = 1 o r 4

From equation (ii) and (iii)
We have common value of

y = 1

So,

x = 3, y = 1

Algebra of Matrices Exercise 4.1 Question 21

Answer:

A = B

Given:

A = [\begin{matrix} a + 4 & 3 b \\ 8 & - 6 \end{matrix}], B [\begin{matrix} 2 a + 2 & b^{2} + 2 \\ 8 & b^{2} - 10 \end{matrix}]

We have to find out the value of

a

and

b

Hint: We will use equality of matrices.
Solution:

A = [\begin{matrix} a + 4 & 3 b \\ 8 & - 6 \end{matrix}], B [\begin{matrix} 2 a + 2 & b^{2} + 2 \\ 8 & b^{2} - 10 \end{matrix}]

∵ A = B

Corresponding elements of two equal matrix are equal.

a + 4 = 2 a + 2

….. (i)

3 b = b^{2} + 2

….. (ii)

b^{2} - 10 = - 6

….. (iii)
Solving equation (i), We get

a + 4 = 2 a + 2 \Rightarrow 2 a - a = 4 - 2 \Rightarrow a = 2

Solving equation (ii), We get

b^{2} - 3 b + 2 = 0 \Rightarrow b^{2} - 2 b - b + 2 = 0 \Rightarrow b (b - 2) - 1 (b - 2) = 0 \Rightarrow (b - 1) (b - 2) = 0 \Rightarrow= 1 o r 2

Solving equation (iii), We get

b^{2} - 10 = - 6 \Rightarrow b^{2} = - 6 + 10 \Rightarrow b^{2} = 4 \Rightarrow b = \pm 2

Here, we have common value of

b = 2

from equation (ii) and (iii)
Hence,

a = 2, b = 2

The class 12 RD Sharma chapter 4 exercise 4.1 solution contains the chapter Algebra of Matrices which explores Order of the matrix, Matrix formation, addition, subtraction, multiplication, etc., of matrices along with types of Matrices like null matrices, diagonal Matrices, and triangular matrices. Exercise 4.1 contains 39 questions including subparts, on two levels based on these concepts.

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The RD Sharma class 12th exercise 4.1 Solutions can be availed online from Career360, which has RD Sharma solutions for all subjects. They can be downloaded free of cost from their website.

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Chapter-wise RD Sharma Class 12 Solutions

Frequently Asked Questions (FAQs)

1. What are the benefits of using RD Sharma class 12th exercise 4.1 Solutions?

RD Sharma class 12th exercise 4.1 Solutions can be extremely helpful to students who will be appearing for their board exams. In addition, students can use these answers to test their knowledge at home and develop their math skills.

2. What does the 4th chapter of the Class 12 Mathematics book contain?

The 4th chapter of the book contains advanced concepts and problems on matrices. For example, there will be sums on addition, subtraction, division, and multiplication of matrices. You will also explore types of Matrices like skew-symmetric matrix, null matrix, symmetric matrix, etc.

3. Who can use class 12 RD Sharma chapter 4 exercise 4.1 solution?

The class 12 RD Sharma chapter 4 exercise 4.1 solution can be used by students, teachers, and parents. Students can use these solutions to practice at home. Teachers can use RD Sharma class 12 chapter 4 exercise 4.1 to give homework to students and mark their performance. Parents can also similarly use these books to help their children test themselves at home.

4. Where can I find the class 12 RD Sharma chapter 4 exercise 4.1 solution book?

You will be able to download the class 12 RD Sharma chapter 4 exercise 4.1 solution book from Career360 at no cost.

5. Does RD Sharma class 12 solutions Algebra of Matrices ex 4.1 have the updated syllabus?

The RD Sharma class 12 solutions Algebra of Matrices ex 4.1 books will have an updated syllabus if you download the newest version of their free pdf.

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