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**NCERT mathematics class 12 chapter Matrices Exercise 3.1 Solutions**. You can find the numerical questions solutions for the**chapter 3/Exercise 3.1**of**NCERT class 12 mathematics**in this page. So is the case if you are looking for**NCERT class 12 Maths**related topic**Matrices Exercise 3.1**solutions. This page contains Exercise 3.1 solutions. If you’re looking for summary of the chapter or other exercise solutions, they’re available at●

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Exercise 3.1 Solutions

1. In the matrix {A = \left[\begin{array}{cccc} 2 & 5 & 19 & -7 \\[5pt] 35 & -2 & \dfrac{5}{2} & 12 \\[10pt] \sqrt{3} & 1 & -5 & 17 \end{array} \right]}, write:

(i) The order of the matrix

(ii) The number of elements,

(iii) Write the elements a_{13}, a_{21}, a_{33}, a_{24}, a_{23}.

(i)

The given matrix A has 3 rows and 4 columns.

So, the order of the matrix A is 3 × 4

(ii)

The number of elements in the given matrix A are 3 × 4 = 12

(iii)

The following are the elements at the given positions in the matrix A

Element

Value

a_{13}

19

a_{21}

35

a_{33}

17

a_{24}

12

a_{23}

\dfrac{5}{2}

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

The order of the matrix containing 24 elements would be m × n with all possible values of m and n as follows.

It can be noted that the order of the matrix can be obtained by finding all those two factors of 24 whose product would yield 24 as follows:

24

=

1 × 24

=

2 × 12

=

3 × 8

=

4 × 6

=

6 × 4

=

8 × 3

=

12 × 2

=

24 × 1

So, the number of possible orders for a matrix that has 24 elements is 8.

Now, if we consider a matrix that has 13 elements, the number of possile orders would be as follows:

13

=

1 × 13

=

13 × 1

So, the number of possible orders for a matrix that has 13 elements is 2.

Note: Any matrix whose number of elements is a prime number would have only 2 possible orders. It can also be noted that these orders would be a row matrix and a column matrix.

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

The order of the matrix containing 18 elements would be m × n with all possible values of m and n as follows.

It can be noted that the order of the matrix can be obtained by finding all those two factors of 18 whose product would yield 18 as follows:

18

=

1 × 18

=

2 × 9

=

3 × 6

=

6 × 3

=

9 × 2

=

18 × 1

So, the number of possible orders for a matrix that has 18 elements is 6.

Now, if we consider a matrix that has 5 elements, the number of possile orders would be as follows:

5

=

1 × 5

=

5 × 1

So, the number of possible orders for a matrix that has 5 elements is 2.

Note: Any matrix whose number of elements is a prime number would have only 2 possible orders. It can also be noted that these orders would be a row matrix and a column matrix.

4. Construct a 2 × 2 matrix, {\text{A} = [a_{ij}]}, whose elements are given by:

(i) {a_{ij} = \dfrac{(i + j)^2}{2}} (ii) {a_{ij} = \dfrac{i}{j}} (iii) {a_{ij} = \dfrac{(i + 2j)^2}{2}}

Any 2 x 2 matrix {\text{A} = [a_{ij}]} can be expressed as

{\text{A} = \left[ \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right]}

(i) To construct A when {a_{ij} = \dfrac{(i + j)^2}{2}}

i

j

a_{ij}

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

1

1

a_{11}

{\dfrac{(1 + 1)^2}{2} = \dfrac{2^2}{2} = 2}

1

2

a_{12}

{\dfrac{(1 + 2)^2}{2} = \dfrac{3^2}{2} = \dfrac{9}{2}}

2

1

a_{21}

{\dfrac{(2 + 1)^2}{2} = \dfrac{3^2}{2} = \dfrac{9}{2}}

2

2

a_{22}

{\dfrac{(2 + 2)^2}{2} = \dfrac{4^2}{2} = 8}

So, the constructed matrix will be

{\text{A} = \left[ \begin{array}{cc} a_{11} & a_{12} \\[5pt] a_{21} & a_{22} \end{array} \right] =} {\left[ \begin{array}{cc} 2 & \dfrac{9}{2} \\[10pt] \dfrac{9}{2} & 8 \end{array} \right]}

(ii) To construct A when {a_{ij} = \dfrac{i}{j}}

i

j

a_{ij}

{\dfrac{i}{j}}

1

1

a_{11}

{\dfrac{1}{1} = 1}

1

2

a_{12}

{\dfrac{1}{2} = \dfrac{1}{2}}

2

1

a_{21}

{\dfrac{2}{1} = 2}

2

2

a_{22}

{\dfrac{2}{2} = 1}

So, the constructed matrix will be

{\text{A} = \left[ \begin{array}{cc} a_{11} & a_{12} \\[5pt] a_{21} & a_{22} \end{array} \right] =} {\left[ \begin{array}{cc} 1 & \dfrac{1}{2} \\[10pt] 2 & 1 \end{array} \right]}

(iii) To construct A when {a_{ij} = \dfrac{(i + 2j)^2}{2}}

i

j

a_{ij}

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

1

1

a_{11}

{\dfrac{(1 + (2 × 1))^2}{2} = \dfrac{3^2}{2} = \dfrac{9}{2}}

1

2

a_{12}

{\dfrac{(1 + (2 × 2))^2}{2} = \dfrac{5^2}{2} = \dfrac{25}{2}}

2

1

a_{21}

{\dfrac{(2 + (2 × 1))^2}{2} = \dfrac{16^2}{2} = 8}

2

2

a_{22}

{\dfrac{(2 + (2 × 2))^2}{2} = \dfrac{6^2}{2} = \dfrac{36}{2} = 18}

So, the constructed matrix will be

{\text{A} = \left[ \begin{array}{cc} a_{11} & a_{12} \\[5pt] a_{21} & a_{22} \end{array} \right] =} {\left[ \begin{array}{cc} \dfrac{9}{2} & \dfrac{25}{2} \\[10pt] 8 & 18 \end{array} \right]}

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

(i) {a_{ij} = \dfrac{1}{2}|-3i + j|} (ii) {a_{ij} = 2i - j}

Any 3 x 4 matrix {\text{A} = [a_{ij}]} can be expressed as

{\text{A} = \left[ \begin{array}{cccc} a_{11} & a_{12} & a_{13} & a_{14} \\[5pt] a_{21} & a_{22} & a_{23} & a_{24} \\[5pt] a_{31} & a_{32} & a_{33} & a_{34} \end{array} \right]}

(i) To construct a matrix when {a_{ij} = \dfrac{1}{2}|-3i + j|}

i

j

a_{ij}

{\dfrac{1}{2}\Big|-3i + j\Big|}

1

1

a_{11}

{\dfrac{1}{2}\Big|(-3 × 1) + 1\Big| = \dfrac{1}{2}\Big|-3 + 1\Big| = \dfrac{1}{2} × 2 = 1}

1

2

a_{12}

{\dfrac{1}{2}\Big|(-3 × 1) + 2\Big| = \dfrac{1}{2}\Big|-3 + 2\Big| = \dfrac{1}{2} × 1 = \dfrac{1}{2}}

1

3

a_{13}

{\dfrac{1}{2}\Big|(-3 × 1) + 3\Big| = \dfrac{1}{2}\Big|-3 + 3\Big| = \dfrac{1}{2} × 0 = 0}

1

4

a_{14}

{\dfrac{1}{2}\Big|(-3 × 1) + 4\Big| = \dfrac{1}{2}\Big|-3 + 4\Big| = \dfrac{1}{2} × 1 = \dfrac{1}{2}}

2

1

a_{21}

{\dfrac{1}{2}\Big|(-3 × 2) + 1\Big| = \dfrac{1}{2}\Big|-6 + 1\Big| = \dfrac{1}{2} × 5 = \dfrac{5}{2}}

2

2

a_{22}

{\dfrac{1}{2}\Big|(-3 × 2) + 2\Big| = \dfrac{1}{2}\Big|-6 + 2\Big| = \dfrac{1}{2} × 4 = 2}

2

3

a_{23}

{\dfrac{1}{2}\Big|(-3 × 2) + 3\Big| = \dfrac{1}{2}\Big|-6 + 3\Big| = \dfrac{1}{2} × 3 = \dfrac{3}{2}}

2

4

a_{24}

{\dfrac{1}{2}\Big|(-3 × 2) + 4\Big| = \dfrac{1}{2}\Big|-6 + 4\Big| = \dfrac{1}{2} × 2 = 1}

3

1

a_{31}

{\dfrac{1}{2}\Big|(-3 × 3) + 1\Big| = \dfrac{1}{2}\Big|-9 + 1\Big| = \dfrac{1}{2} × 8 = 4}

3

2

a_{32}

{\dfrac{1}{2}\Big|(-3 × 3) + 2\Big| = \dfrac{1}{2}\Big|-9 + 2\Big| = \dfrac{1}{2} × 7 = \dfrac{7}{2}}

3

3

a_{33}

{\dfrac{1}{2}\Big|(-3 × 3) + 3\Big| = \dfrac{1}{2}\Big|-9 + 3\Big| = \dfrac{1}{2} × 6 = 3}

3

4

a_{34}

{\dfrac{1}{2}\Big|(-3 × 3) + 4\Big| = \dfrac{1}{2}\Big|-9 + 4\Big| = \dfrac{1}{2} × 5 = \dfrac{5}{2}}

So, the constructed matrix will be

{\text{A} = \left[ \begin{array}{cccc} a_{11} & a_{12} & a_{13} & a_{14} \\[5pt] a_{21} & a_{22} & a_{23} & a_{24} \\[5pt] a_{31} & a_{32} & a_{33} & a_{34} \end{array} \right] =} {\left[ \begin{array}{cccc} 1 & \dfrac{1}{2} & 0 & \dfrac{1}{2} \\[10pt] \dfrac{5}{2} & 2 & \dfrac{3}{2} & 1 \\[10pt] 4 & \dfrac{7}{2} & 3 & \dfrac{5}{2} \end{array} \right]}

(ii) To construct a matrix when {a_{ij} = 2i - j}

i

j

a_{ij}

{2i - j}

1

1

a_{11}

(2 × 1) – 1 = 2 – 1 = 1

1

2

a_{12}

(2 × 1) – 2 = 2 – 2 = 0

1

3

a_{13}

(2 × 1) – 3 = 2 – 3 = -1

1

4

a_{14}

(2 × 1) – 4 = 2 – 4 = -2

2

1

a_{21}

(2 × 2) – 1 = 4 – 1 = 3

2

2

a_{22}

(2 × 2) – 2 = 4 – 2 = 2

2

3

a_{23}

(2 × 2) – 3 = 4 – 3 = 1

2

4

a_{24}

(2 × 2) – 4 = 4 – 4 = 0

3

1

a_{31}

(2 × 3) – 1 = 6 – 1 = 5

3

2

a_{32}

(2 × 3) – 2 = 6 – 2 = 4

3

3

a_{33}

(2 × 3) – 3 = 6 – 3 = 3

3

4

a_{34}

(2 × 3) – 4 = 6 – 4 = 2

So, the constructed matrix will be

{\text{A} = \left[ \begin{array}{cccc} a_{11} & a_{12} & a_{13} & a_{14} \\[5pt] a_{21} & a_{22} & a_{23} & a_{24} \\[5pt] a_{31} & a_{32} & a_{33} & a_{34} \end{array} \right] =} {\left[ \begin{array}{cccc} 1 & 0 & -1 & -2 \\[5pt] 3 & 2 & 1 & 0 \\[5pt] 5 & 4 & 3 & 2 \end{array} \right]}

6. Find the values of x, y and z from the following equations:

(i) {\left[ \begin{array}{cc} 4 & 3 \\[5pt] x & 5 \end{array} \right] = \left[ \begin{array}{cc} y & z \\[5pt] 1 & 5 \end{array} \right]}

(ii) {\left[ \begin{array}{cc} x + y & 2 \\[5pt] 5 + z & xy \end{array} \right] = \left[ \begin{array}{cc} 6 & 2 \\[5pt] 5 & 8 \end{array} \right]}

(iii) {\left[ \begin{array}{c} x + y + z \\[5pt] x + z \\[5pt] y + z \end{array} \right] = \left[ \begin{array}{cc} 9 \\[5pt] 5 \\[5pt] 7 \end{array} \right]}

(i) To find the values of x, y and z from the equation {\left[ \begin{array}{cc} 4 & 3 \\[5pt] x & 5 \end{array} \right] = \left[ \begin{array}{cc} y & z \\[5pt] 1 & 5 \end{array} \right]}

As the given matrices are equal, therefore, their corresponding elements must be equal. Comparing the corresponding elements, we get

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

(ii) To find the values of x, y and z from the equation {\left[ \begin{array}{cc} x + y & 2 \\[5pt] 5 + z & xy \end{array} \right] = \left[ \begin{array}{cc} 6 & 2 \\[5pt] 5 & 8 \end{array} \right]}

As the given matrices are equal, therefore, their corresponding elements must be equal. Comparing the corresponding elements, we get

x + y

=

6 ———-❶

5 + z

=

5 ———-❷

xy

=

8 ———-❸

From equation ❸ we have {y = \dfrac{8}{x}}

Substituting this in equation ❶, we get

{x + \dfrac{8}{x} = 6}

{⇒ \dfrac{x^2 + 8}{x} = 6}

{⇒ x^2 + 8 = 6x}

{⇒ x^2 - 6x + 8 = 0}

{⇒ x^2 - 4x - 2x + 8 = 0}

{⇒ x(x - 4) - 2(x - 4) = 0}

{⇒ (x - 2)(x - 4) = 0}

{⇒ (x - 2) = 0} or {x - 4 = 0}

{⇒ x = 2} or {x = 4}

Now,

when {x = 2}

y

=

{\dfrac{8}{x}}

=

{\dfrac{8}{2}}

=

4

when {x = 4}

y

=

{\dfrac{8}{x}}

=

{\dfrac{8}{4}}

=

2

From equation ❷, we have

{z = 0}

So, the values of {x,} y and z are

{x = 2,} {y = 4} and {z = 0}

OR

{x = 4,} {y = 2} and {z = 0}

(iii) To find the values of x, y and z from the equation {\left[ \begin{array}{c} x + y + z \\[5pt] x + z \\[5pt] y + z \end{array} \right] = \left[ \begin{array}{cc} 9 \\[5pt] 5 \\[5pt] 7 \end{array} \right]}

As the given matrices are equal, therefore, their corresponding elements must be equal. Comparing the corresponding elements, we get

x + y + z

=

9 ———-❶

x + z

=

5 ———-❷

y + z

=

7 ———-❸

Substituting ❸ into ❶, we get

{x + 7 = 9}

{⇒ x = 2}

Substituting the value of x into ❷, we get

{2 + z = 5}

{⇒ z = 3}

Substituting the value of z into ❸, we get

{y + 3 = 7}

{⇒ y = 4}

So, the values of {x,} y and z are

{x = 2,} {y = 4} and {z = 3}

7. Find the value of {a,} {b,} c and d from the eqution:

{\left[\begin{array}{cc} a - b & 2a + c \\[5pt] 2a - b & 3c + d \end{array}\right] = \left[ \begin{array}{cc} -1 & 5 \\[5pt] 0 & 13 \end{array}\right]}

a - b

=

-1 ———-❶

2a + c

=

5 ———-❷

2a - b

=

0 ———-❸

3c + d

=

13 ———-❹

Subtracting equation ❶ from ❸, we get

{(2a - b) - (a - b) = 0 -(-1)}

{⇒ 2a - b - a + b = 0 + 1}

{⇒ a = 1}

Substituting {a = 1} into the equation ❶, we have

{1 - b = -1}

{⇒ -b = -1 - 1}

{⇒ -b = -2}

{⇒ b = 2}

Similarly, substituting {a = 1} into the equation ❷, we have

{(2 × 1) + c = 5}

{⇒ c = 5 - 2}

{⇒ c = 3}

Now, substituting {c = 3} into the equation ❹, we have

{(3 × 3) + d = 13}

{⇒ d = 13 - 9}

{⇒ d = 4}

So, the values of {a,} {b,} c and d are

{a = 1,} {b = 2,} {c = 3} and {d = 4}

8. {\text{A} = {[a_{ij}]}_{m × n}} is a square matrix, if

(A) {m \lt n}

(B) {m \gt n}

(C) {m = n} ✔

(D) None of these

The following is the definition of a square matrix.

Square Matrix: A matrix in which the number of rows are equal to the number of columns, is said to be square matrix. Thus an {m × n} matrix is said to be a square matrix if {m = n} and is known as a square matrix of the order {'n'}

This implies that option C is the correct answer.

9. Which of the given values of x and y make the following pair of matrices equal {\left[ \begin{array}{cc} 3x + 7 & 5 \\[5pt] y + 1 & 2 - 3x \end{array}\right],} {\left[ \begin{array}{cc} 0 & y - 2 \\[5pt] 8 & 4 \end{array}\right]}

(A) {x = \dfrac{-1}{3}, y = 7}

(B) Not possible to find ✔

(C) {y = 7, x = \dfrac{-2}{3}}

(D) {x = \dfrac{-1}{3}, y = \dfrac{-2}{3}}

For the given matrices to be equal, their corresponding elements must be equal. Comparing the corresponding elements, we get

3x + 7

=

0 ———-❶

y - 2

=

5 ———-❷

y + 1

=

8 ———-❸

2 - 3x

=

4 ———-❹

On solving all the equations,

On solving ❶, we get {x = -\dfrac{7}{3}}

On solving ❷, we get {y = 7}

On solving ❸, we get {y = 7}

On solving ❹, we get {x = -\dfrac{2}{3}}

The results obtained by equations ❷ and ❸ are consistent. However, the results obtained by solving both the equations ❶ and ❹ are not consistent.

So, it is not possible to find the values of both x and y which would make both the matrices equal.

Hence the option B which states that it is “Not possible to find” the values of x and y is the correct answer.

10. The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is:

(A) 27

(B) 18

(C) 81

(D) 512 ✔

The number of elements in a 3 × 3 matrix are 3 × 3 = 9

So, there are 9 positions to be filled in with either 0 or 1.

So, the number of possibilities are:

2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

=

{2^9}

=

512

So, option D is the correct answer.