# NCERT Solutions Class 12 maths Chapter-3 (matrices)

**NCERT Solutions Class 12 Maths**from class 12th Students will get the answers of

**Chapter-3 (matrices)Exercise 3.1**This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.

We have given the answers of all the questions of

**NCERT Board Mathematics Textbook**in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

**NCERT Question-Answer**

**Class 12 ****Mathematics**

**Chapter-3 ****(****matrices****)**

**Questions and answers given in practice**

**Chapter****-3 ****(matrices)**

### Exercise 3.1

**Question 1. In the matrix A= , write:**

**(i)The order of the matrix **

**Solution:**

We can see that matrix contains 3 rows and 4
columns So, the order of this matrix is 3×4

**(ii) The number of elements **

**Solution:**

We know that number of elements in the matrix =
product of number of rows and number of columns in matrix So, number of
elements = 3 x 4 =12.

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

**Solution:**

a_{13 }= Element in first row and third column i.e, 19_{ }

a_{21 }= Element in second row and first column i.e, 35_{ }

a_{33} = Element in third row and third column i.e, -5

a_{24} = Element in second row and fourth column i.e, 12

a_{23 }= Element in second row
and third column i.e, 5/2

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

**Solution:**

We know that number of
elements in the matrix is the product of number

of rows and number of
columns in the matrix .

If matrix has order mxn
then number of elements are mn in that matrix.

So we have to find the
ordered pairs of natural number whose product is 24.

The ordered pairs are:
(1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6), (6, 4)

Hence possible orders
are: 1×24, 24×1, 2×12, 12×2, 3×8, 8×3, 4×6, and 6×4

If matrix has 13
elements then ordered pairs will be (1, 13) and (13, 1)

Hence possible orders are: 1×13 and 13×1

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

**Solution:**

We know that number of
elements in the matrix is the product of number

of rows and number of
columns in the matrix .

If matrix has order mxn
then number of elements are mn in that matrix.

So we have to find the
ordered pairs of natural number whose product is 18.

The ordered pairs
are:(1, 18), (18, 1), (2, 9), (9, 2), (3, 6), and (6, 3)

Hence possible orders
are: 1×18, 18×1, 2×9, 9×2, 3×6, and 6×3

If matrix has 5 elements
then ordered pairs will be (1, 5) and (5, 1)

Hence possible orders are: 1×5 and 5×1

**Question
4. Construct a 2×2 matrix , A = [a _{ij}]
whose elements are given by :**

**(i)
a _{ij }= (i + j)^{2}**

**/2**

**Solution:**

Elements in this 2×2
matrix = a_{11 }, a_{12 },a_{21 },a_{22}

a_{11} => i = 1 and j = 1 => (1 + 1)^{2}/2 = 4/2 = 2

a_{12 }=> i = 1 and j = 2 => (1 + 2)^{2}/2 = 9/2

a_{21} => i = 2 and j = 1 => (2 + 1)^{2}/2 = 9/2

a_{22} => i = 2 and j = 2 =>(2 + 2)^{2}/2 = 16/2 = 8

Resultant Matrix is:

**(ii)
a**_{ij}** = i/j**

**Solution:**

Elements in this 2×2
matrix = a_{11} , a_{12} ,a_{21} ,a_{22}

a_{11} => i = 1 and j = 1 = 1/1 = 1

a_{12} => i = 1 and j = 2 = 1/2

a_{21} => i = 2 and j = 1 = 2/1 = 2

a_{22} => i = 2 and j = 2 = 2/2 = 1

Resultant Matrix is:

**(iii)
a**_{ij}** = (i + 2j)**^{2}**/2**

**Solution:**

Elements in this 2×2
matrix = a_{11 }, a_{12 }, a_{21 }, a_{22}

a_{11} => i = 1 and j = 1 => (1 + 2 x 1)^{2}/2 = 9/2

a_{12} => i = 1 and j = 2 => (1 + 2 x 2)^{2}/2 = 25/2

a_{21} => i = 2 and j = 1 =>(2 + 2 x 1)^{2}/2 = 16/2 = 8

a_{22} => i = 2 and j = 2 =>(2 + 2 x 2)^{2}/2 = 36/2 = 18

Resultant Matrix is:

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

**(i)
a**_{ij }**= 1/2 {|-3i + j|}**

**Solution:**

Elements in this 3 x 4
matrix are a_{11} , a_{12} , a_{13} , a_{14} , a_{21} , a_{22}, a_{23 }, a_{24 } , a_{31 }, a_{32} , a_{33} , a_{34}

a_{11} => i = 1 and j = 1 => 1/2 (|-3 x 1 + 1|) = 1

a_{12 }=> i = 1 and j = 2 => 1/2 (|-3 x 1 + 2|) = 1/2

a_{13} => i = 1 and j = 3 => 1/2 (|-3 x 1 + 3) = 0

a_{14} => i = 1 and j = 4 => 1/2 (|-3 x 1 + 4|) = 1/2

a_{21} => i = 2 and j = 1 => 1/2 (|-3 x 2 + 1|) = 5/2

a_{22} => i = 2 and j = 2 => 1/2 (|-3 x 2 + 2|) = 2

a_{23} => i = 2 and j = 3 => 1/2 (|-3 x 2 + 3|) = 3/2

a_{24 }=> i = 2 and j = 4 => 1/2 (|-3 x 2 + 4|) = 1

a_{31} => i = 3 and j = 1 => 1/2 (|-3 x 3 + 1|) = 4

a_{32} => i = 3 and j = 2 => 1/2 (|-3 x 3 + 2|) = 7/2

a_{33} => i = 3 and j = 3 => 1/2 (|-3 x 3 + 3|) = 3

a_{34} => i = 3 and j = 4 => 1/2 (|-3 x 3 + 4|) = 5/2

Resultant matrix is:

**(ii)
a**_{ij }**= 2i – j**

**Solution:**

Elements in this 3 x 4
matrix are a_{11} , a_{12} , a_{13} , a_{14} , a_{21} , a_{22} , a_{23} , a_{24} , a_{31} , a_{32} , a_{33} , a_{34}

So,

a_{11} => i = 1 and j = 1 => 2 x 1 – 1 = 1

a_{12 }=> i = 1 and j = 2 => 2 x 1 – 2 = 0

a_{13} => i = 1 and j = 3 => 2 x 1 – 3 = -1

a_{14} => i = 1 and j = 4 => 2 x 1 – 4 = -2

a_{21} => i = 2 and j = 1 => 2 x 2 – 1 = 3

a_{22} => i = 2 and j = 2 => 2 x 2 – 2 = 2

a_{23} => i = 2 and j = 3 => 2 x 2 – 3 = 1

a_{24 }=> i = 2 and j = 4 => 2 x 2 – 4 = 0

a_{31} => i = 3 and j = 1 => 2 x 3 – 1 = 5

a_{32} => i = 3 and j = 2 => 2 x 3 – 2 = 4

a_{33} => i = 3 and j = 3 => 2 x 3 – 3 = 3

a_{34} => i = 3 and j = 4 => 2 x 3 – 4 = 2

Resultant matrix is:

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

**(i) **

**Solution:**

We can compare or equate
both the matrices because both are equal

So on equating both the
matrices we get

x = 1; y = 4; z = 3

** (ii) **

**Solution:**

We can compare or equate
both the matrices because both are equal

So, on equating both the
matrices. we get

x + y = 6
-(1)

5 + z = 5
-(2)

xy = 8
-(3)

Now, we can solve these
equations

z = 0 from eq(2)

x = 6 – y
-(4)

Now putting value of x
from eq(4) in eq(3)

(6 – y)(y) = 8

6y – y^{2 }= 8

y^{2 }– 6y + 8 = 0 -(5)

Now we have to factorize
this equation

(y – 4)(y – 2) = 0

either y – 4 = 0 or y –
2 = 0

so, y = 2 or y = 4

Put these values in
eq(4) we get

x = 4 and x = 2

Therefore, the value of x = 2 , y = 4 , z = 0

**(iii) **

**Solution:**

We can compare or equate
both the matrices because both are equal

So, on equating both the
matrices, we get

x + y + z = 9
-(1)

x + z = 5
-(2)

y + z = 7
-(3)

If we put the value of
eq(2) in eq(1)

we get, 5 + y = 9

y = 4

On putting value of y in
eq(3)

4 + z = 7

z = 3

On putting value of z in
eq(2)

x + 3 = 5

x = 2

So, the value of x = 2; y = 4; z = 3

**Question 7. Find the value of a, b, c, and d from the equation:**

**Solution:**

We can compare or equate
both the matrices because both are equal

So, on equating both the
matrices, we get

a – b = -1
-(1)

2a – b = 0
-(2)

2a + c= 5
-(3)

3c + d = 13
-(4)

On solving eq(1) and
eq(2) we get

a = 1

On putting a = 1 in
eq(3) we get

c = 3

On putting a = 1 in
eq(2) we get

b = 2

On putting c = 3 in
eq(4) we get

d = 4

So, the value of a = 1; b = 2; c = 3; d = 4

**Question
8. ****A = [a**_{ij}**]**_{mxn }**is
a square matrix, if **

**(A) m < n (B) m > n
(C) m = n (D) None of these**

**Solution:**

This will be square
matrix if number of rows = number of columns

So , m = n is correct
option.

Hence, the option answer is C.

**Question 9. Which of the given values of x and y
make the following pair of **

**matrices equal**

**(A) x = -1/3, y = 7 (B) Not possible
to find (C) y = 7, x = -2/3 (D) x = -1/3, y = -2/3
**

**Solution:**

We can compare or equate
both the matrices because both are equal

So on equating both the
matrices we get;

3x + 7 = 0
-(1)

y + 1 = 8
-(2)

2 – 3x = 4
-(3)

y – 2 = 5
-(4)

From eq(2) and eq(4) we
get same value of y i.e, y=7

but on solving eq(1) we
get value of x = -7/3 and on solving eq(3) we get value of x = -2/3

Both the values of x are
different for the value of y. So, it is not possible to find.

Hence, the correct option is B

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

**(A) 27 (B) 18 (C) 81
(D) 512**

**Solution:**

We know that number of
elements in a matrix of order mxn is mn.

So number of elements in
matrix of 3 x 3 is 9.

For each element we have
two choices either 0 or 1

So, total number of
possible matrices of order 3 x 3 with each entry 0 or 1 = 2^{9 }= 512

Correct option is D

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