# Class 12 NCERT Solutions- Mathematics Part I – 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, -5a

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

_{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 = 2a

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

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

_{22}=> i = 2 and j = 2 =>(2 + 2)^{2}/2 = 16/2 = 8Resultant 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 = 1a

_{12}=> i = 1 and j = 2 = 1/2a

_{21}=> i = 2 and j = 1 = 2/1 = 2a

_{22}=> i = 2 and j = 2 = 2/2 = 1Resultant 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 = 9/2a

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

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

_{22}=> i = 2 and j = 2 =>(2 + 2 x 2)^{2}/2 = 36/2 = 18Resultant 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|) = 1a

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

_{22}=> i = 2 and j = 2 => 2 x 2 – 2 = 2a

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

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

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

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

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

_{34}=> i = 3 and j = 4 => 2 x 3 – 4 = 2Resultant 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 }= 8y

^{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 }= 512Correct option is D