# Plus Two Math's Solution Ex 3.1 Chapter 3 Matrices

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Discover how to improve your maths grades with our high-quality, easy-to-understand maths textbook solutions. Our solutions are packed with information to ensure you get the best out of your studying Here is the solution for Exercise 3.1 Chapter 3 Matrices of NCERT plus two maths. Here we have given a detailed explanation of each and every exercise so that students can understand the concepts easily without any difficulty. The solution to each and every question is provided here so you can solve them by yourself if you don’t get the answer here.

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 Board SCERT, Kerala Text Book NCERT Based Class Plus Two Subject Math's Textbook Solution Chapter Chapter 3 Exercise Ex 3.1 Chapter Name Matrices Category Plus Two Kerala

## Kerala Syllabus Plus Two Math's Textbook Solution Chapter  3 Matrices Exercises 3.1

### Chapter 3 Matrices Textbook Solution

Kerala plus two maths NCERT textbooks, we provide complete solutions for the exercise and answers provided at the end of each chapter. We also cover the entire syllabus given by the Board of secondary education, Kerala state.

### Chapter 3 Matrices Exercise   3.1

Question 1:

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

24 = 1 X 24

= 2 X 12

= 3 X 8

= 4 X 6

= 6 X 4

= 8 X 3

= 12 X 2

= 24 X 1

So, there are 8 orders possible.

Similarly 13 = 1 X 13 and 13 X1, So, only 2 orders are possible.

In the matrix, write:

(i) The order of the matrix (ii) The number of elements,

(iii) Write the elements $a_{13},&space;a_{21},&space;a_{33},&space;a_{23}$

i) Order is 3×4 because there are 3 rows and 4 columns in A.

ii) Number of elements is 12. as there are 3 × 4 = 12 elements in it.

(iii) a13 = 19, a21 = 35, a33 = −5, a24 = 12, a23 =

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

18= 1 × 18, 18 × 1, 2 × 9, 9 × 2, 3 × 6, and 6 × 3

Therefore, there are 6 posiible orders.

And for 5 it is 1 X 5 and 5 X 1, So, it has 2 possible orders

Construct a 2 × 2 matrix,, whose elements are given by:

(i)

(ii)

(iii)

For 2 X 2 matrix the elements are a11, a12, a21 and $a_{22}$

(i)

Therefore, the required matrix is

(ii)

Therefore, the required matrix is

(iii)

Therefore, the required matrix is

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

(i)  (ii)

In general, a 3 × 4 matrix is given by

(i)

Therefore, the required matrix is

(ii)

Therefore, the required matrix is

Find the value of abc, and d from the equation:

As the two matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements, we get:

a − b = −1 … (1)

2a − b = 0 … (2)

2a + c = 5 … (3)

3c + d = 13 … (4)

a - b = -1 and 2a - b = 0; solving a =1 and b = 2

2a + c = 5, substituting a = 1, c =  $\frac{3}{2}$

3c + d = 13, substituting c =  $\frac{3}{2}$;  d = $\frac{17}{2}$

$A=\left&space;[&space;a_{ij}&space;\right&space;]_{m\times&space;n}$is a square matrix, if

(A) m < n

(B) m > n

(C) m = n

(D) None of these

(C)

If m and n are the rows and columns, then in a square matrix m=n.

Find the value of xy, and z from the following equation:

(i)  (ii)

(iii)

(i)

As the given matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements, we get:

x = 1, y = 4, and z = 3

(ii)

As the given matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements, we get:

x + y = 6, xy = 8, 5 + = 5

Now, 5 + z = 5 ⇒ z = 0

We know that:

(x − y)2 = (x + y)2 − 4xy

⇒ (x − y)2 = 36 − 32 = 4

⇒ x − y = ±2

Now, when x − y = 2 and x + y = 6, we get x = 4 and y = 2

When x − = − 2 and x + y = 6, we get x = 2 and = 4

x = 4, y = 2, and z = 0 or x = 2, y = 4, and z = 0

(iii)

As the two matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements, we get:

x + y + z = 9 … (1)

x + z = 5 … (2)

y + z = 7 … (3)

From (1) and (2), we have:

+ 5 = 9

⇒ y = 4

Then, from (3), we have:

4 + z = 7

⇒ z = 3

∴ x + z = 5

⇒ x = 2

∴ x = 2, y = 4, and z = 3

Which of the given values of x and y make the following pair of matrices equal

(A)

(B) Not possible to find

(C)

(D)

It is given that

Equating the corresponding elements, we get:

We find that on comparing the corresponding elements of the two matrices, we get two different values of x, which is not possible.

Hence, it is not possible to find the values of x and y for which the given matrices are equal.

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 given matrix of the order 3 × 3 has 9 elements and each of these elements can be either 0 or 1.

Now, each of the 9 elements can be filled in two possible ways.

Therefore, by the multiplication principle, the required number of possible matrices is 29 = 512

#### Chapter 3 Matrices Excercise 3.1 Textbook Solution - Preview

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