Matrices solved ex 3.1

Chapter - 3 MATRICES (Ex – 3.1)

PCM ENCYCLOPEDIA (www.pcmpedia.blogspot.in)

1

1. In the Matrix A=[

⁄

√

], Write:

(i) The order of the matrix,

(ii) The number of elements,

(iii) Write the elements of a13, a21, a33, a24, a23.

Solution 1:

(i) In the given matrix, the number of rows is 3 and the number of

columns is 4. Therefore, the order of the matrix is 3 × 4. (ii) Since the order of the matrix is 3 × 4, there are 3 × 4 = 12 elements

in it. (iii) a13 = 19, a21 = 35, a33 = −5, a24 = 12, a23 = 5/2

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

What, if it has 13 elements?

Solution 2:

We know that if a matrix is of the order m × n, it has mn elements.So, here we have to find out the possible orders of a matrix having 24 elements. Hence, the possible orders of a matrix having 24 elements are: 1 × 24, 24 × 1, 2 × 12, 12 × 2, 3 × 8, 8 × 3, 4 × 6, and 6 × 4 and the possible orders of a matrix having 13 elements are 1 × 13 and 13 × 1.

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

What, if it has 5 elements?

Solution 3:

We know that if a matrix is of the order m × n, it has mn elements.So, here we have to find out the possible orders of a matrix having 18



2

elements. Hence, the possible orders of a matrix having 18 elements are: 1 × 18, 2 × 9, 3 × 6, 6 × 3, 9 × 2, and 18 × 1 and the possible orders of a matrix having 5 elements are 1 × 5 and 5 × 1.

4. Construct a 2X2 matrix, A=[aij], whose elements are given by:

(i) aij = ( )

(ii) aij =

(iii) aij =

( )

Solution 4:

(i) Given aij = ( )

We Know, a 2 x 2 matrix is given by A = [

]

So,

a11 = ( )

=

= 2 a12 =

( )

=

a21 = ( )

=

a22 =

( )

=

= 8

Therefore, the required Matrix is

A = [

]

(ii) Given aij =


]

So,

a11 =

= 1 a12 =

a21 =

= 2 a22 =

= 1



3


A = [

]

(iii) Given aij = ( )


]

So,

a11 = ( )

=

a12 =

( )

=

a21 = ( )

=

= 8 a22 =

( )

=

= 18


A = [

]

5. Construct a 3X4 Matrix, whose elements are given by:

(i) aij =

| | (ii) aij = 2i – j

Solution 5:

(i) aij =

| |


]

a11 =

| | = 1 a12 =

| | =

a13 =

| | = 0 a14 =

| | =



4

a21 =

| | =

a22 =

| | = 2

a23 =

| | =

a24 =

| | = 1

a31 =

| | = 4 a32 =

| | =

a33 =

| | = 3 a34 =

| | =

Hence, the required matrix is:

A =

[

]

(ii) aij = 2i – j


]

a11 = 2 × 1 − 1 = 1 a12 = 2 × 1 − 2 = 0

a13 = 2 × 1 − 3 = −1 a14 = 2 × 1 − 4 = −2

a21 = 2 × 2 − 1 = 3 a22 = 2 × 2 − 2 = 2

a23 = 2 × 2 − 3 = 1 a24 = 2 × 2 − 4 = 0

a31 = 2 × 3 − 1 = 5 a32 = 2 × 3 − 2 = 4

a33 = 2 × 3 − 3 = 3 a34 = 2 × 3 − 4 = 2

Hence, the required matrix is:

A = [

]



5

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

(i) [

]=[

] (ii) [

]=[

]

(ii) [

]=[ ]

Solution 6:

(i) [

] = [

]

Since both are matrices are equal, their corresponding elements are also

equal.

Comparing corresponding elements, we get

y = 4, z = 3 and x = 1

(ii) [

]=[

]


equal.


x + y = 6, 5 + z = 5 and xy = 8

Now, 5 + z = 5 z = 0

Here, x + y = 6 ----------------------- eq. 1

xy = 8 -------------------------- eq. 2

From eq. 1

y = 6 – x -------------------------------- eq .3

Put this value of y in eq. 2, we get

x (6 – x) = 8

6x – x2 = 8 or x2 6x + 8 = 0

By Factorization,

x2 2x – 4x + 8 = 0

x(x 2) 4(x 2) = 0

(x 4)(x – 2) = 0



6

x = 4 or x = 2

Put these values in eq. 3

From x = 4, we get From x = 2, we get

y = 6 4 = 2 y = 6 – 2 = 4

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

(iii) [

]=[ ]


equal.


x y z = ……………….. ( )

x z = ……………………….( )

y z = ……………………….( )

From (1) and (2), we have

y + 5 = 9

⇒ y = 4

Put this value of y in (3)

4 + z = 7

⇒ z = 3

Also,

x + z = 5

x + 3 = 5

⇒ x = 2

∴ x = 2, y = 4 and z = 3

7. Find the Value of a, b, c and d from the equation:

[

] = [

]



7

Solution 7:


equal.

Comparing corresponding elements, we get:

= ……………( )

= …………… ( )

= ……………( )

= ……………( )

From (2) we have:

= …………… . ( )

From (1) & (5),

=

=

= and =

Now from (3), we have:

=

⇒ =

Now from (4), we get:

=

=

Hence, = = = =

8. = is a square matrix, if

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

Solution 8:

Correct Option is (C).

To be a Square Matrix m should be equal to n. i.e No. of rows should be

equal to No. of Columns.

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

equal

[

] = [

]



8

(A) =

= (B) Not Possible to find

(C) = =

(D) =

=

Solution 9:

Correct Option is (B).

Given that [

] = [

]

Equating the Corresponding element of the given Matrix.

3 = =

and 5 = = 7

= = 7 and = =

On Comparing the Corresponding elements of the two matrices, we get

two different values of , which is not possible.

Hence, it is not possible to find the values of and y for which the given

matrices are equal.

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 10:

Correct Option is (D).

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.

Education

Matrices solved ex 3.1