The availability of ISC Mathematics Class 12 Solutions Chapter 3 Matrices Ex 3.1 encourages students to tackle difficult exercises.

## ML Aggarwal Class 12 Maths Solutions Section A Chapter 3 Matrices Ex 3.1

Question 1.

If A is a matrix of type p × q and R is a row of A, then what is the type of R as a matrix ?

Solution:

Since A be a matrix of order p × q i.e. A contains p rows and q columns.

Also it is given that R is a row of A and containing q elements.

Thus order of R be 1 × q,

Question 2.

If A is a column matrix with 5 rows, then what type of matrix is a row of A ?

Solution:

Given A be a column matrix with 5 rows so order of matrix A be 5 × 1.

[since every column matrix contains only one column].

Question 3.

If a matrix has 18 elements, what the possible orders it can have ? What, if it has 11 elements ?

Solution:

We know that, if a matrix of order p × q then it has pq elements.

Now to find the possible orders of matrix containing 18 elements we shall find all ordered pairs of natural numbers whose product is 18.

Such ordered pairs are (1, 18), (2, 9), (3, 6), (6, 3), (9, 2) and (18, 1).

Thus, the possible orders are 1 × 18, 2 × 9, 3 × 6, 6 × 2, 9 × 2 and 18 × 1.

Now if a matrix contains 11 elements.

Then we shall find all ordered pairs of natural numbers whose product is 11.

Such ordered pairs are (1, 11) and (11, 1).

Thus the possible orders are 1 × 11 and 11 × 1.

Question 3 (old).

(i) If a matrix has 5 elements, write all possible orders it can have.

(ii) If a matrix has 8 elements, what are the possible orders it can have ? (NCERT)

Solution:

(i) We know that, if a matrix of order p × q then it contains pq elements.

Now to find all possible orders of matrix containing 5 elements we shall find all ordered pairs of natural numbers whose product is 5.

Such ordered pairs are (1, 5) and (5, 1).

Thus, the possible orders are 1 × 5 and 5 × 1.

(ii) We know that a matrix of order m × n have mn elements.

Since given matrix has 8 elements and we have to find its possible orders.

Thus we have to find all ordered pairs (a, b) where a and b are factors of 8.

Thus such ordered pairs are ; (1, 8), (2, 4), (4, 2) and (8, 1). ‘

Hence the possible orders of matrix are ; 1 × 8, 2 × 4, 4 × 2 and 8 × 1.

Question 4.

If A is a 3 × 3 matrix whose elements are given by a_{ij} = \(\frac{1}{3}\) |- 3i + j| write the value of a_{23}.

Solution:

Given A = [a_{ij}]_{3 × 3} where 1 ≤ i ≤ 3 and 1 ≤ j ≤ 3

Also, a_{ij} = \(\frac{1}{3}\) |- 3i + j|

When i = 2, j = 3 ;

a_{23} = \(\frac{1}{3}\) |- 6 + 3| = 1

Thus a_{23} = 1.

Question 5.

Construct a 2 × 2 matrix A = [a_{ij}] whose y elements a_{ij} are given by :

(i) a_{ij} = 2i – j

(ii) a_{ij} = \(\frac{(i-2 j)^2}{2}\) (NCERT Exemplar)

(iii) a_{ij} = | 2i – 3j|.

Solution:

(i) Given A = [a_{ij}]_{2 × 2} ; 1 ≤ i, j ≤ 2

where a_{ij} = 2i – j

When i = 1 = j ;

a_{11} = 2 – 1 = 1 ;

When i = 2 = j;

a_{22} = 4 – 2 = 2

When i = 1, j = 2 ;

a_{12} = 2 – 2 = 0

When i = 2, j = 1 ;

a_{21} = 4 – 1 = 3

∴ A = \(\left[\begin{array}{ll}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{array}\right]\)

= \(\left[\begin{array}{ll}

1 & 0 \\

3 & 2

\end{array}\right]\)

(ii) Given A = [a_{ij}]_{2 × 2} where 1 ≤ i, j ≤ 2

where a_{ij} = | 2i – 3 j |

When i = j = 1 ;

a_{11} = |2 – 3 | = |- 1 | = 1

When i = 1, j = 2 ;

a_{12} = |2 — 6| = | — 4| = 4

When i = 2, j = 1 ;

a_{21} = | 4 – 3 | = | 1 | = 1

When i = 2, j = 2 ;

a_{22} = |4 – 6| = |- 2| = 2

Question 5 (old).

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

Solution:

Given matrix has 24 elements

We know that, if a matrix of order p × q then it has pq elements.

Now to find the possible orders of matrix containing 24 elements we shall find all ordered pairs of natural numbers whose product is 24.

Such ordered pairs are 1 × 24, 24 × 1, 2 × 12, 12 × 2, 3 × 8, 8 × 3, 4 × 6, 6 × 4

If it has 13 elements

.’. possible orders are 1 × 13, 13 × 1.

Question 6.

Conxtruct a 2 × 3 matrix B = [b_{ij}] whose elements b_{ij} are given by :

(i) b_{ij} = i – 3j

(ii) b_{ij} = (i + 2j)^{2}

Solution:

(i) Given B = [b_{ij}]_{2 × 3} where 1 ≤ i ≤ 2 and 1 ≤ j ≤ 3

Also, b_{ij} = i – 3j

When i = 1 = j ;

b_{11} = 1 – 3 = – 2

When i = 1, j = 2

b_{12} = 1 – 6 = – 5

When i = 1, j = 3

b_{13} = 1 – 9 = – 8

When i = 2, j = 1

b_{21} = 2 – 3 = – 1

When i = 2, j = 2

b_{22} = 2 – 6 = – 4

When i = 2, j = 3

b_{23} = 2 – 9 = – 7

Thus, B = \(\left[\begin{array}{lll}

b_{11} & b_{12} & b_{13} \\

b_{21} & b_{22} & b_{23}

\end{array}\right]\)

= \(\left[\begin{array}{lll}

-2 & -5 & -8 \\

-1 & -4 & -7

\end{array}\right]_{2 \times 3}\)

(ii) Given B = [b_{ij}]_{2 × 3}

where 1 ≤ i ≤ 2 ; 1 ≤ j ≤ 3

Given b_{ij} = (i + 2j)^{2}

When i = 1 = j ;

b_{11} = (1 + 2)^{2} = 9

When i = 1, j = 2

b_{12} = (1 + 4)^{2} = 25

When i = 1, j = 3

b_{12} = (1 + 6)^{2} = 49

When i = 2, j = 1

b_{12} = (2 + 2)^{2} = 16

When i = 2, j = 2

b_{12} = (2 + 4)^{2} = 36

When i = 2, j = 3

b_{12} = (2 + 6)^{2} = 64

Thus, B = \(\left[\begin{array}{lll}

b_{11} & b_{12} & b_{13} \\

b_{21} & b_{22} & b_{23}

\end{array}\right]\)

= \(\left[\begin{array}{ccc}

9 & 25 & 49 \\

16 & 36 & 64

\end{array}\right]_{2 \times 3}\)

Question 6 (old).

(i) For a 2 × 2 matrix, A = |a_{ij}|, whose elements are given by a_{ij} = \(\frac{i}{j}\) value of a_{12}.

(ii) Write the element a_{23} of a 3 × 3 matrix A = [a_{ij}] whose elements are given by a_{ij} = \(\frac{|i-j|}{2}\).

(iii) If A is a 3 × 3 matrix whose elements are given by a_{ij} = \(\frac{1}{3}\) |- 3i + j|, write the value of a_{23}.

(iv) Construct a 2 × 2 matrix A = [a_{ij}] whose elements a_{ij} are given by a_{ij} = i + 2j.

Solution:

(i) Given a_{ij} = \(\frac{i}{j}\),

where i ≤ i ≤ 2 ; 1 ≤ j ≤ 2

a_{ij} = \(\frac{1}{2}\) ;

(ii) Given A = [a_{ij}]_{3 × 3}

where 1 ≤ i ≤ 3 and 1 ≤ j ≤ 3

Also, a_{ij} = \(\frac{|i-j|}{2}\)

When i = 2, j = 3 ;

a_{23} = \(\frac{|2-3|}{2}=\frac{1}{2}\)

(iii) Given A = [a_{ij}]_{3 × 3} where 1 ≤ i ≤ 3, and 1 ≤ j ≤ 3

Also, a_{ij} = \(\frac{1}{3}\) |- 3i + j|

When i = 2, j = 3 ;

a_{23} = \(\frac{1}{3}\) |- 6 + 3| = 1

Thus a_{23} = 1

(iv) Given A = [a_{ij}]_{2 × 2} where 1 ≤ i, j ≤ 2

where a_{ij} = i + 2j

When i = 1 = j ;

a_{11} = 1 + 2 = 3

When i = 1 = j = 2 ;

a_{11} = 1 + 4 = 5

When i = 2 = j = 1 ;

a_{11} = 2 + 2 = 4

When i = j = 2 ;

a_{11} = 2 + 4 = 6

∴ A = \(\left[\begin{array}{ll}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{array}\right]\)

= \(\left[\begin{array}{ll}

3 & 5 \\

4 & 6

\end{array}\right]\)

Question 7.

(i) If \(\left[\begin{array}{cr}

x+2 y & -y \\

3 x & 4

\end{array}\right]=\left[\begin{array}{rr}

-4 & 3 \\

6 & 4

\end{array}\right]\), find the values of x and y.

(ii) If \(\left[\begin{array}{cc}

x+3 y & y \\

7-x & 4

\end{array}\right]=\left[\begin{array}{rr}

4 & -1 \\

0 & 4

\end{array}\right]\), find the values of x and y.

(iii) Find the value of y, if \(\left[\begin{array}{cc}

x-y & 2 \\

x & 5

\end{array}\right]=\left[\begin{array}{ll}

2 & 2 \\

3 & 5

\end{array}\right]\).

(iv) Find the value of x, if \(\left[\begin{array}{cc}

3 x+y & -y \\

2 y-x & 3

\end{array}\right]=\left[\begin{array}{rc}

1 & 2 \\

-5 & 3

\end{array}\right]\).

(v) If \(\left[\begin{array}{cc}

x-y & z \\

2 x-y & w

\end{array}\right]=\left[\begin{array}{rr}

-1 & 4 \\

0 & 5

\end{array}\right]\) find the value of x – y.

(vi) If \(\left[\begin{array}{cc}

x-y & z \\

2 x-y & w

\end{array}\right]=\left[\begin{array}{rr}

-1 & 4 \\

0 & 5

\end{array}\right]\) find the value of (x + y).

(vii) If \(\left[\begin{array}{cc}

x+y & x+2 \\

2 x-y & 16

\end{array}\right]=\left[\begin{array}{cc}

8 & 5 \\

1 & 3 y+1

\end{array}\right]\), then write the value of (y- x).

Solution:

(i) Given \(\left[\begin{array}{cr}

x+2 y & -y \\

3 x & 4

\end{array}\right]=\left[\begin{array}{rr}

-4 & 3 \\

6 & 4

\end{array}\right]\)

On comparing,

x + 2y = – 4 ………….(1)

– y = 3

⇒ y = – 3

(ii) Given \(\left[\begin{array}{cc}

x+3 y & y \\

7-x & 4

\end{array}\right]=\left[\begin{array}{rr}

4 & -1 \\

0 & 4

\end{array}\right]\)

On comparing,

∴ x + 3y = 4 ………………(1)

y = – 1 ……………..(2)

7 – x = 0 …………….(3)

From (3) ; 7 – x = 0

⇒ x = 7 and y = – 1

Further the values of x = 7 and y = – 1.

(iii) Given, \(\left[\begin{array}{cc}

x-y & 2 \\

x & 5

\end{array}\right]=\left[\begin{array}{ll}

2 & 2 \\

3 & 5

\end{array}\right]\)

Thus, their corresponding elements are equal.

x – y = 2 ………..(1)

x = 3 ………..(2)

From (1) and (2) ;

x = 3 ;

y = 3 – 2 = 1

(iv) Given, \(\left[\begin{array}{cc}

3 x+y & -y \\

2 y-x & 3

\end{array}\right]=\left[\begin{array}{rc}

1 & 2 \\

-5 & 3

\end{array}\right]\)

On comaring ;

3x + y = 1 ……………..(1)

and – y = 2

⇒ y = – 2

2y – x = – 5

∴ from (1) ;

3x = 3

⇒ x = 1

(v) Given, \(\left[\begin{array}{cc}

x-y & z \\

2 x-y & w

\end{array}\right]=\left[\begin{array}{rr}

-1 & 4 \\

0 & 5

\end{array}\right]\)

On comparing,

x + 2y = 0 …………….(1)

3y = – 3

⇒ y = – 1

∴ from (1) ;

x = 2

Thus x – y = 2 – (- 1) = 3.

(vi) Given \(\left[\begin{array}{cc}

x-y & z \\

2 x-y & w

\end{array}\right]=\left[\begin{array}{rr}

-1 & 4 \\

0 & 5

\end{array}\right]\)

Thus their corresponding elements are equal.

∴ x – y = – 1 ………….(1)

z = 4 …………(2)

2x – y = 0 ………..(3)

and w = 5

eqn. (3) – eqn. (1) gives ;

2x – x = 0 + 1

⇒ x = 1 ; y = 2

Thus, x = 1 ; y = 2 ; z = 4 and w = 5

Thus x + y = 1 + 2 = 3

(vii) Given \(\left[\begin{array}{cc}

x+y & x+2 \\

2 x-y & 16

\end{array}\right]=\left[\begin{array}{cc}

8 & 5 \\

1 & 3 y+1

\end{array}\right]\)

Thus, their corresponding elements are equal.

x + y = 8 …………(1)

x + 2 = 5 …………..(2)

2x – y = 1 …………..(3)

3y + 1 = 16 ………….(4)

From (2) ; x = 5 – 2 = 3

From (4) ; 3y = 15

y = 5

Now x = 3 and y = 5 satisfies eqn. (1) and eqn. (3)

Thus y – x = 5 – 3 = 2.

Question 8.

If \(\left[\begin{array}{cc}

x & 3 x-y \\

2 x+z & 3 y-w

\end{array}\right]=\left[\begin{array}{ll}

3 & 2 \\

4 & 7

\end{array}\right]\), find x, y, z and w.

Solution:

\(\left[\begin{array}{cc}

x & 3 x-y \\

2 x+z & 3 y-w

\end{array}\right]=\left[\begin{array}{ll}

3 & 2 \\

4 & 7

\end{array}\right]\)

Thus, their corresponding elements are equal.

∴ x = 3 ; 3x – y = 2

⇒ 9 – y = 2

⇒ y = 9 – 2 = 7

and 2x + z = 4

⇒ 6 + z = 4

z = 4 – 6 = – 2

Also 3y – w = 7

⇒ w = 3 × 7 – 7 = 14

Hence, x = 3 ; y = 7 ; z = + 2 and w = 14.

Question 9.

Find the values of a, b, c and d from the following equations:

\(\left[\begin{array}{cc}

2 a+b & a-2 b \\

5 c-d & 4 c+3 d

\end{array}\right]=\left[\begin{array}{rr}

4 & -3 \\

11 & 24

\end{array}\right]\)(NCERT)

Solution:

Given \(\left[\begin{array}{cc}

2 a+b & a-2 b \\

5 c-d & 4 c+3 d

\end{array}\right]=\left[\begin{array}{rr}

4 & -3 \\

11 & 24

\end{array}\right]\)

Thus, their corresponding elements are equal.

∴ 2a + b = 4 ………………(1)

a – 2b = – 3 …………….(2)

5c – d = 11 ……………..(3)

4c + 3d = 24 …………….(4)

Multiplying eqn. (1) by 2 and adding to eqn. (2) ; we get

5a = 5

⇒ a = 1

∴ From (1) ;

b = 4 – 2 = 2

Multiplying eqn. (3) by 3 and adding eqn. (4) ; we get

19c = 57

⇒ c = 3

∴ From (3) ;

d = 15 – 11 = 4

Hence, a = 1 ; b = 2 ; c = 3 and d = 4.

Question 9 (old).

(i) If \(\left[\begin{array}{cc}

x+3 & 4 \\

y-4 & x+y

\end{array}\right]=\left[\begin{array}{ll}

5 & 4 \\

3 & 9

\end{array}\right]\) find x and y.

Solution:

Given \(\left[\begin{array}{cc}

x+3 & 4 \\

y-4 & x+y

\end{array}\right]=\left[\begin{array}{ll}

5 & 4 \\

3 & 9

\end{array}\right]\)

On comparing,

x + 3 = 5

x = 2

and y – 4 = 3

⇒ y = 7

(vi) Find the value of x, if \(\left[\begin{array}{cc}

2 x-y & 5 \\

3 & y

\end{array}\right]=\left[\begin{array}{rr}

6 & 5 \\

3 & -2

\end{array}\right]\).

Solution:

Given, \(\left[\begin{array}{cc}

2 x-y & 5 \\

3 & y

\end{array}\right]=\left[\begin{array}{rr}

6 & 5 \\

3 & -2

\end{array}\right]\)

On comparing ;

2x – y = 6

and y = – 2

∴ x = 2.

Question 10.

Find x, y, a and b if \(\left[\begin{array}{ccc}

2 x-3 y & a-b & 3 \\

1 & x+4 y & 3 a+4 b

\end{array}\right]=\left[\begin{array}{crr}

1 & -2 & 3 \\

1 & 6 & 29

\end{array}\right]\).

Solution:

Given, \(\left[\begin{array}{ccc}

2 x-3 y & a-b & 3 \\

1 & x+4 y & 3 a+4 b

\end{array}\right]=\left[\begin{array}{crr}

1 & -2 & 3 \\

1 & 6 & 29

\end{array}\right]\)

Thus, their corresponding elements are equal.

∴ 2x – 3y = 1 …………..(1)

x + 4y = 6 ……………(2)

a – b = – 2 ……………(3)

3a + 4b = 29 …………….(4)

Multiply eqn. (2) by 2 and subtract it from (1) we get

– 11y = – 11

⇒ y = 1

∴ From (1) ;

2x – 3 = 1

⇒ x = 2

Multiplying eqn. (3) by 4 and add to eqn. (4); we have

7a = 21

⇒ a = 3

∴ b = 5

Hence x = 2; y = 1; a = 3; b = 5.

Question 11.

(i) Write the number of all possible matrices of order 2 × 3 with each entry 1 or 2.

(ii) Write the number of all possible matrices of order 2 × 2 with each entry 1, 2 or 3.

Solution:

(i) Since a matrix of order 2 × 3 containing 2 × 3 = 6 elements i.e. 6 entries and each entry can be filled in 2 ways 1 or 2.

Thus the total number of ways of filling all the entries of matrix be 2^{6} i.e. 64.

Also corresponding to every diffèrent entry we have different matrix.

Hence the required iiumberof all possible matrices of order 2 × 3 with each entry 1 or 2 be 64.

(ii) Since a matrix of order 2 × 2 containing 2 × 2 = 4 elements

i.e. 4 entries and each entry can be filled in 3 ways i.e. 1. 2 or 3.

Thus, the total number of ways of filling all the entries of matrix be 3^{4} i.e. 81.

Also corresponding to every different entry we have a different matrix.

Hence, the required number of all possible matrices of order 2 × 2 with each entry 1,2 or 3 be 3^{4} i.e. 81.