NCERT Solutions for Class 12 Maths Chapter 3: Matrices

(i)Given: 

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

Comparing the corresponding elements, we get:

x = 1, y = 4 = 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 and 5 + z = 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, y = 2

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

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

(iii)

 

(i) There are 3 horizontal lines (rows) and 4 vertical lines (columns) in the given matrix A.

Therefore, Order of the matrix is 3 x 4.

(ii) The number of elements in the matrix A is 3 x 4 = 12.

(iii) a13 Element in first row and third column = 19

a21 Element in second row and first column = 35

a33 Element in third row and third column = -5

a24 Element in second row and fourth column = 12

a23 Element in second row and third column = 5/2

 

We know that if a matrix is of the order m × n, it has mn elements. Thus, to find all the possible orders of a matrix having 24 elements, we have to find all the ordered pairs of natural numbers whose product is 24.

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

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

(1, 13) and (13, 1) are the ordered pairs of natural numbers whose product is 13.

Hence, the possible orders of a matrix having 13 elements are 1 × 13 and 13 × 1.

 

We know that if a matrix is of the order m × n, it has mn elements. Thus, to find all the possible orders of a matrix having 18 elements, we have to find all the ordered pairs of natural numbers whose product is 18.

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

Hence, the possible orders of a matrix having 18 elements are:

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

(1, 5) and (5, 1) are the ordered pairs of natural numbers whose product is 5.

Hence, the possible orders of a matrix having 5 elements are 1 × 5 and 5 × 1.

 

(i)

(ii)

(iii)

 

(i)

(ii)

 

 

 

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)

From (2), we have:

b = 2a

Then, from (1), we have:

a − 2a = −1

⇒ a = 1

⇒ b = 2

Now, from (3), we have:

2 ×1 + c = 5

⇒ c = 3

From (4) we have:

3 ×3 + d = 13

⇒ 9 + d = 13 ⇒ d = 4

∴a = 1, b = 2, c = 3, and d = 4

 

It is known that a given matrix is said to be a square matrix if the number of rows is equal to the number of columns. A = [aij]m x n

option (C) is correct.

 

 

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.

Therefore, option (B) is correct.

 

The correct answer 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

 

(i) 

(ii)

(iii)

(iv)

(v) 

 

 

(ii)

(iii)

(iv)

 

(i)

(ii)

(iii) =

(iv)=

(v)=

(vi)=

L.H.S. = R.H.S.  Proved.

 

 

(i) 

(ii)

 

 

 

Comparing the corresponding elements of these two matrices, we get:

2x − y = 10 and 3x + y = 5

Adding these two equations, we have:

5x = 15

⇒ x = 3

Now, 3x + y = 5

⇒ y = 5 − 3x

⇒ y = 5 − 9 = −4

∴x = 3 and y = −4

 

 

 

 

(i)

(ii)

 

L.H.S. = A3 – 6A2 + 7A + 2I

 

= R.H.S.  Proved.

 

(a) It is given that Rs.30,000 must be invested into two types of bonds with 5% and 7% interest rates.

Let Rs. x be invested in bonds of the first type. Thus, Rs. (30,000 − x) will be invested in the other type.

Hence, the amount invested in each type of the bonds can be represented in matrix form with each column corresponding to a different type of bond as:

X = [x ​30,000 − x​]

Annual interest obtained is Rs. 1800. We know, Interest = PTR/100

Here, the time is one year and thus T = 1.

Hence, the interest obtained after one year can be expressed in matrix representation as-

⇒  5x + 210000 − 7x = 180000

⇒ – 2x = – 30000

∴  x = 15000

Amount invested in the first bond = x = Rs. 15000

⇒ Amount invested second bond = Rs. (30000 − x) = Rs. (30000 − 15000) = Rs. 15000

∴ The trust has to invest Rs. 15000 each in both the bonds in order to obtain an annual interest of Rs. 1800.

(b) It is given that Rs.30,000 must be invested into two types of bonds with 5% and 7% interest rates.

Let Rs. x be invested in bonds of the first type. Thus, Rs. (30,000 − x) will be invested in the other type.

Hence, the amount invested in each type of the bonds can be represented in matrix form with each column corresponding to a different type of bond as:

X = [x ​30,000 − x​]

Annual interest obtained is Rs. 2000.

Hence, the interest obtained after one year can be expressed in matrix representation as-

⇒  5x + 210000 − 7x = 200000

⇒  −2x = − 10000

∴  x = 5000

Amount invested in the first bond = x = Rs. 5000

⇒ Amount invested second bond = Rs. (30000 − x) = Rs. (30000 − 5000) = Rs. 25000

∴ The trust has to invest Rs.5000 in the first bond and Rs.25000 in the second bond in order to obtain an annual interest of Rs.2000.

 

The bookshop has 10 dozen chemistry books, 8 dozen physics books, and 10 dozen economics books.

The selling prices of a chemistry book, a physics book, and an economics book are respectively given as Rs 80, Rs 60, and Rs 40.

The total amount of money that will be received from the sale of all these books can be represented in the form of a matrix as:

Thus, the bookshop will receive Rs 20160 from the sale of all these books.

 

Matrices P and Y are of the orders p × k and 3 × k respectively.

Therefore, matrix PY will be defined if k = 3. Consequently, PY will be of the order p × k.

Matrices W and Y are of the orders n × 3 and 3 × k respectively.

Since the number of columns in W is equal to the number of rows in Y, matrix WY is well-defined and is of the order n × k.

Matrices PY and WY can be added only when their orders are the same.

However, PY is of the order p × k and WY is of the order n × k. Therefore, we must have p = n.

Thus, k = 3 and p = n are the restrictions on n, k, and p so that PY + WY will be defined.

Therefore, option (A) is correct.

 

Matrix X is of the order 2 × n.

Therefore, matrix 7X is also of the same order.

Matrix Z is of the order 2 × p, i.e., 2 × n [Since n = p]

Therefore, matrix 5Z is also of the same order.

Now, both the matrices 7X and 5Z are of the order 2 × n.

Thus, matrix 7X − 5Z is well-defined and is of the order 2 × n.

The correct answer is B.

 

 

(i) Let A =

Transpose of A = A’ or AT = [ 5 1/2 -1]

(ii)

Transpose of A = A’ or AT=

(iii)

Transpose of A = A’ or AT=

 

 

(i) Given: A =

Changing rows of matrix A as the columns of new matrix A’ = = A

∴ A’ = A

Therefore, by definitions of symmetric matrix, A is a symmetric matrix.

(ii) Given: A =

A’ = = – = – A

∴ A’ = – A

Therefore, by definition matrix A is a skew-symmetric matrix

 

 

(iii)

(iv) 

 

Given: A and B are symmetric matrices ∴ A = A’ and B = B’

Now, (AB – BA)’ = (AB)’ – (BA)’  ∴ (AB – BA)’ = B’A’ – A’B’ [Reversal law]

∴ (AB – BA)’ = BA – AB [From eq. (i)] ∴ (AB – BA)’ = – (AB – BA)

∴  (AB – BA) is a skew matrix.

Therefore, option (A) is correct.

 

Therefore, option (B) is correct.