# NCERT Solution Exercise 2

**Question 1:**

Let

Find each of the following

(i) A+B (ii) A-B (iii) 3A-C (iv) AB (v) BA

**Answer:**

(iv) Matrix A has 2 columns. This number is equal to the number of rows in matrix B. Therefore, AB is defined as:

(v) Matrix B has 2 columns. This number is equal to the number of rows in matrix A.

Therefore, BA is defined as:

**Question 2:** Compute the following:

**Answer**

(iii)

(v)

**Question 3**: Compute the indicated products

**Answer**

**Qu**estion 4:

Compute (A+B) and (B-C). Also, verify that

**Answer:**

Hence, we have verified that A+(B-C)=(A+B)-C.

**Question:5: **

**Answer:**

**Question:6**

**Answer**:

**Question 7:**

Find X and Y, if

**Answer:**

…………………(2)

Adding equations (1) and (2), we get:

(ii)

Multiplying equation (3) with (2), we get:

Multiplying equation (4) with (3), we get:

From (5) and (6), we have:

Now,

**Question 8:**

Find x, if and

**Answer**

**Question 9:**

Find x and y, if** **

**Answer:**

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

2+y=5

⇒ Y=2

2x+2=8

⇒ x=3

x = 3 and y = 3

**Question 10:** Solve the equation for x, y, z and t if

**Answer:**

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

2x+3=9

⇒ 2x = 6

⇒ x=3

2y=12

⇒ y=6

2z-3=15

⇒ 2z=18

⇒ z=9

2t + 6 =18

⇒ 2t=12

⇒ t=6

∴ x=3, y=6, z=9, and t=6

**Question 11:**

If , find values of x and y.

**Answer**

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

**Question 12:**

Given, find the values of x, y, z and w.

**Answer**

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

⇒3x = x+4

⇒2x =4

⇒ x =2

3y = 6+x+y

⇒ 2y=6+x = 6+2 = 8

⇒ y=4

3w = 2w+3

⇒ w=3

3z = -1+z+w

⇒ 2z=-1+w = -1+3 = 2

⇒ z = 1

∴ x = 2, y = 4, z = 1, and w = 3

**Question 13:**

If

Show that

F(x)F(y) = F(x+y).

**Answer:**

F(x)F(y)

= F(x+y)

∴ F(x)F(y) = F(x+y)

**Question 14: **

Show that

**Answer**

**Question 15:**

If

Find

A^{2}-5A+6I

**Answer**

We have A^{2} = A x A

∴ A2 – 5A + 6I

**Question 16:**

If A, prove that A^{3} – 6A^{2} + 7A + 2I = )

**Answer**

Now A^{3} = A^{2}.A

A^{3} - 6^{2} + 7A + 21

∴ A^{3} – 6A^{2} + 7A + 21 = 0

**Question 17:**

If and , find k so that A^{2} = kA – 21

**Answer:**

Now A^{2} = kA - 21

Comparing the corresponding elements, we have:

3K – 2 = 1

⇒ 3k = 3

⇒ k = 1

Thus, the value of k is 1.

**Question 18:**

If and I is identity matrix of order 2, show that

**Answer:**

**On the L.H.S**.

I + A

**On the R.H.S.**

Thus, from (1) and (2), we get L.H.S. = R.H.S.

**Question 19**: A trust fund has Rs 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of: (a) Rs 1,800 (b) Rs 2,000

**Answer **

(a) Let Rs x be invested in the first bond. Then, the sum of money invested in the second bond will be Rs (30000 − x). It is given that the first bond pays 5% interest per year and the second bond pays 7% interest per year. Therefore, in order to obtain an annual total interest of Rs 1800, we have:

⇒ (5x/100) + ((7(30000-x)) / 100) = 1800

⇒ 5x + 210000 – 7x = 180000

⇒ 210000 – 2x = 180000

⇒ 2x = 210000 – 180000

⇒2x = 30000

⇒ x = 15000

Thus, in order to obtain an annual total interest of Rs 1800, the trust fund should invest Rs 15000 in the first bond and the remaining Rs 15000 in the second bond. (b) Let Rs x be invested in the first bond. Then, the sum of money invested in the second bond will be Rs (30000 − x).

Therefore, in order to obtain an annual total interest of Rs 2000, we have:

⇒ (5x/100) + ((7(30000-x)) / 100) = 2000

⇒ 5x + 210000 – 7x = 200000

⇒ 210000 – 2x = 200000

⇒ 2x = 210000 – 200000

⇒2x = 10000

⇒ x = 5000

Thus, in order to obtain an annual total interest of Rs 2000, the trust fund should invest Rs 5000 in the first bond and the remaining Rs 25000 in the second bond.

**Question 20:** The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are Rs 80, Rs 60 and Rs 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

**Answer**

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:

= 12[10 x 80 x 8 x 60 + 10 x 40]

= 12(800 + 480 + 400)

=12(1680)

=20160

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

**Question 21**:

Assume X, Y, Z, W and P are matrices of order 2 x n,3 x k,2 x p,n x 3 and p x k respectively. The restriction on n, k and p so that PY + WY will be defined are:

- k = 3, p = n
- k is arbitrary, p=2
- p is arbitrary, k=3
- k = 2, p = 3

**Answer**

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.

**Question 22**:

Assume X, Y, Z, W and P are matrices of order 2xn, 3xk, 2xp, nx3 and pxk respectively. If n = p, then the order of the matrix is 7X – 5Z is

**A **p x 2

**B** 2 x n

**C** n x 3

**D** p x n

**Answer **

The correct answer is B.

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.