Learn Exercise 3.2 with Free Lessons & Tips

Question 2:

Simplify

Solution 2:

Comments

Harshit Bhardwaj

Question 3:

If, find values of x and y.

Solution 3:

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

Comments

Harshit Bhardwaj

Question 4:

Let

Find each of the following

(i) (ii) (iii)

(iv) (v)

Solution 4:

(i)

(ii)

(iii)

(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:

Comments

Harshit Bhardwaj

Question 5:

Compute the following:

(i) (ii)

(iii)

(iv)

Solution 5:

In the addition of matrices , we add similar row and column matrix with another matrix . For example in first part a will be added with a , b with b , -b with b and a with a .

Comments

Kush Prakash Srivastava

Question 6:

Compute the indicated products

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Solution 6:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Comments

Harshit Bhardwaj

Question 7:

If, and, then compute and. Also, verify that

Solution 7:

Comments

Harshit Bhardwaj

Question 8:

Find X and Y, if

(i) and

(ii) and

Solution 8:

(i)

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,

Comments

Harshit Bhardwaj

Question 9:

Find X, if and #mew_question#Find x and y, if

Solution 9:

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

∴x = 3 and y = 3

Comments

Harshit Bhardwaj

Question 10:

Solve the equation for x, y, z and t if

Solution 10:

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

Comments

Harshit Bhardwaj

Question 11:

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

Solution 11:

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

Comments

Harshit Bhardwaj

Question 12:

If, show that.

Solution 12:

Comments

Tasneem

Question 13:

(i)

(ii)

Solution 13:

(i)

(ii)

Comments

Harshit Bhardwaj

Question 14:

Find if

Solution 14:

We have A² = A × A

Comments

Harshit Bhardwaj

Question 15:

If, prove that

Solution 15:

$A^{3}-6A^{2}+7A+2I=egin{bmatrix} 1 &0 &2 \ 0 &2 &1 \ 2 &0 &3 \end{bmatrix}^{3}-6egin{bmatrix} 1 &0 &2 \ 0 &2 &1 \ 2 &0 &3 \end{bmatrix}^{2}+7egin{bmatrix} 1 &0 &2 \ 0 &2 &1 \ 2 &0 &3 \end{bmatrix}+2egin{bmatrix} 1 &0 &0 \ 0 &1 &0 \ 0 &0 &1 \end{bmatrix} =egin{bmatrix} 21 &0 &34 \ 12 &8 &23 \ 34 &0 &55 \end{bmatrix}-6egin{bmatrix} 5 &0 &8 \ 2 &4 &5 \ 8 &0 &13 \end{bmatrix}+7egin{bmatrix} 1 &0 &2 \ 0 &2 &1 \ 2 &0 &3 \end{bmatrix}+2egin{bmatrix} 1 &0 &0 \ 0 &1 &0 \ 0 &0 &1 \end{bmatrix} =egin{bmatrix} 21 &0 &34 \ 12 &8 &23 \ 34 &0 &55 \end{bmatrix}-egin{bmatrix} 30 &0 &48 \ 12 &24 &30 \ 48 &0 &78 \end{bmatrix}+egin{bmatrix} 7 &0 &14 \ 0 &14 &7 \ 14 &0 &21 \end{bmatrix}+egin{bmatrix} 2 &0 &0 \ 0 &2 &0 \ 0 &0 &2 \end{bmatrix} =egin{bmatrix} 0 &0 &0 \ 0 &0 &0 \ 0 &0 &0 \end{bmatrix} =O$

Comments

Ashish Jain

Question 16:

If and, find k so that

Solution 16:

$A^{2}=kA-2I$

$\Rightarrow egin{bmatrix} 3 &-2 \ 4 &-2 \end{bmatrix}^{2}=kegin{bmatrix} 3 &-2 \ 4 &-2 \end{bmatrix}-2egin{bmatrix} 1 &0 \ 0 &1 \end{bmatrix}$

$\Rightarrow egin{bmatrix} 1 &-2 \ 4 &-4 \end{bmatrix}=egin{bmatrix} 3k-2 &-2k \ 4k &-2k-2 \end{bmatrix}$

$\Rightarrow 3k-2=1$

$\Rightarrow k=1$

Comments

Ashish Jain

Question 17:

Ifand I is the identity matrix of order 2, show that

Solution 17:

Comments

Harshit Bhardwaj

Question 18:

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

Solution 18:

(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:

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:

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.

Comments

Harshit Bhardwaj

Question 19:

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.

Solution 19:

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.

Comments

Harshit Bhardwaj

Question 20:

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

A. k = 3, p = n

B. k is arbitrary, p = 2

C. p is arbitrary, k = 3

D. k = 2, p = 3

Solution 20:

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 will be defined.

Comments

Harshit Bhardwaj

Question 21:

Assume X, Y, Z, W and P are matrices of order, and respectively. If n = p, then the order of the matrix is

(A) p × 2 (B) 2 × n (C) n × 3 (D) p × n

Solution 21:

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.

Comments

Abhay

All Exercises - Chapter 3 - Matrices

Exercise 3.1

Exercise 3.2

Exercise 3.3

Exercise 3.4

Miscellaneous Exercise 3

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Learn Exercise 3.2 with Free Lessons & Tips

All Exercises - Chapter 3 - Matrices

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