Note: If two matrices have different orders then we can neither add nor subtract the matrices but we can multiply the matrices either the orders are same or number of columns of first matrix is equal to the number of rows of second matrix.

By definition of equality of matrix as the given matrices are equal, then corresponding elements are equal. Comparing the corresponding elements, we get

2+y = 5

and 2x+2 = 8

→ y = 5-2 = 3 and 2x=8 – 2 → y=3 and x = 6/2 = 3

2+y = 5

and 2x+2 = 8

→ y = 5-2 = 3 and 2x=8 – 2 → y=3 and x = 6/2 = 3

By definition of equality of matrix as the given matrices are equal, then corresponding elements are equal. Comparing the corresponding elements, we get

2x-y = 10 …… (i)

and 3x+y = 5 ….(ii)

Adding Eq. (i) and (ii), we get

5x = 15 → x = 3

Substituting x = 3 in Eq. (i), we get

2×3 – y=10 → y=6-10 = -4

2x-y = 10 …… (i)

and 3x+y = 5 ….(ii)

Adding Eq. (i) and (ii), we get

5x = 15 → x = 3

Substituting x = 3 in Eq. (i), we get

2×3 – y=10 → y=6-10 = -4

By definition of equality of matrix as the given matrices are equal, then corresponding elements are equal. Comparing the corresponding elements, we get

3x = x+4 → 2x = 4 → x = 2

Hence, the values of x, y, z and w are 2, 4, 1 and 3

3x = x+4 → 2x = 4 → x = 2

Hence, the values of x, y, z and w are 2, 4, 1 and 3

By definition of equality of matrix as the given matrices are equal, their corresponding elements are equal. Comparing the corresponding elements, we get

3k-2 = 1 → k=1

-2k = -2 → k=1

4k = 4 → k=1

-4 = -2k-2 → k=1

Hence k=1

3k-2 = 1 → k=1

-2k = -2 → k=1

4k = 4 → k=1

-4 = -2k-2 → k=1

Hence k=1

20.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

Rs. 1800

Rs. 2000

Rs. 1800

Rs. 2000

Sol. Let the amount invested in first type of bonds is Rs. x, then that invested in second type of bonds will be Rs. (30,000-x).

According to given condition,

Hence, the amounts invested in the two types of bonds are respectively Rs. 15000 and Rs. (30000-15000) = Rs. 15000

Hence, the amounts invested in two types of bonds are respectively Rs. 5000 and Rs. (30000-5000) = Rs. 25000

According to given condition,

Hence, the amounts invested in the two types of bonds are respectively Rs. 15000 and Rs. (30000-15000) = Rs. 15000

Hence, the amounts invested in two types of bonds are respectively Rs. 5000 and Rs. (30000-5000) = Rs. 25000

21.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.

Amount received by the bookseller on selling the types of books can be computed by evaluating the product AB.

Direction (Q. No. 22 and 23) Assume X, Y, Z, W and P are matrices of orders 2 x n, 3 x k, 2 x p, n x 3 and p x k respectively. Choose the correct answer in Q. 22 and Q. 23

22.The restrictions 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=2, p=3

22.The restrictions 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=2, p=3

Sol. Matrices P and Y are of the orders p x k and 3 x k respectively. Therefore, matrix PY will be defined if k=3. Consequently, PY will be of the order p x k. Matrices W and Y are of the orders n x 3 and 3 x k respectively.

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

Matrices PY and WY can be added only when their orders are the same. However, PY is of the order p x k and WY is of the order n x 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. So, correct option is (a).

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

Matrices PY and WY can be added only when their orders are the same. However, PY is of the order p x k and WY is of the order n x 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. So, correct option is (a).

23.If n = p, then the order of the matrix 7X – 5Z is

P x 2

2 x n

n x 3

p x n

P x 2

2 x n

n x 3

p x n

Sol. Matrix X is of the order 2 x n

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

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

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

Now both the matrices 7X and 5Z are of the same order 2 x n.

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

So, correct option is (b)

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

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

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

Now both the matrices 7X and 5Z are of the same order 2 x n.

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

So, correct option is (b)