# Class 12 NCERT Solutions- Mathematics Part I – Chapter 3 Matrices – Exercise 3.2 | Set 2

Last Updated : 03 Apr, 2024

### Question 11. If , find the values of x and y.

Solution:

Given:

Equating corresponding entries, we have

2x – y = 10           -(1)

3x + y = 5           -(2)

Adding eq.(1) and (2), we have 5x = 15 â‡’ x = 3

Putting x = 3 in eq.(2)

9 + y = 5 â‡’ y = -4

Therefore, x = 3 and y = -4

### Question 12. Given , find the values of x, y, z and w.

Solution:

Given:

Equating corresponding entries, we have

3x = x + 4 â‡’ 2x = 4 â‡’ x = 2

and 3y = 6 + x + y

â‡’ 2y = 6 + 2

â‡’ 2y = 8

â‡’ y = 4

and 3z = -1 + z + w â‡’ 2z – w = – 1           -(1)

and 3w = 2w + 3 â‡’ w = 3

Putting w = 3 in eq(i), 2z – 3 = -1

â‡’ 2z = 2 â‡’ z = 1

Therefore, x = 2, y = 4, z = 1, w = 3

### Question 13. If , show that F(x) F(y) = F(x + y).

Solution:

= F(x + y)

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

### Question 14. Show that

Solution:

(i) L.H.S =

R.H.S =

Therefore, from (1) and (2), we get

i.e. L.H.S. â‰  R.H.S

(ii) L.H.S =

Multiply both the matrices

R.H.S.=

Therefore,

L.H.S. â‰  R.H.S.

i.e.

Solution:

### Question 16. If , prove that A3 â€“ 6A2 + 7A + 2I = 0

Solution:

= 0 (Zero matrix)

= R.H.S.

Hence Proved

### Question 17. If , find k so that A2 = kA â€“ 2I

Solution:

Given:

Equating corresponding entries, we have

3k – 2 = 1

3k = 3

k = 1

and 4k = 4

k = 1

and -4 = -2k – 2

2k = 2

k = 1

Therefore, k = 1

Solution:

L.H.S. = R.H.S.

Hence, Proved.

### (b) Rs.2000

Solution:

Let invested in the first bond = Rs x

Then, the sum of money invested in the second bond = â‚¹(30000 â€“ x)

It is given that the first bond pays 5% interest per year, and the second bond pays 7% interest per year.

Thus, in order to obtain an annual total interest of â‚¹1800, we get:

â‡’ 5x/100 + 7(30000 âˆ’ x)/100 = 1800

â‡’ 5x + 210000 -7x = 180000

â‡’ 210000 -2x = 180000

â‡’ 2x = 210000 â€“ 180000

â‡’ 2x = 30000

â‡’ x = 15000

Therefore, in order to obtain an annual total interest of â‚¹1800, the trust fund should invest â‚¹15000 in the first bond and the remaining â‚¹15000 in the second bond.

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 =

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

â‡’ 5x/100 + 7(30000 âˆ’ x)/100 = 2000

â‡’ 5x + 210000 âˆ’ 7x = 200000

â‡’ 210000 âˆ’ 2x = 200000

â‡’ 2x = 210000 â€“ 200000

â‡’ 2x = 10000

â‡’ x = 5000

Therefore, in order to obtain an annual total interest of â‚¹2000, the trust fund should invest â‚¹5000 in the first bond and the remaining â‚¹(30000 âˆ’ 5000) = â‚¹25000 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.

Solution:

Let the number of books as 1 Ã— 3 matrix = B =

Let the selling prices of each book is a 3 Ã— 1 matrix S =

Therefore, Total amount received by selling all books = BS =

Therefore, Total amount received by selling all the books = Rs 20,160

### Question 21. The restriction on n, k and p so that PY + WY 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:

Since, Matrices P and Y are of the orders p Ã— k and 3 Ã— k respectively.

Therefore, matrix PY will be defined if k = 3.

Then, PY will be of the order p Ã— k = p Ã— 3.

Matrices W and Y are of the orders n Ã— 3 and 3 Ã— k = 3 Ã— 3 respectively.

As, 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 Ã— 3.

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

Therefore, PY is of the order p Ã— 3 and WY is of the order n Ã— 3.

Thus, we must have p = n.

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

### Question 22. If n = p, then the order of the matrix 7X â€“ 5Z is:

(A) p Ã— 2                (B) 2 Ã— n

(C) n Ã— 3                (D) p Ã— n

Solution:

Matrix X is of the order 2 Ã— n.

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

Matrix Z is of order 2 Ã— p = 2 Ã— n               -(âˆµ p = n)

Then, Matrix 5Z is also of the same order.

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

Thus, matrix 7X â€“ 5Z is well- defined and is of the order 2 Ã— n.

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