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# Solutions 3. Matrices - Exercise 3.2 | Class 12 Mathematics-I - Toppers Study

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## Solutions 3. Matrices - Exercise 3.2 | Class 12 Mathematics-I - Toppers Study

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### You can Find Mathematics-I solution Class 12 Chapter 3. Matrices

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Chapter 3 Mathematics-I class 12

### Exercise 3.2 class 12 Mathematics-I Chapter 3. Matrices

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- Solutions 3. Matrices - Exercise 3.2 | Class 12 Mathematics-I - Toppers Study
- Class 12 Ncert Solutions
- Solution Chapter 3. Matrices Class 12
- Solutions Class 12
- Chapter 3. Matrices Exercise 3.2 Class 12

## 3. Matrices

### | Exercise 3.2 |

## Solutions 3. Matrices - Exercise 3.2 | Class 12 Mathematics-I - Toppers Study

**Exercise 3.2**

**Ques.1. Let A = B = C = . Find each of the following:**

**(i) A + B**

**(ii) A – B**

**(iii) 3A – C**

**(iv) AB**

**(v) BA**

**Ans. (i)** A + B = =

**(ii)** A – B = =

**(iii)** 3A – C = =

**(iv)** AB = =

**(v)** BA = =

**Ques.****2****. Compute the following:**

**(i) **

**(ii) **

**(iii) **

**(iv) **

**Ans. (i) ** =

**(ii)**

=

**(iii)** =

**(iv)** =

**Ques.3. Compute the indicated products:**

**(i) **

**(ii) **

**(iii) **

**(iv) **

**(v) **

**(vi) **

**Ans. (i)** =

**(ii)** =

**(iii)** =

**(iv)**

=

=

**(v)**

=

**(vi)**

=

**Ques.4. If A = B = and C = then compute (A + B) and (B – C). Also, verify that A + (B – C) = (A + B) – C.**

**Ans. **A + B = = =

B – C = = =

Now, A + (B – C) = (A + B) – C

=

=

=

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

hance Proved.

**Ques.5. If A = and B = then compute 3A – 5B.**

**Ans. **3A – 5B =

=

=

**Ques.6. Simplify: **

**Ans. **Given:

=

=

**Ques.7. Find X and Y, if:**

**(i) X + Y = and X – Y = **

**(ii) 2X + 3Y = and 3X + 2Y = **

**Ans. (i)** Given: X + Y = …..(i)

and X – Y = …..(ii)

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

2X =

X =

Subtracting eq. (i) and (ii), we get

2Y =

Y =

**(ii)** Given: 2X + 3Y = …..(i)

and 3X + 2Y = …..(ii)

Multiplying eq. (i) by 2, 4X + 6Y = ……….(iii)

Multiplying eq. (ii) by 3, 9X + 6Y = ………(iv)

subtracting Eq. (iii) from Eq. (iv)

5X =

=

X =

Now, From eq. (i), 3Y = 2X =

3Y = =

Y =

**Ques.8. Fin X if Y = and 2X + Y = **

**Ans. **2X + Y =

2X = – Y

2X =

2X =

X = =

**Ques.9. Find and if **

**Ans. **Given:

Equating corresponding entries, we have

and

and

and

and

**Ques.10. Solve the equation for and if **

**Ans. **Given:

Equating corresponding entries, we have

And

And

And

, , ,

**Ques.11. If find the values of and **

**Ans. **Given:

Equating corresponding entries, we have

……….(i) and ……….(ii)

Adding eq. (i) and (ii), we have

Putting in eq. (ii),

**Ques.12. Given: find the values of and **

**Ans. **Given:

Equating corresponding entries, we have

And

And ……….(i)

And

Putting in eq. (i),

, , ,

**Ques.13. If show that **

**Ans. **Given: ……….(i)

Changing to in eq. (i),

L.H.S. =

=

=

=

= R.H.S. [changing to in eq. (i)]

**Ques.14. Show that:**

**(i) **

**(ii) **

**Ans. (i)** L.H.S. = = =

R.H.S. = = =

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

**(ii)** L.H.S. =

=

=

R.H.S. =

=

=

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

**Ques.15. Find A2 – 5A + 6I if A = .**

**Ans. **A2 – 5A + 6I =

=

= =

=

**Ques.16. If A = prove that A3 – 6A2 + 7A + 2I = 0.**

**Ans. **L.H.S. = A3 – 6A2 + 7A + 2I

=

=

=

= =

= =

= = 0 (Zero matrix)

= R.H.S.

hance Proved.

**Ques.17. If A = and I = find so that **

**Ans. **Given: A = and I =

Equating corresponding entries, we have

And and

**Ques.18. If A = and I is the identity matrix of order 2, show that **

**Ans. **L.H.S. = I + A =

and, I – A =

R.H.S. = =

=

=

=

= = =

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

hance Proved.

**Ques.19. A trust fund has ` 30,000 that must be invested in two different types of bond. The first bond pays 5% interest per year and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide ` 30,000 in two types of bonds, if the trust fund must obtain an annual interest of (a) ` 1800, (b) ` 2000.**

**Ans. **Let the investment in first bond = ,

investment in the second bond = `

Interest paid by first bond = 5% = per rupee and

interest paid by second bond = 5% = per rupee.

Matrix of investment is A =

Matrix of annual interest per rupee B =

Matrix of total annual interest is AB =

=

=

=

Total annual interest = `

**(a)** According to question,

hance, Investment in first bond = ` 15,000

And Investment in second bond = ` (30000 – 15000) = ` 15,000

**(b)** According to question,

hance, Investment in first bond = ` 5,000

And Investment in second bond = ` (30000 – 15000) = ` 25,000

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

**Ans. **Let the number of books as a 1 x 3 matrix

B =

Let the selling prices of each book as a 3 x 1 matrix S =

Total amount received by selling all books =

=

=

hance, Total amount received by selling all the books = ` 20160

**Ques.21. The restriction on and so that PY + WY will be define are:**

**(A) **

**(B) is arbitrary, **

**(C) is arbitrary, **

**(D) **

**Ans. **Given:

Now,

On comparing, and

hance, option (A) is correct.

**Ques.22. If then order of matrix 7X – 5Z is:**

**(A) **

**(B) **

**(C) **

**(D) **

**Ans. **Here (given), the order of matrices X and Z are equal.

7X – 5Z is well defined and the order of 7X – 5Z is same as the order of X and Z.

The order of 7X – 5Z is either equal to or

But it is given that

hance, the option (B) is correct.

##### Other Pages of this Chapter: 3. Matrices

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