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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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Chapter 3 Mathematics-I class 12

Exercise 3.2 class 12 Mathematics-I Chapter 3. Matrices

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.

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