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NCERT Solutions class 12 Maths Matrices

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 =  =

2. Compute the following:

(i)

(ii)

(iii)

(iv)

Ans. (i)   =

(ii)

=

(iii)   =

(iv)  =

3. Compute the indicated products:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Ans. (i)  =

(ii)  =

(iii)  =

(iv)

=

=

(v)

=

(vi)

=

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.  Proved.

Ans. 3A – 5B =

=

=

6. Simplify:

Ans. Given:

=

=

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)

Eq. (iv) – Eq. (iii) = 5X =  =

X =

Now, From eq. (i),  3Y =  2X =

3Y =  =

Y =

Ans. 2X + Y =

2X =  – Y

2X =

2X =

X =  =

9. Find  and  if

Ans. Given:

Equating corresponding entries, we have

and

and

and

and

10. Solve the equation for  and  if

Ans. Given:

Equating corresponding entries, we have

And

And

And

, , ,

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),

12. Given:  find the values of  and

Ans. Given:

Equating corresponding entries, we have

And

And      ……….(i)

And

Putting  in eq. (i),

, , ,

13. If  show that

Ans. Given:    ……….(i)

Changing  to  in eq. (i),

L.H.S. =

=

=

=

= R.H.S. [changing  to  in eq. (i)]

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.

15. Find A2 – 5A + 6I if A = .

Ans. A2 – 5A + 6I =

=

=  =

=

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

Ans. L.H.S. = A3 – 6A2 + 7A + 2I

=

=

=

=  =

=  =

=  = 0 (Zero matrix)

= R.H.S.      Proved.

17. If A =  and I =  find  so that

Ans. Given:  A =  and I =

Equating corresponding entries, we have

And       and

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

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

Now, I – A =

R.H.S. =  =

=

=

=

=  =  =

L.H.S. = R.H.S.         Proved.

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 be  , then the 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,

Therefore, Investment in first bond =  15,000

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

(b) According to question,

Therefore, Investment in first bond =  5,000

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

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 = BS =

=  =

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

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

Therefore, option (A) is correct.

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

Therefore, the option (B) is correct.

NCERT Solutions class 12 Maths Exercise 3.2

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