Matrices(Miscellaneous Ex.)

Miscellaneous Exercise Chapter 3

Solun:- Given Eq. is (aI+bA)n=anI+nan-1bA

Using mathematical induction:-

Put n=1:-

Thus, given eq. is true for every n∈N.

Solun:- Using mathematical induction:-

Put n=1:-

is true for every n∈N.

Solun:- Using mathematical induction:-

Thus, given eq. is true for every positive integer.

4. If A and B are symmetric matrices, prove that AB-BA is a skew-symmetric matrix.

Solun:- Given A and B are symmetric matrix

Then A=A’

⇒ B=B’

⇒ (AB-BA)’ = (AB)’ - (BA)’  {We know (A-B)’=A’ - B’ }

⇒ (AB-BA)’ = B’A’ - A’B’      {We know (AB)’ = B’A’}

⇒ (AB-BA)’ = BA - AB   (Given)

⇒ (AB-BA)’ = - (AB-BA)

So, (AB - BA) is the skew-symmetric matrix.

5. Show that the matrix B’AB is symmetric or skew-symmetric matrix according as A is symmetric or skew-symmetric.

Solun:- Given matrix is B’AB

⇒ (B’AB)’ = (AB)’(B’)’

We know (AB)’ = B’A’ and (B’)’ = B

⇒ (B’AB)’ = B’ A’ B

(i) If A is symmetric matrix then A=A’

⇒ (B’AB)’ = B’AB 

So, if A is a symmetric matrix then (B’AB)’ is also a symmetric matrix.

(ii) If A is skew-symmetric matrix then A’ = - A

⇒ (B’AB)’ = B’(- A)B

⇒ (B’AB)’ = -B’AB

So, if A is a skew-symmetric matrix then (B’AB)’ is also a skew-symmetric matrix.

6. Find the values of x, y, z if the matrix 

The order of both the matrices is the same then compare corresponding elements.

⇒ 4y2+z2 = 1   ………(1)

⇒ 2y2-z2 = 0    (By elimination method)

⇒ 6y2 =1

⇒ y2 = 1/6

Put the value of y in eq. 1:-

⇒ 4/6+z2 = 1

⇒ z2 = 1/3

And x2+y2+z2 = 1

Put the values of y and z:-

⇒ x2 + 1/6 + 1/3 =1

⇒ x2 = 1/2

⇒ 4x+4 = 0

⇒ x = -4/4 = -1

L.H.S = R.H.S (Hence Proved……)

10. A manufacturer produces three products x,y,z which he sells in two markets. Annual sales are indicated below:

Market	Products
I	10,000	2,000	18,000
II	6,000	20,000	8,000

(a) If unit sales prices of x,y, and z  are Rs 2.50, Rs 1.50 and Rs 1.00 respectively, find the total revenue in each market with the help of matrix algebra.

(b) If the unit costs of the above three commodities are Rs 2.00, Rs 1.00, and 50 paise respectively. Find the gross profit.

Solun:- Total revenue is:-

Total revenue:-Market I = Rs 46000

Market II = Rs 53000

Gross Profit:-

Market I = Rs 15000

Market II = Rs 17000

Solun:- For multiplication of X and left side matrix:-

No. of columns of X = No. of rows of left side matrix

⇒ No. of columns of X = 2

⇒ No. of rows of X = No. of rows of right side matrix

⇒ No. of rows of X = 2

⇒ a+4b = -7 ………....(1)

⇒ 3a+6b = -9

⇒ a+2b = -3     (By elimination method)

⇒ 2b = -4

⇒ b = -2

Put the value of b in eq. 1:-

⇒ a = 1

⇒ c+4d = 2 ……....(2)

⇒ 3c+6d = 6

⇒ c+2d=2     (By elimination method)

⇒ 2d = 0

⇒ d =0

Put the value of d in eq. 2:-

⇒ c=2

⇒ a=1, b=-2, c=2, d=0

12. If A and B are square matrices of the same order such that AB = BA, then prove

by induction that ABn = BnA. Further prove that (AB)n = AnBn for all n∈N.

Solun:- Given AB=BA

Using mathematical induction:-

Put n=1:-

⇒ ABn=BnA

⇒ AB=BA (True)

Put n=k:-

⇒ ABk=BkA (True) ………(1)

Prove:- Put n = k+1

⇒ ABk+1=Bk+1A ………....(2)

Taking L.H.S:

= ABk+1

= ABkxB   (From Eq. 1)

= BkAxB

= BkBA   (Given AB = BA)

⇒ Bk+1A=R.H.S  (Hence Proved) 

⇒ (AB)n=(AnBn)

Using mathematical induction:-

Put n=1:-

⇒ AB=BA (True)

Put n=k:-

⇒ (AB)k=AkBk (True) ………(3)

Prove:- Put n = k+1

⇒ (AB)k+1=Ak+1Bk+1

Taking L.H.S:

= (AB)k+1

= (AB)k x AB   (From Eq. 3)

= AkBk x AB

= AkBkx BA  (Given AB = BA)

= AkBk+1A  (From eq. 2)

= AkABk+1

= Ak+1Bk+1 (Hence Proved……....)

Hence these equations satisfy n∈N.

(Given A2 = I)

The order of both the matrices is the same then compare corresponding elements.

The answer is ...C

14. If the matrix A is both symmetric and skew-symmetric, then:

(A) A is a diagonal matrix  

(B) A is a zero matrix

(C) A is a square matrix  

(D) None of these

Solun:- Given A is symmetric matrix A’=A …….......(1)

⇒ A is the skew-symmetric matrix A’=-A  …………....(2)

From Eq. 1 and 2:-

⇒ A = -A

⇒ 2A = 0

⇒ A = 0

Then A is a zero matrix.

Answer is…B

15. If A is square matrix such that A2 = A, then (I+A)3-7A is equal to:-

(A) A

(B) I-A
(C) I
(D) 3A

Solun:- Given eq. is (I+A)3-7A

We know (A+B)3 = A3+B3+3AB(A+B)

= (I+A)3-7A

= I3+A3+3AI(A+I)-7A

= I+A2A+3A(A+I)-7A  (We know I3=I and AI = A) 

= I+AA+3A2+3AI -7A   (Given A2=A)

= I+A+3A+3A-7A

= I+7A-7A

=I

Answer is……C

Download Pdf of Miscellaneous Ex.

See also:- Exercise 3.3

Exercise 3.2

Exercise 3.1

Check notes of chapter 3

If you have any queries, you can ask me in the comment section

And you can follow/subscribe me for the latest updates on your e-mails

For subscribing me follow these instructions:-

1. Fill your E-mail address

2. Submit Recaptcha

3. Go to your email and then click on the verify link

Then you get all update on your email

Thanks for Reading ......