Question 1: Let , show that (aI + bA)ⁿ = aⁿ I + na^n-1 bA, where I is the identity matrix of order 2 and n ∈ N

Answer:

It is given that

To show:            P(n) : (aI + bA)ⁿ = aⁿ I + na^n-1 bA, n ∈ N

We shall prove the result by using the principle of mathematical induction.

For n = 1, we have:

P(1) : (aI + bA) = aI + ba⁰ A = aI + bA

Therefore, the result is true for n = 1.

Let the result be true for n = k.

That is,

P(k) : (aI + bA)^k = a^k I + ka^k-1 bA

Now, we prove that the result is true for n = k + 1. Consider

(aI + bA)^k+1 = (aI + bA)^k (aI + bA)

=(a^kI + ka^k-1 bA)(aI + bA)

=a^k+1 + ka^k bAI + a^k bIA + ka^k-1 b² A²

=a^k+1 + (k + 1) a^k bA + ka^k-1 b²A²          …………(1)

From (1), we have:

(aI + bA)^k+1 = a^k+1I + (k + 1) a^k bA + O

= a^k+1 I + (k + 1) a^kbA

Therefore, the result is true for n = k + 1.

Thus, by the principle of mathematical induction, we have:

Question 2: 

Answer:

We shall prove the result by using the principle of mathematical induction.

For n = 1, we have:

Therefore, the result is true for n = 1.

Let the result be true for n = k.

Now, we prove that the result is true for n = k + 1.

Now, A^k+1 = A.A^k

Therefore, the result is true for n = k + 1.

Thus by the principle of mathematical induction, we have:

Question 3:

Answer

We shall prove the result by using the principle of mathematical induction.

For n = 1, we have:

Therefore, the result is true for n = 1.

Let the result be true for n = k.

That is,

Now, we prove that the result is true for n = k + 1.

Consider

A^k+1 = A^k . A

Therefore, the result is true for n = k + 1.

Thus, by the principle of mathematical induction, we have:

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

Answer

It is given that A and B are symmetric matrices. Therefore, we have:

A’ = A and B’ = B                   ………….(1)

Now,  (AB - BA)’ = (AB)’ – (BA)’                   [(A-B)’ = A’ – B’]

= B’’ – A’B’                                                      [(AB)’ = B’A’ ]

= BA – AB                                                        [Using (1)]

= - (AB - BA)

∴ (AB - BA)’= -(AB - BA)

Thus, (AB − BA) is a skew-symmetric matrix.

 

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

Answer

We suppose that A is a symmetric matrix, then A’= A         … (1)

Consider  

(B’AB)’ = {B’(AB)}’

= (AB)’ (B’)’                      [(AB)’ = B’A’ ]

= B’A’ (B)                          [(B’)’ = B]

= B’ (A’B)

= B’ (AB)                            [Using (1)]

∴ (B’AB)’ = B’AB

Thus, if A is a symmetric matrix, then B’AB is a symmetric matrix.

Now, we suppose that A is a skew-symmetric matrix.

Then, A’ = A

Consider

(B’AB)’ = [B’ (AB)]’ = (AB)’ (B’)’

= (B’A’)B = B’ (-A) B

= - B’AB

∴ (B’AB)’ = B’AB

Thus, if A is a skew-symmetric matrix, then B’AB is a skew-symmetric matrix.

Hence, if A is a symmetric or skew-symmetric matrix, then B’AB is a symmetric or skew-symmetric matrix accordingly.

 

Question 6: Solve system of linear equations, using matrix method.

2x – y = -2

3x + 4y = 3

Answer

 The given system of equations can be written in the form of AX = B, where

Now,

|A| = 8+3 = 11 ≠ 0

Thus, A is non-singular. Therefore, its inverse exists.

Now,

Hence, x = (-5/11) and y = (12/11).

 

Question 7: For what values of

Answer

We have:

⇒  [6(0) + 2(2) + 4(x)] = O

⇒  [4 + 4x] = [0]

∴ 4 + 4x = 0

⇒ x = -1

Thus, the required value of x is −1.

 

Question 8:

If  , show that A² – 5A + 7I = 0

Answer

∴ L.H.S. = A² – 5A + 7I

= O = R.H.S.

∴ A² – 5A + 7I = 0

 

Question 9:

Answer

We have:

⇒ [x(x - 2) -40 + 2x - 8] = O

⇒ [x² - 2x -  40 + 2x - 8] = O

⇒ [x² - 48] = [0]

∴ x² - 48 = 0

⇒ x² = 48

Question 10: A manufacturer produces three products x, y, z which he sells in two markets.

Annual sales are indicated below:

Market	Products
I	10000	2000	18000
II	6000	20000	8000

(a) If unit sale 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.

Answer

(a) The unit sale prices of x, y, and z are respectively given as Rs 2.50, Rs 1.50, and Rs

1.00.

Consequently, the total revenue in market I can be represented in the form of a matrix as:

= 10000 x 2.50 + 2000 x 1.50 + 18000 x 1.00

= 25000 + 3000 + 18000

= 46000

The total revenue in market II can be represented in the form of a matrix as:

= 6000 x 2.50 + 20000 x 1.50 + 8000 x 1.00

= 15000 + 30000 + 8000

= 53000

Therefore, the total revenue in market I isRs 46000 and the same in market II isRs 53000.

(b) The unit cost prices of x, y, and z are respectively given as Rs 2.00, Rs 1.00, and 50 paise.

Consequently, the total cost prices of all the products in market I can be represented in the form of a matrix as:

= 10000 x 2.00 + 2000 x 1.00 + 18000 x 0.50

= 20000 + 2000 + 9000

= 31000

Since the total revenue in market I isRs 46000, the gross profit in this marketis (Rs 46000 − Rs 31000) Rs 15000.

The total cost prices of all the products in market II can be represented in the form of a matrix as:

= 6000 x 2.00 + 20000 x 1.00 + 8000 x 0.50

= 12000 + 20000 + 4000

= RS 36000

Since the total revenue in market II isRs 53000, the gross profit in this market is (Rs 53000 − Rs 36000) Rs 17000.

 

Question 11:

Find the matrix X so that

Answer

The matrix given on the R.H.S. of the equation is a 2 × 3 matrix and the one given on the L.H.S. of the equation is a 2 × 3 matrix. Therefore, X has to be a 2 × 2 matrix.

Now, let

Therefore, we have:

Equating the corresponding elements of the two matrices, we have:

a + 4c = -7,         2a + 5c = -8,       3a + 6c = -9

b + 4d = 2,          2b + 5d = 4,        3b + 6d = 6

Now, a + 4c = -7 ⇒ a = -7 – 4c

∴ 2a + 5c = -8 ⇒ -14 – 8c + 5c = -8

⇒ -3c = 6

⇒ c = -2

∴ a = -7 – 4(-2)= -7 + 8 = 1

Now, b + 4d = 2 ⇒ b = 2 – 4d

∴ 2b + 5d = 4 ⇒ 4 – 8d + 5d = 4

⇒ -3d = 0

⇒ d = 0

∴ b = 2 – 4(0) = 2

Thus, a = 1, b = 2, c = −2, d = 0

Hence, the required matrix X is

Question 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

Answer

A and B are square matrices of the same order such that AB = BA.

To prove:           P(n) : ABⁿ = Bⁿ A, n ∈ N

For n = 1, we have:

P(1) : AB = BA                   [Given]

⇒ AB¹ = B¹A

Therefore, the result is true for n = 1.

Let the result be true for n = k.

P(k) : AB^k = B^k A               ….(1)

Now, we prove that the result is true for n = k + 1.

AB^k+1 = AB^k . B

= (B^k A)B                            [By (1)]

= B^k(AB)                             [Associative law]

= B^k (BA)                            [AB = BA (given)]

= (B^k B)A                            [Associative law]

= B^k+1 A

Therefore, the result is true for n = k + 1.

Thus, by the principle of mathematical induction, we have

Now, we prove that (AB)ⁿ = AⁿBⁿ for all n ∈ N

For n = 1, we have:

(AB)¹ = A¹B¹ = AB

Therefore, the result is true for n = 1.

Let the result be true for n = k.

(AB)^k = A^kB^k                       ….(2)

Now, we prove that the result is true for n = k + 1.

(AB)^k+1 = (AB)^k . (AB)

= (A^kB^k) . (AB)                   [by (2)]

= A^k (B^kA)B                        [Associative law]

= A^k (AB^k)B                        [ABn = BnA for all n ∈ N]

=  (AkA) . (Bk B)               [Associative law]

= A^k+1 B^k+1

Therefore, the result is true for n = k + 1.

Thus, by the principle of mathematical induction, we have (AB)n = Aⁿ Bⁿ, for all natural numbers.

 

Question 13: Choose the correct answer in the following questions:

If  is such that A² = I then

A. 1 + α² + βγ = 0

B. 1 - α² + βγ = 0

C. 1 - α² - βγ = 0

D. 1 + α² - βγ = 0

Answer

Answer: C

On comparing the corresponding elements, we have:

α² + βγ = 1

⇒ α² + βγ – 1 = 0 

⇒ 1 - α² + βγ = 0

 

Question 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 Answer

Answer: B

If A is both symmetric and skew-symmetric matrix, then we should have

A’ = A and A’ = -A

⇒ A = - A

⇒ A + A = O

⇒ 2A = O

⇒ A = O

Therefore, A is a zero matrix.

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

A. A

B. I − A

C. I

D. 3A

Answer

Answer: C

(I + A)³ – 7A = I³ + A³ + 3I²A + 3A²I – 7A

= I + A³ + 3A + 3A² – 7A

= I + A² .A + 3A + 3A – 7A             [A² = A]

= I + A .A – A

= I + A² – A

= I + A – A

= I

∴ (I + A)³ – 7A = I