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Miscellaneous

Matrices

**Question-1 :-**
where I is the identity matrix of order 2 and n ∈ N.

Given that : To Show : (aI + bA)ⁿ = aⁿI + naⁿ¯¹bA, where I is the identity matrix of order 2 and n ∈ N. By using Principal of Mathematical Induction, we get P(n): (aI + bA)ⁿ = aⁿI + naⁿ¯¹bA, n∈N P (1): (aI + bA)¹ = a¹I + 1.a¹¯¹bA ; (aI + bA) = aI + bA Therefore, the result is true for n = 1. Let the result be true for n = k. So P(k): (aI + bA)ᴷ = aᴷI + kaᴷ¯¹bA Now, we prove that the result holds for n = k+1 (aI + bA)ᴷ⁺¹ = (aI + bA)ᴷ.(aI + bA) = (aᴷI + kaᴷ¯¹bA)(aI + bA) = aᴷ⁺¹I² + aᴷbA + kaᴷbAI + kaᴷ¯¹b²A² = aᴷ⁺¹I + aᴷbA(k+1) + kaᴷ¯¹b²A² ......(I) Now, From (1), we get (aI + bA)ᴷ⁺¹ = aᴷ⁺¹I + aᴷbA(k+1) + O = aᴷ⁺¹I + aᴷbA(k+1) Therefore, the result is true for n = k + 1. Thus by principle of mathematical induction, we have (aI + bA)ⁿ = aⁿI + naⁿ¯¹bA, n∈N.

**Question-2 :-**

**Question-3 :-**
, where n is any positive integer.

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

It is given that A and B are symmetric matrices. Therefore, we have: A' = A and B' = B ......(I) Now, (AB - BA)' = (AB)' - (BA)' [(A - B)' = A' - B'] = B'A' - A'B' [(AB)' = B'A'] = BA - AB [Using I] = -(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.

We suppose that A is a symmetric matrix, then A'= A ...... (I) Now, (B'AB)' = {B'(AB)}' = (AB)'(B')' [(AB)' = B'A'] = B'A'(B) [(B')' = B] = B'(A'B) = B'(AB) [Using I] 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 .......(II) Now, (B'AB)' = {B'(AB)}' = (AB)'(B')' [(AB)' = B'A'] = B'A'(B) [(B')' = B] = B'(A'B) = B'(-AB) [Using II] = -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 :-**
Find the values of x, y, z if the matrix
satisfy the equation A′A = I.

**Question-7 :-**

**Question-8 :-**

**Question-9 :-**

**Question-10 :-**
A manufacturer produces three products x, y, z which he sells in two markets. Annual sales are indicated below:
(a) If unit sale prices of x, y and z are ₹ 2.50, ₹ 1.50 and ₹ 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 ₹ 2.00, ₹ 1.00 and 50 paise respectively. Find the gross profit.

(a) The unit sale prices of x, y, and z are respectively given as ₹ 2.50, ₹ 1.50, and ₹ 1.00. Consequently, the total revenue in market I can be represented in the form of a matrix as: The total revenue in market II can be represented in the form of a matrix as: Therefore, the total revenue in market I is ₹ 46000 and the same in market II is ₹ 53000.

(b) ) The unit cost prices of x, y, and z are respectively given as ₹ 2.00, ₹ 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: Since the total revenue in market I is ₹ 46000, the gross profit in this marketis (₹ 46000 − ₹ 31000) ₹ 15000. The total cost prices of all the products in market II can be represented in the form of a matrix as: Since the total revenue in market II is ₹ 53000, the gross profit in this market is (₹ 53000 − ₹ 36000) = ₹ 17000.

**Question-11 :-**

**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.

A and B are square matrices of the same order such that AB = BA. To Prove :- P(n) : ABⁿ = BⁿA for all n ∈ N For n = 1, we get 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) : ABᴷ = BᴷA ......(I) Now, we prove that the result is true for n = k + 1. ABᴷ⁺¹ = ABᴷ.B = (BᴷA)B [By I] = Bᴷ(AB) [By Associative Law] = Bᴷ(BA) [AB = BA Given] = (BᴷB)A [By Associative Law] = Bᴷ⁺¹A Therefore, the result is true for n = k + 1. Thus, by the principle of mathematical induction, we have ABⁿ = BⁿA, for all n ∈ N Now, we prove that (AB)ⁿ = AⁿBⁿ for all n ∈ N For n = 1, we get P(1) : (AB)¹ = A¹B¹ = AB Therefore, the result is true for n = 1. Let the result be true for n = k. P(k) : (AB)ᴷ = AᴷBᴷ ......(II) Now, we prove that the result is true for n = k + 1. (AB)ᴷ⁺¹ = (AB)ᴷ.(AB) = AᴷBᴷ(AB) = Aᴷ(BᴷA)B = Aᴷ(ABᴷ)B = (AᴷA)(BBᴷ) = Aᴷ⁺¹Bᴷ⁺¹ Therefore, the result is true for n = k + 1. Thus, by the principle of mathematical induction, we have (AB)ⁿ = AⁿBⁿ, for all natural numbers.

**Question-13 :-**
(A) 1 + α² + βγ = 0 (B) 1 - α² + βγ = 0 (C) 1 - α² - βγ = 0 (D) 1 + α² - βγ = 0

The correct answer is C.

**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

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. The correct answer is B.

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

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

(I + A)³ - 7A = I³ + A³ + 3I²A + 3A²I - 7A = I + A³ + 3A + 3A² - 7A = I + A².A + 3A + 3A - 7A = I + A.A - A = I + A² - A = I + A - A = I The correct answer is C.

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