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Miscellaneous

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

Solution :-
  Given that : matrices
  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, matrices
  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 :-  matrices

Solution :-
    matrices
    

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

Solution :-
    matrices
    

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

Solution :-
  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.

Solution :-
  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 matrices satisfy the equation A′A = I.

Solution :-
   matrices
    

Question-7 :-  matrices

Solution :-
 matrices
    

Question-8 :-  matrices

Solution :-
    matrices
    

Question-9 :-  matrices

Solution :-
    matrices
    

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

Solution :-
(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:
    matrices
  The total revenue in market II can be represented in the form of a matrix as:
    matrices
  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:
    matrices
  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:
    matrices
  Since the total revenue in market II is ₹ 53000, the gross profit in this market is (₹ 53000 − ₹ 36000) = ₹ 17000.
    

Question-11 :-  matrices

Solution :-
 matrices
    

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.

Solution :-
  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 :-  matrices (A) 1 + α² + βγ = 0   (B) 1 - α² + βγ = 0   (C) 1 - α² - βγ = 0   (D) 1 + α² - βγ = 0

Solution :-
  matrices
  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

Solution :-
  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

Solution :-
  (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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