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TOPICS
Miscellaneous

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

Solution :-
```  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.
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Question-2 :-

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

Solution :-
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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.
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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.
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Question-6 :-  Find the values of x, y, z if the matrix satisfy the equation A′A = I.

Solution :-
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Question-7 :-

Solution :-
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Question-8 :-

Solution :-
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Question-9 :-

Solution :-
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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.

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:

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.
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```(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.
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Question-11 :-

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

Solution :-
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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.
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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