Class 12

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.2

Question 1.
Find the rank of the following matrices by the minor method:


Solution:

Question 2.
Find the rank of the folowing matrices by row reduction method:

Solution:
(i) Let

The last equivalent matrix is in row-echelon form. It has three non zero rows. So ρ(A) = 3
(ii) Let

The last equivalent matrix is in row-echelon form. It has three non zero rows. ρ(A) = 3
(iii) Let

The last equivalent matrix is in row-echelon form. It has three non zero rows. ρ(A) = 3

Question 3.
Find the inverse of each of the following by Gauss-Jordan method:

Solution:
(i) Let \(A=\left(\begin{array}{cc}{2} & {-1} \\ {5} & {-2}\end{array}\right)\)
Applying Gauss-Jordan method we get

(ii) Let


(iii) Let

Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.2 Additional Problems

Question 1.
Find the rank of the following matrices.
Solution:

A has at least one non-zero minor of order 2. \(\rho(\mathrm{A})\) = 2

Question 2.
Find the rank of the following matrices.
Solution:

The last equivalent matrix is in the echelon form. It has three non-zero rows.
∴ \(\rho(\mathrm{A})\) = 3; Here A is of order 3 × 4

Question 3.
Find the rank of the following matrices.
Solution:


The last equivalent matrix is in the echelon form. The number of non-zero rows in this matrix is two. A is a matrix of order 3 × 4. ∴ \(\rho(\mathrm{A})\) = 2

Question 4.
Using elementary transformations find the inverse of the following matrix
Solution:

Question 5.
Using elementary transformations find the inverse of the following matrices
Solution:

Question 6.
Using elementary transformations find the inverse of the following matrices
Solution:

Question 7.
Using elementary transformations, find the inverse of the following matrices
Solution:

Question 8.
Using elementary transformations, find the inverse of the following matrices
Solution:

Question 9.
Using elementary transformations, find the inverse of the following matrices
Solution:

Question 10.
Using elementary transformations, find the inverse of the following matrices
Solution:

Since R₂ has all numbers zero, Thus inverse of matrix A does not exist.

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 12 Discrete Mathematics Ex 12.3

Choose the correct or the most suitable answer from the given four alternatives.

Question 1.
A binary operation on a set S is a function from …….
(a) S ➝ S
(b) (S × S) ➝ S
(c)S ➝ (S × S)
(d) (S × S) ➝ (S × S)
Solution:
(b) (S × S) ➝ S

Question 2.
Subtraction is not a binary operation in
(a) R
(b) Z
(c) N
(d) Q
Solution:
(c) N
Hint:
For example 2, 5 ∈ N but 2 – 5 = 3 ∉ N

Question 3.
Which one of the following is a binary operation on N?
(a) Subtraction
(b) Multiplication
(c) Division
(c) All the above
Solution:
(b) Multiplication

Question 4.
In the set R of real numbers ‘*’ is defined as follows. Which one of the following is not a binary operation on R?
(a) a * b = min (a, b)
(b) a * b = max (a, b)
(c) a * b = a
(d) a * b = a^b
Solution:
(d) a * b = a^b
Hint:
a^b ∉ R

Question 5.
The operation * defined by a * b = \(\frac{a b}{7}\) is not a binary operation on ……….
(a) Q⁺
(b) Z
(c) R
(c) C
Solution:
(b) Z
Hint:
Since 3, 5 ∈ Z, but \(\frac{3 \times 5}{7} \notin\) Z.

Question 6.



Solution:
(b) y = \(\frac{-2}{3}\)

Question 7.
If a * b = \(\sqrt{a^{2}+b^{2}}\) on the real numbers then * is ……..
(a) commutative but not associative
(b) associative but not commutative
(c) both commutative and associative
(d) neither commutative nor associative
Let, a, b ∈ R

(1) = (2) = * is associative
So * is both commutative and associative
Solution:
(c) both commutative and associative

Question 8.
Which one of the following statements has the truth value T?
(a) sin x is an even function.
(b) Every square matrix is non-singular.
(c) The product of a complex number and its conjugate is purely imaginary.
(d) √5 is an irrational number.
Solution:
(d) √5 is an irrational number.

Question 9.
Which one of the following statements has truth value F ?
(a) Chennai is in India or \(\sqrt{2}\) is an integer
(b) Chennai is in India or \(\sqrt{2}\) is an irrational number
(c) Chennai is in China or \(\sqrt{2}\) is an integer
(d) Chennai is in China or \(\sqrt{2}\) is an irrational number
Solution:
(c) Chennai is in China or \(\sqrt{2}\) is an integer

Question 10.
If a compound statement involves 3 simple statements, then the number of rows in the truth table is
(a) 9
(b) 8
(c) 6
(d) 3
Solution:
(b) 8
Hint:
Number of statements = 3
No of rows in truth table = 2³ = 8 rows

Question 11.
Which one is the inverse of the statement

Solution:
(a) \((\neg p \wedge \neg q) \rightarrow(\neg p \vee \neg q)\)

Question 12.
Which one is the contrapositive of the statement \((p \vee q) \rightarrow r\)?

Solution:
(a) \(\neg r \rightarrow(\neg p \wedge \neg q)\)

Question 13.
The truth table for \((p \wedge q) \vee \neg q\) is given below

Which one of the following is true?

Hint: The truth table for \((p \wedge q) \vee \neg q\)

Solution:
(3) T T F T

Question 14.
In the last column of the truth table for \(\neg(p \vee \neg q)\) the number of final outcomes of the truth value ‘F’ are
(a) 1
(b) 2
(c) 3
(d) 4
Hint:
The truth table for \(\neg(p \vee \neg q)\)

Solution:
(c) 3

Question 15.
Which one of the following is incorrect? For any two propositions p and q, we have …….

Solution:
(c) \(\neg(p \vee q) \equiv \neg p \vee \neg q\)

Question 16.

Which of the following is correct for the truth \((p \wedge q) \rightarrow \neg p\) ?

Hint:

Solution:
(2) F T T T

Question 17.


Solution:
(d) \(\neg(p \wedge q) \wedge | p \wedge(p \vee \neg r)]\)

Question 18.
The proposition p ∧ (¬p v q) is
(a) a tautology
(b) a contradiction
(c) logically equivalent to p ∧ q
(d) logically equivalent to p v q
Solution:
(c) logically equivalent to p ∧ q
Hint:
p ∧ (¬p v q) ≡ p ∧ q

Question 19.
Determine the truth value of each of the following statements:
(a) 4 + 2 = 5 and 6+ 3 = 9
(b) 3 + 2 = 5 and 6 + 1 = 7
(c) 4 + 5 = 9 and 1 + 2 = 4
(d) 3 + 2 = 5 and 4 + 7 = 11

Solution:
(1) F T F T

Question 20.
Which one of the following is not true?
(a) Negation of a statement is the statement itself.
(b) If the last column of the truth table contains only T then it is a tautology.
(c) If the last column of its truth table contains only F then it is a contradiction.
(d) If p and q are any two statements then p ↔ q is a tautology.
Solution:
(d) If p and q are any two statements then p ↔ q is a tautology.
Hint:
If p and q are any two statements, then P ↔ q is a tautology.
P ↔ q is actually a contingency.

Samacheer Kalvi 12th Maths Solutions Chapter 12 Discrete Mathematics Ex 12.3 Additional Problems

Choose the correct or the most suitable answer from the given four alternatives.

Question 1.
Which of the following are statements?
(i) May God bless you
(ii) Rose is a flower
(iii) milk is white
(iv) 1 is a prime number
(a) (i), (ii), (iii)
(b) (i), (ii), (iv)
(c) (i), (iii), (iv)
(d) (ii), (iii), (iv)
Hint:
Sentence (ii), (iii) and (iv) are statements
(ii) Rose is a flower – True
(iii) Milk is white – True
(iv) 1 is a prime number
∴ (ii), (iii), (iv) are statements
(i) May god bless you. This statement can not be assigned True or False.
∴ (i) is not a statements
Solution:
(d) (ii), (iii), (iv)

Question 2.
If a compound statement is made up of three simple statements, then the number of rows in the truth table is …….
(a) 8
(b) 6
(c) 4
(d) 2
Hint:
The number of rows in truth table = 2ⁿ = 2³ = 8
Solution:
(a) 8

Question 3.
If p is T and q is F, then which of the following have the truth value T? ………

(a) (i), (ii), (iii)
(b) (i), (ii), (iv)
(c) (i), (iii), (iv)
(d) (ii), (iii), (iv)
Hint:
p is T then ~p is F
q is F then ~ q is T

Solution:
(c) (i), (iii), (iv)

Question 4.
The number of rows in the truth table is ……..
(a) 2
(b) 4
(c) 6
(d) 8
Hint:
Number of simple statements given is 2. i.e., p and q.
Number of rows in the truth table of \(\sim[p \wedge(\sim q)]\) = 2² = 4
Solution:
(b) 4

Question 5.
The conditional statement p ➝ q is equivalent to ……..


The truth table for p ➝ q and \((\sim p \vee q)\) having the last column identical.
∴ p ➝ q is equivalent to \((\sim p \vee q)\)
Solution:
(3) \(\sim p \vee q\)

Question 6.
Which of the following is a tautology?

Solution:
(c) \(\boldsymbol{p} \vee \sim \boldsymbol{p}\)
Hint:
A statement is said to be a tautology if the last column of its truth table contains only T.

Question 7.
In the set of integers with operation * defined by a * b = a + b – ab, the value of 3 * (4 * 5) is ……..
(a) 25
(b) 15
(c) 10
(d) 5
Hint:
a * b = a + b – ab
3 * (4 * 5) = 3 * (4 + 5 – 4(5))
= 3 * (9 – 20)
= 3 * (-11)
= 3 + (-11) – 3(-11)
= 3 – 11 + 33
= -8 + 33 = 25
Solution:
(a) 25

Question 8.
In the multiplicative group of the cube root of unity, the order of \(\omega^{2}\) is ……
(a) 4
(b) 3
(c) 2
(d) 1
Solution:
(b) 3
Hint:

Question 10.

(a) 5
(b) 5\(\sqrt{2}\)
(c) 25
(d) 50
Hint:

Solution:
(b) 5\(\sqrt{2}\)

Question 11.
The order of -i in the multiplicative group of 4^th roots of unity is ……..
(a) 4
(b) 3
(c) 2
(d) 1
Hint:
The roots of fourth roots of unity are 1, -1, i, -i
The identity element is 1
(-i)⁴ = i⁴ = 1
Order of (-i) = 4.
Solution:
(a) 4

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.1

Question 1.
Find the adjoint of the following:

Solution:

Question 2.
Find the inverse (if it exists) of the following:

Solution:
For a matrix A, \(\mathrm{A}^{-1}=\frac{1}{|\mathrm{A}|}(\mathrm{adj} \mathrm{A})\). Where |A| ≠ 0. If |A| = 0 then A is called a singular matrix and so \(\mathrm{A}^{-1}\) does not exist.

Question 3.
If F(α) = \(\left[\begin{array}{ccc}{\cos \alpha} & {0} & {\sin \alpha} \\ {0} & {1} & {0} \\ {-\sin \alpha} & {0} & {\cos \alpha}\end{array}\right]\) show that \([\mathrm{F}(\alpha)]^{-1}=\mathrm{F}(-\alpha)\)
Solution:
Let A = F (α)
So \([\mathrm{F}(\alpha)]^{-1}=\mathrm{A}^{-1}\)
Now

Question 4.
If A = \(\left[\begin{array}{cc}{5} & {3} \\ {-1} & {-2}\end{array}\right]\) show that A² – 3A – 7I₂ = O₂. Hence find A^-1.
Solution:
A = \(\left[\begin{array}{cc}{5} & {3} \\ {-1} & {-2}\end{array}\right]\)

To Find A-1
Now we have proved that A² – 3A – 7I₂ = O₂
Post multiply by A^-1 we get
A – 3I – 7A^-1 = O₂

Question 5.
If \(\mathbf{A}=\frac{1}{9}\left[\begin{array}{ccc}{-8} & {1} & {4} \\ {4} & {4} & {7} \\ {1} & {-8} & {4}\end{array}\right]\) prove that A^-1 = A^T
Solution:

Question 6.
If \(\mathbf{A}=\left[\begin{array}{rr}{8} & {-4} \\ {-5} & {3}\end{array}\right]\), verify that A(adj A) = (adj A)A = |A| I₂
Solution:

Question 7.
If \(\mathbf{A}=\left[\begin{array}{ll}{3} & {2} \\ {7} & {5}\end{array}\right]\), and \(\mathbf{B}=\left[\begin{array}{cc}{-1} & {-3} \\ {5} & {2}\end{array}\right]\) verify that (AB)^-1 = B^-1 A^-1.
Solution:

Question 8.
If adj (A) = \(\left[\begin{array}{ccc}{2} & {-4} & {2} \\ {-3} & {12} & {-7} \\ {-2} & {0} & {2}\end{array}\right]\) find A
Solution:

Question 9.
If adj(A) = \(\left[\begin{array}{ccc}{0} & {-2} & {0} \\ {6} & {2} & {-6} \\ {-3} & {0} & {6}\end{array}\right]\) find A^-1
Solution:

Question 10.
Find adj(adj(A)) if adj A = \(\left[\begin{array}{ccc}{1} & {0} & {1} \\ {0} & {2} & {0} \\ {-1} & {0} & {1}\end{array}\right]\)
Solution:

Question 11.

Solution:

Question 12.
Find the matrix A for which A \(\left[\begin{array}{cc}{5} & {3} \\ {-1} & {-2}\end{array}\right]=\left[\begin{array}{cc}{14} & {7} \\ {7} & {7}\end{array}\right]\)
Solution:
Given A \(\left[\begin{array}{cc}{5} & {3} \\ {-1} & {-2}\end{array}\right]=\left[\begin{array}{cc}{14} & {7} \\ {7} & {7}\end{array}\right]\)
Let \(\mathrm{B}=\left(\begin{array}{cc}{5} & {3} \\ {-1} & {-2}\end{array}\right) \text { and } \mathrm{C}=\left(\begin{array}{cc}{14} & {7} \\ {7} & {7}\end{array}\right)\)
Given AB = C, To find A
Now AB = C
Post multiply by B^-1 on both sides
ABB^-1 = CB^-1 (i.e) A (BB^-1) = CB^-1
⇒ A(I) = CB^-1 (i.e) A = CB^-1
To find B^-1:

Question 13.
Given \(\mathbf{A}=\left[\begin{array}{cc}{1} & {-1} \\ {2} & {0}\end{array}\right], \mathbf{B}=\left[\begin{array}{cc}{3} & {-2} \\ {1} & {1}\end{array}\right] \text { and } \mathbf{C}\left[\begin{array}{ll}{1} & {1} \\ {2} & {2}\end{array}\right]\), find a matrix X such that AXB = C.
Solution:
A × B = C
Pre multiply by A^-1 and post multiply by B^-1 we get
A^-1 A × BB^-1 = A^-1CB^-1 (i.e) X = A^-1CB^-1

Question 14.

Solution:

Question 15.
Decrypt the received encoded message \(\left[\begin{array}{cc}{2} & {-3}\end{array}\right]\left[\begin{array}{ll}{20} & {4}\end{array}\right]\) with the encryption matrix \(\left[\begin{array}{cc}{-1} & {-1} \\ {2} & {1}\end{array}\right]\) and the decryption matrix as its inverse, where the system of codes are described by the numbers 1-26 to the letters A- Z respectively, and the number 0 to a blank space.
Solution:
Let the encoding matrix be \(\left[\begin{array}{cc}{-1} & {-1} \\ {2} & {1}\end{array}\right]\)


So the sequence of decoded matrices is [8 5], [12 16].
Thus the receivers read this message as HELP.

Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.1 Additional Problems

Question 1.
Using elementary transformations find the inverse of the following matrix
Solution:

Question 2.
Using elementary transformations find the inverse of the matrix
Solution:

Question 3.
Using elementary transformation find the inverse of the matrix
Solution:

Question 4.
Using elementary transformations find the inverse of the matrix
Solution:

Question 5.
Using elementary transformation, find the inverse of the following matrix
Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Students can Download Samacheer Kalvi 12th English Expand the News Headlines, Notes, Samacheer Kalvi 12th English Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th English Expand the News Headlines

1. Saina Receives Khel Ratna Award
New Delhi: Badminton ace Saina Nehwal, whose phenomenal rise was highlighted by three top-flight international titles, recently, was on Sunday conferred the coveted Rajiv Gandhi Khel Ratna award at a glittering ceremony here on Sunday. Saina’s achievements in the past one year made her an automatic choice for the prestigious award, the highest sporting honour in the country.

2. Forgotten Brother Appears
Jharkhand: A forgotten brother has appeared after a long period of time.Mr.Ranjan and Mr. Anjan were reunited after the latter went missing in a Mela at Jharkand. They are united after 43 long years. Mr. Ranjan was taken care of by a local NGO and has a success story to narrate.

3. Maid Wins Maruti Car as Prize
Trichy: The fortune of a house-maid today changed when she won Maruti car as a prize in the local mela raffle. As she has no one to drive the car in her family, she decided to sell it. However, Mr. Ramakrishnan in whose house she works as a house maid, suggested that she rents it out to earn an extra amount for her family.

4. Crores Approved
Trichy:The municipality has launched a comprehensive drive, at a cost of Rs. 1.41 crore, to clean drainage-canal-tumed irrigation canals, meandering through the town, ahead of the Mahamaham festival.The effort would provide a fresh lease of life to five canals that were once irrigation channels but over the decades got transformed into drainage channels. A lot of development works were being undertaken; the town was being given a facelift ahead of the festival.Public sanitation has been accorded priority, she added.

5. Passerby Sees Woman Jump
ChennaiiA passerby reportedly stopped a woman from jumping off a pedestrian bridge at the Kotturpuram bus stop on Saturday. It is understood that some family duel is the reason for her drastic decision. The woman climbed over the guardrail and sat on it. Some passersby joined hands and forced her to come down.Suicidal Prevention Centre was informed.

6. Difficult Times Ahead
New Delhi: The Finance Minister has predicted difficult times for the Indian Economy in the years ahead. As he enters the final year of the government’s term, he is probably facing as many – if not more – economic problems as he did when he took charge. In fact, one would say if the BJP returns to power, Prime Minister Modi will have to work even harder than he did in his first term to keep the economy on an even keel.

7. Professors Protest Pay Cuts
New Delhi: Days after warning them, the administration of the Jawaharlal Nehru University (JNU) has decided to deduct the salaries of the teachers for participating in protests. The teachers protested against the administration’s decision to scrap the old recruitment and admission policy as well as the multiple show-cause notices against students and teachers. The university’s administration has banned any kind of protest within 20 meters of the building.

8. Tommy The Dog Named Hero
New Delhi: Tommy,the popular dog from Chennai was declared the hero at The Canine Meet held at Delhi. The owner Vijay Prabhakar, says that he has hired Dog trainers from far and wide to train his dogs and this is the second National award for Tommy. Tommy is expected to take part in the International Canine meet in 2020.

9. President Declares Celebration
Mumbai: The President of India, Shri Ram NathKovind, graced and addressed a function organised to declare urban Maharashtra Open. He also inaugurated the centenary year celebrations of the Samadhi of Shri Sai Baba in Shirdi. He was felicitated by the local party members.

10. Security Alert In Bhopal
Bhopal: In the city of Bhopal, after receiving two anonymous calls, police department is on the alert day and night.Bhopal is on the radar of the terrorist outfits. Taking serious note of the intelligence given previously, the State police headquarters has sounded a high alert in the city.

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.1

Question 1.
For each of the following differential equations, determine its order, degree (if exists)

Solution:
Order = 1,
Degree = 1

(ii)
Solution:

(iii)
Solution:

(iv)
Solution:

(v)
Solution:

(vi)
Solution:

(vii)
Solution:

(viii)
Solution:
Order = 2

(ix)
Solution:

(x)
Solution:

Order = 1,
degree = Not exist

Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.1 Additional Problems

Question 1.
Find the order and degree of the following differential equations:

Solution:
Order = 1,
Degree = 1

(ii) y’ + y² + y³ = 0
Solution:
Order = 1,
Degree = 1

(iii)
Solution:

(iv)
Solution:

∴ Order = 2
Degree = 2

(v)
Solution:

(vi)
Solution:

(vii)
Solution:

(viii)
Solution:

(ix)
Solution:

(x) sin x(dx + dy) = cos x(dx – dy)
Solution:
Order = 1;
Degree = 1

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8

Question 1.
If to ω ≠ 1 is a cube root of unity, then show that \(\frac{a+b \omega+c \omega^{2}}{b+c \omega+a \omega^{2}}+\frac{a+b \omega+c \omega^{2}}{c+a \omega+b \omega^{2}}=1\)
Solution:
Since ω is a cube root of unity, we have ω³ = 1 and 1 + ω + ω² = 0

Question 2.
Show that \(\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^{5}+\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^{5}=-\sqrt{3}\)
Solution:

Question 3.
Find the value of \(\left(\frac{1+\sin \frac{\pi}{10}+i \cos \frac{\pi}{10}}{1+\sin \frac{\pi}{10}-i \cos \frac{\pi}{10}}\right)^{10}\)
Solution:

Question 4.
If 2 cos α = x + \(\frac{1}{x}\) and 2 cos β = y + \(\frac{1}{y}\), show that

Solution:
(i) 2 cos α = x + \(\frac{1}{x}\)
⇒ 2 cos α = \(\frac{x^{2}+1}{x}\)
⇒ 2x cos α = x² + 1
⇒ x² – 2x cos α + 1 = 0

Question 5.
Solve the equation z³ + 27 = 0
Solution:
z³ + 27 = 0
⇒ z³ = -27
⇒ z³ = 3³(-1)
⇒ z = 3(-1)^1/3

Question 6.
If ω ≠ 1 is a cube root of unity, show that the roots of the equation (z – 1)³ + 8 = 0 are -1, 1 – 2ω, 1 – 2ω²
Solution:
(z – 1)³ + 8 = 0
⇒ (z – 1 )³ = -8

The roots are -1, 1 – 2ω, 1 – 2ω²

Question 7.
Find the value of \(\sum_{k=1}^{8}\left(\cos \frac{2 k \pi}{9}+i \sin \frac{2 k \pi}{9}\right)\)
Solution:
\(\sum_{k=1}^{8}\left(\cos \frac{2 k \pi}{9}+i \sin \frac{2 k \pi}{9}\right)\)
We know that 9th roots of unit are 1, ω, ω², ……., ω⁸
Sum of the roots:
1 + ω + ω² + …. + ω⁸ = 0 ⇒ ω + ω² + ω³ + …… + ω⁸ = -1

The sum of all the terms \(\sum_{k=1}^{8}\left(\cos \frac{2 k \pi}{9}+i \sin \frac{2 k \pi}{9}\right)\) = -1

Question 8.
If ω ≠ 1 is a cube root of unity, show that
(i) (1 – ω + ω²)⁶ + (1 + ω – ω²)⁶ = 128
(ii) (1 + ω)(1 + ω²)(1 + ω⁴)(1 + ω⁸)……(1 + ω²ⁿ) = 1
Solution:
(i) (1 – ω + ω²)⁶ + (1 + ω – ω²)⁶
= (-ω + ω)⁶ + (-ω – ω²)⁶
= (-2ω)⁶ + (-2ω²)⁶
= 2⁶ [ω⁶ + ω¹²] [∵ ω⁶ = 1]
= 64 [1 + 1]
= 64 × 2
= 128
(ii) (1 – ω) (1 + ω²) (1 + ω) (1 + ω²) ….. 2n factors
(-ω²) (-ω) (-ω²) (-ω) ……. 2n factor
(ω³) (ω³) …… 2n factor
= 1.1 …….. 2n factor
= 1

Question 9.
If z = 2 – 2i, find the rotation of z by θ radians in the counter clockwise direction about the origin when
(i) θ = \(\frac{\pi}{3}\)
(ii) θ = \(\frac{2 \pi}{3}\)
(iii) θ = \(\frac{3 \pi}{2}\)
Solution:
(i) z = 2 – 2i = 2 (1 – i) = r(cos θ + i sin θ)
\(r=\sqrt{x^{2}+y^{2}}=2 \sqrt{1+1}=2 \sqrt{2}\)
\(\alpha=\tan ^{-1}=\left|\frac{y}{x}\right|=\tan ^{-1}|1|=\frac{\pi}{4}\)
(1 – i) lies in IV quadrant
θ = -α = \(-\frac{\pi}{4}\)
\(\Rightarrow z=2 \sqrt{2}\left[\cos \left(\frac{-\pi}{4}\right)+i \sin \left(\frac{-\pi}{4}\right)\right]\)
z is rotated by θ = \(\frac{\pi}{3}\) in the counter clock wise direction.

(ii) z is rotated by θ = \(\frac{2 \pi}{3}\) in the counter clockwise direction.

(iii) z is rotated by θ = \(\frac{3 \pi}{2}\) in the counter clockwise direction.

Question 10.
Prove that the values of \(\sqrt[4]{-1} \text { are } \pm \frac{1}{\sqrt{2}}(1 \pm i)\)
Solution:

The roots are \(\pm \frac{1}{\sqrt{2}}(1 \pm i)\)

Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 Additional Problems

Question 1.

Solution:

Question 2.

Solution:

Question 3.
Prove that: (1 + i)⁴ⁿ and (1 + i)^{4n + 2} are real and purely imaginary respectively.
Solution:
Let z = 1 + i

= 2i (-1)ⁿ which is purely imaginary.

Question 4.

Solution:

Question 5.
If a = cos 2α + i sin 2α, b = cos 2β + i sin 2β and c = cos 2γ + i sin 2γ, prove that

Solution:

Question 6.
Solve: x⁴ + 4 = 0
Solution:

Question 7.
If x = a + b, y = aω + bω², z = aω² + bω, show that
(i) xyz = a³ + b³
(ii) x³ + y³ + z³ = 3(a³ + b³)
Solution:
(i) x = a + A; y = aω + bω², z = aω² + bω
Now xyz = (a + b) (aω + bω²) (abω² + bω) = (a+A) [aω³ + abω² + abω + b²ω³]
= (a + b) (a² – ab + b²) = a³ + b³
xyz = a³ + b³

(ii) x = a + b, y = aω + bω², z = aω² + bω
x + y + z = (a + aω + aω²) + (b + bω² + bω)
= a (1 + ω + ω²) + b (1 + ω + ω²) = a(0) + b(0) = 0
Now x + y + z = 0 ⇒ x³ + y³ + z³ = 3xyz
Here xyz = a³ + b³
∴ x³ + y³ + z³ = 3 (a³ + b³)

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.7

Question 1.
Solve the following system of homogeneous equations.
(i) 3x + 2y + 7z = 0, 4x – 3y – 2z = 0, 5x + 9y + 23z = 0
(ii) 2x + 3y – z = 0, x – y – 2z = 0, 3x + y + 3z = 0
Solution:
(i) The matrix form of the above equations is

The above matrix is in echelon form.
Here ρ(A, B) = ρ(A) < number of unknowns.
The system is consistent with infinite number of solutions. To find the solutions.
Writing the equivalent equations.
We get 3x + 2y + 7z = 0 ……. (1)
-17y – 34z = 0 ……. (2)
Taking z = t in (2) we get -17y – 34t = 0
⇒ -17y = 34t
⇒ y = -2t
Taking z = t, y = -2t in (1) we get
3x + 2(-2t) + 7t = 0
⇒ 3x – 4t + 7t = 0
⇒ 3x = -3t
⇒ x = -t
So the solution is x = -t; y = -2t; and z = t, t ∈ R
(ii) The matrix form of the equations is
\(\left(\begin{array}{ccc}{2} & {3} & {-1} \\ {1} & {-1} & {-2} \\ {3} & {1} & {3}\end{array}\right)\left(\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right)=\left(\begin{array}{l}{0} \\ {0} \\ {0}\end{array}\right)\)
(i.e) AX = B
The augmented matrix [A, B] is

The above matrix is in echelon form also ρ(A, B) = ρ(A) = 3 = number of unknowns
The system is consistent with unique solution, x = y = z = 0
(i.e) The system has trivial solution only.

Question 2.
Determine the values of λ for which the following system of equations.
x + y + 3z = 0, 4x + 3y + λz = 0, 2x +y + 2z = 0 has
(i) a unique solution
(ii) a non-trivial solution
Solution:
The matrix form of the equation is
\(\left(\begin{array}{lll}{1} & {1} & {3} \\ {4} & {3} & {\lambda} \\ {2} & {1} & {2}\end{array}\right)\left(\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right)=\left(\begin{array}{l}{0} \\ {0} \\ {0}\end{array}\right)\)
(i.e) AX = B
The augmented matrix [A, B] is

The above matrix is in echelon form
Case 1: When λ ≠ 8, ρ(A, B) = ρ(A) = 3 = number of unknowns
The system is consistent with a unique solution.
Case 2: When λ = 8, ρ(A, B) = ρ(A) = 2 < number of unknowns.
The system is consistent with non-trivial solutions.

Question 3.
By using the Gaussian elimination method, balance the chemical reaction equation:
C₂H₆ + O₂ → H₂O + CO₂
Solution:
We are searching for positive integers x₁, x₂, x₃ and x₄ such that
x₁C₂H₆ + x₂O₂ = x₃H₂O + x₄CO₂ ……. (1)
The number of carbon atoms on LHS of (1) should be equal to the number of carbon atoms on the RHS of (1).
So we get a linear homogeneous equation.
2x₁ = x₄
⇒ 2x₁ – x₄ = 0 …… (2)
Similarly considering hydrogen and oxygen atoms we get respectively.
2x₂ = x₃ + 2x₄
⇒ 2x₂ – x₃ – 2x₄ = 0 …… (3)
and -2x₃+ 3x₄ = 0 …… (4)
Equations (2), (3) and (4) constitute a homogeneous system of linear equations in four unknowns.
The augmented matrix (A, B) is

Now ρ(A, B) = ρ(A) = 3 < number of unknowns.
So the system is consistent and has an infinite number of solutions.
Writing the equations using the echelon form we get
2x₁ – x₄ = 0 …… (5)
2x₂ – x₃ – 2x₄ = 0 ……. (6)
-2x₃ + 3x₄ = 0 ……. (7)

Since x₁, x₂, x₃ and x₄ are positive integers. Let us choose t = 4t.
Then we get x₁ = 2, x₂ = 7, x₃ = 6, and x₄ = 4
So the balanced equation is 2C₂H₆ + 7O₂ → 6H₂O + 4CO₂.

Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.7 Additional Problems

Question 1.
Solve the following homogeneous linear equations.
x + 2y – 5z = 0,
3x + 4y + 6z = 0,
x + y + z = 0
Solution:
The given system of equations can be written in the form of matrix equation

AX = B
The augmented matrix is

This is in the echelon form.
Clearly ρ[A, B] = 3 and ρ(A) = 3
∴ ρ(A) = ρ[A, B] = 3 = number of unknowns
∴ The given system of equations is consistent and has a unique solution. i.e., trivial solution
∴ x = 0, y = 0 and z = 0
Note: Since ρ(A) = 3, | A | ≠ 0 i.e. A is non-singular;
∴ The given system has only trivial solution x = 0, y = 0, z = 0

Question 2.
For what value of n the equations.
x + y + 3z = 0,
4x + 3y + µz = 0,
2x + y + 2z = 0 have a
(i) trivial solution,
(ii) non-trivial solution.
Solution:
The system of equations can be written as AX = B

Case (i): If µ ≠ 8 then 8 – µ ≠ 0 and hence there are three non-zero rows.
∴ ρ[A] = ρ[A, B] = 3 = the number of unknowns.
∴ The system has the trivial solution x = 0, y = 0, z = 0
Case (ii): If µ = 8 then.
ρ[A, B] = 2 and ρ(A) = 2
∴ ρ(A) = ρ[A, B] = 2 < number of unknowns.
The given system is equivalent to
x + y + 3z = 0; y + 4z = 0
∴ y= – 4z ; x = z
Taking z = k, we get x = k,y = – 4k, z = k [k ∈ R – {0}] which are non-trivial solutions.
Thus the system is consistent and has infinitely many non-trivial solutions.
Note: In case (ii) the system also has trivial solution. For only non-trivial solutions we removed k = 0.

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.3

Question 1.
If z₁ = 1 – 3i, z₂ = -4i, and z₃ = 5, show that
(i) (z₁ + z₂) + z₃ = z₁ + (z₂ + z₃)
(ii) (z₁ z₂) z₃ = z₁ (z₂ z₃)
Solution:
(i) Given z₁ = 1 – 3i, z₂ = -4i, z₃ = 5
z₁ + z₂ = (1 – 3i) + (-4i) = 1 – 7i
(z₁ + z₂) + z₃ = 1 – 7i + 5 = 6 – 7i ……..(1)
z₂ + z₃ = (-4i) + (5) = 5 – 4i
z₁ + (z₂ + z₃) = (1 – 3i) + (5 – 4i)
= 6 – 7i ………. (2)
From (1) and (2)
(z₁ + z₂) + z₃ = z₁ + (z₂ + z₃)
Hence proved.

(ii) z₁ z₂ = (1 – 3i) (-4i)
= -4i – 12i²
= -12 – 4i
(z₁ z₂)z₃ = (-12 – 4i)(5)
= – 60 – 20 i …………. (1)
z₂ z₃ = (-4i) (5) = -20 i
z₁(z₂ z₃) = (1 – 3i) (-20i) = – 20i + 60 i²
= -60 – 20 i …………. (2)
∴ from 1 and 2
(z₁ z₂)z₃ = z₁(z₂ z₃)

Question 2.
If z₁ = 3, z₂ = -7i, and z₃ = 5 + 4i, show that
(i) z₁ (z₂ + z₃) = z₁ z₂ + z₁ z₃
(ii) (z₁ z₂) z₃ = z₁ z₃ + z₂ z₃
Solution:
(i) z₁ = 3, z₂ = -7i, z₃ = 5 + 4i
z₁ (z₂ + z₃) = 3 (-7i + 5 + 4i)
= 3 (5 – 3i)
= 15 – 9i …… (1)
z₁ z₂ + z₁ z₃ = 3 (-7i) + 3 (5 + 4i)
= -21i + 15 + 12i
= 15 – 9i …… (2)
from (1) & (2), we get
∴ z₁ (z₂ + z₃) = z₁ z₂ + z₁ z₃

(ii) (z₁ + z₂) z₃ = (3 – 7i) (5 + 4i)
= 15 + 12i – 35i – 28 i²
= 15 – 23i + 28
= 43 – 23i ……. (1)
z₁ z₃ + z₂ z₃ = 3(5 + 4i) – 7i(5 + 4i)
= 15 + 12i – 35i – 28 i²
= 15 – 23i + 28
= 43 – 23i ….. (2)
from (1) & (2), we get
∴ (z₁ + z₂) z₃ = z₁ z₃ + z₂ z₃

Question 3.
If z₁ = 2 + 5i, z₂ = -3 – 4i, and z₃ = 1 + i, find the additive and multiplicative inverse of z₁, z₂, and z₃.
Solution:
z₁ = 2 + 5i
(а) Additive inverse of z₁ = -z₁ = -(2 + 5i) = -2 – 5i
(b) Multiplicative inverse of
\(z_{1}=\frac{1}{z_{1}}=\frac{1}{2+5 i} \times \frac{2-5 i}{2-5 i}=\frac{(2-5 i)}{4+25}=\frac{2-5 i}{29}\)

z₂ = -3 – 4i
(a) Additive inverse of z₂ = -z₂ = -(-3 – 4i) = 3 + 4i
(b) Multiplicative inverse of
\(z_{2}=\frac{1}{z_{2}} \frac{1}{-3-4 i} \times \frac{-3+4 i}{-3+4 i}=\frac{-3+4 i}{9+16}=\frac{-3+4 i}{25}\)

z₃ = 1 + i
Additive inverse z₃ is -z₃
⇒ -(1 + i) = -1 – i
Multiplicative inverse z₃ is (z₃)^-1
We know
z₃ z₃^-1 = 1
⇒ (1 + i) (u + iv) = 1 [∵ z₃^-1 = u + iv]
u + iv + iu – v = 1
(u – v) + i (u + v) = 1 + i 0
Equating real and imaginary parts
u – v = 1
u + v = 0
Solving them, we get u = \(\frac {1}{2}\) and v = –\(\frac {1}{2}\)
∵ z₃^-1 = \(\frac {1}{2}\) (1 – i)

Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.3 Additional Problems

Question 1.
If z₁ = 4 – 7i, z₂ = 2 + 3i and z₃ = 1 + i show that.
(i) z₁ + (z₂ + z₃) = (Z₁ + z₂) + Z₃
(ii) (Z₁ z₂) z₃ = Z₁ (z₂ z₃)
Solution:
(i) Z₁ + (z₂ + z₁₃) = 4 – 7i + (2 + 3i + 1 + i) = 4 – 7i + (3 + 4i) = 7 – 3i
(z₁ + z₂) + z₃ = (4 – 7i + 2 + 3i) + (1 + i) = (6 – 4i) + (1 + i)
= 7 – 3i …(2)
From (1) and (2) we get, z₁ + (z₂ + z₃) = (z₁ + z₂) + z₃

(ii) (z₁z₂) z₃ = (4 – 7i) + (2 + 3i) (1 + i) = (8 + 12i – 14 i + 21) (1 + i)
= (29 – 2i)(1 + i) = 29 + 29i – 2i + 2
= 31 + 27i ….(1)
Z₁ (z₂z₃) = (4 – 7i) [(2 + 3i) (1 + i)] = 4 – 7i [2 + 2i + 3i – 3]
= 4 – 7i[5i – 1] = 20i – 4 + 35 + 7i
= 31 + 27i …(2)
From (1) and (2) we get, (Z₁Z₂) z₃ = z₁ (z₂z₃)

Question 2.
Given z₁ = 1 + i, z₂ = 4 – 3i and z₃ = 2 + 5i verify that.
Z₁(Z₂ Z₃) = Z₁ Z₂ – z₁z₃
Solution:
Z₁ = 1 + i,
z₂ = 4 – 3i,
z₃ = 2 + 5i
Z₁ (z₂ – z₃) = 1 + i[(4 – 3i) – (2 + 5i)] = 1 + i[2 – 8i] = 2 – 8i + 2i + 8
= 10 – 6i …(1)
Z₁ z₂ = (1 + i) (4 – 3i) = 4 – 3i + 4i + 3 = 7 + i
Z₁Z₃= (1 + i) (2 + 5i) = 2 + 5i + 2i – 5 = 7i – 3
Z₁Z₂ – Z₁ z₃ = 7 + i – (7i – 3) = 7 + i – 7i + 3
= 10 – 6i …(2)
From (1) and (2) we get, Z₁ (z₂ – z₃) = z₁ z₂ – z₁ z₃

Question 3.
Given z₁ = 4 – 7i and z₂ = 5 + 6i find the additive and multiplicative inverse of z₁ + z₂ and Z₁ – Z₂.
Solution:
Z₁ = 4 – 7i, z₂ = 5 + 6i
Z₁ + z₂ = 4 – 7i + 5 + 6i = 9 + i
Additive inverse of z₁ + z₂ = -(z₁ + z₂) = -(9 – i) = i – 9

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.2

Question 1.
Evaluate the following if z = 5 – 2i and w = -1 + 3i
(i) z + w
(ii) z – iw
(iii) 2z + 3w
(iv) zw
(v) z² + 2zw+ w²
(vi) (z + w)²
Solution:
(i) z + w = (5 – 2i) + (-1 + 3i)
= 4 + i
(ii) z – iw= (5 – 2i) -i (-1 + 3i)
= 5 – 2i + i + 3i² (∵ i² = -1)
= 5 – 2i + i – 3(-1)
= 5 – 2i + i + 3
= 8 – i
(iii) 2z + 3w = 2 (5 – 2i) + 3 (- 1 +3i)
= 10 – 4i – 3 + 9i
= 7 + 5i
(iv) zw = (5 – 2i) (- 1 + 3i)
= -5 + 2i + 15i – 6i²
= -5 + 17i + 6
= 1 + 17i
(v) z² + 2zw + w²
= (z + w)²
= (4 + i)²
= (4) + 2(4)(i) + (i)²
= 16 + 8i + i²
= 15 + 8i
(vi) (z + w)² = z² + 2zw + w²
= (z + w)² = (4 + i)²
= 16 + 8i + i²
= 15 + 8i

Question 2.
Given the complex number z = 2 + 3i, represent the complex numbers in the Argand diagram.
(i) z, iz, and z + iz
(ii) z, -iz, and z – iz
Solution:
(i) z, iz and z + iz.
z = 2 + 3i
iz = i(2 + 3i) = -3 + 2i
z + iz = 2 + 3i – 3 + 2i = -1 + 5i

(ii) z = 2 + 3i
-iz = -i(2 + 3i)
= -2i – 3i²
= (3 – 2i)
z – iz = (2 + 3i) + (3 – 2i)
= 5 + i

Question 3.
Find the values of the real numbers x and y, if the complex numbers.
(3 – i) x – (2 – i) y + 2i + 5 and 2x + (-1 + 2i) y + 3 + 2i are equal
Solution:
Given that the complex numbers are equal
(3 – i) x -(2 – i) y + 3 + 2i + 5
= 2x + (-1 + 2i)y + 3 + 2i
3x – ix – 2y + iy + 2i +5
= 2x – y + 2iy + 3 + 2i
(3x – 2y + 5) + i(y – x + 2)
= (2x – y + 3) + i(2y + 2)
Equating real and imaginary parts separately
3x – 2y + 5 = 2x – y + 3
x – y = -2 ………. (1)
y – x + 2 = 2y +
-x – y = 0 ………. (2)
solving 1 and 2

y = 1
Substituting y = 1 in (1)
x – 1 = -2
x = -2 + 1 = -1
values of x and y are -1, 1

Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.2 Additional Problems

Question 1.
Find the real values of x and y, if
(i) (3x – 7) + 2 iy = -5y + (5 + x)i
(ii) (1 – i)x + (1 + i)y = 1 – 3i
(iii) (x + iy)(2 – 3i) = 4 + i
(iv)
Solution:
(i) We have (3x – 7) + 2 iy = 5y + (5 + x)i
⇒ 3x – 7 = 5y and 2y = 5 + x
⇒ 3x + 5y = 7 and x – 2y = -5
⇒ x = -1 y = 2

(ii) We have, (1 – i) x + (1 + i)y = 1 – 3i
⇒ (x + y) + i(-x + y) = 1 – 3i
⇒ x + y = 1
and -x + y = 3 [On equating real and imaginary parts]
⇒ x = 2 and y = -1

⇒ x + y – 2 = 0 and y – x = 10
⇒ x = -4 , y = 6.

Question 2.
Find the real values of x and y for which the complex numbers -3 + ix²y and x² + y + 4i are conjugate of each other.
Solution:
Since -3 + ix²y and x² + y + 4i are complex conjugates.
∴ -3 + ix²y = x² + y + 4i ….. (i) and, x² y = -4 …… (ii)

Question 3.
Given x = 2 – 3i and y = 4 + i

Solution:
Do it yourself

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.1

Simplify the following:

Question 1.
i¹⁹⁴⁷ + i¹⁹⁵⁰
Solution:

Question 2.
i¹⁹⁴⁸ – i^-1869
Solution:

Question3.
\(\sum_{n=1}^{12} i^{n}\)
Solution:

Question 4.
\(i^{59}+\frac{1}{i^{59}}\)
Solution:

Question 5.
i i² i³ …….. i²⁰⁰⁰
Solution:

Question 6.
\(\sum_{n=1}^{10} i^{n+50}\)
Solution:

Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 2.1 Additional Problems

Question 1.
Evaluate the following:

Solution:
135 leaves remainder as 3 when it is divided by 4.
∴ i¹³⁵ = i³ = -i

(ii) The remainder is 3 when 19 is divided by 4.
∴ i¹⁹ = i³ = -i

(iii) We have, i⁹⁹⁹ = 1/i⁹⁹⁹
On dividing 999 by 4, we obtain 3 as the remainder.

(iv)

Question 2.
Show that:

Solution:

 

Question 3.

Solution: