Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.6

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 1 Applications of Matrices and Determinants Ex 1.6 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.6

Question 1.
Test for consistency and if possible, solve the following systems of equations by rank method.
(i) x – y + 2z = 2, 2x + y + 4z = 7, 4x – y + z = 4
Solution:
Matrix form
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.6 1
The system is consistent.
ρ(A) ρ[A|B] = 3 = n
it has unique solution.
Writing the equivalent equations from echelon form
x – y + 2z = 2 ………… (1)
3y = 3 ⇒ y = 1
-7z = -7
z = 1
(1)⇒ x – y + 2z = 2
x – 1 + 2 = 2
x = 1
∴ Solution is x = 1, y = 1, z = 1

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.6

(ii) 3x + y + z = 2, x – 3y + 2z = 1, 7x – y + 4z = 5
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.6 2
ρ(A) = 2 ρ[A | B] = 2
ρ(A) = ρ[A | B] = 2 < n
The system is consistent. It has infinitely many solution.
Writing the equivalent equations from echelon form.
x – 3y + 2z = 1 ………. (1)
10y – 5z = -1 ………. (2)
Put z = t.
(2) ⇒ 10y – 5z = -1
10y = -1 + 5z = 5t – 1
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.6 3

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.6

(iii) 2x + 2y + z = 5, x – y + z = 1, 3x + y + 2z = 4
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.6 4
ρ(A) = 2 ρ[A | B] = 3
ρ(A) ≠ ρ[A | B] = 2 < n
∴ The system is inconsistent. It has no solution.

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.6

(iv) 2x – y + z = 2, 6x – 3y + 3z = 6, 4x – 2y + 2z = 4
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.6 5
ρ(A) = 1 ρ[A | B] = 1
ρ(A) = ρ[A | B] = 1 < n.
∴ The system reduces into a single equation.
∴ It is consistent and has infinitely many solutions.
Writing the equivalent equations from echelon form
2x – y + z = 2
Put y = s, z = t
2x – s + t = 2
2x = 2 + s – t
x = \(\frac {2+s-t}{2}\)
(x, y, z) = (\(\frac {2+s-t}{2}\), s, t) ∀ s, t ∈ R

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.6

Question 2.
Find the value of k for which the equations kx – 2y + z = 1, x – 2ky + z = -2, x – 2y + kz = 1 have
(i) no solution
(ii) unique solution
(iii) infinitely many solution.
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.6 6
[∵ 2 – k – k² = -(k² + k – 2)
= -(k + 2)(k – 1)
= (k + 2)(1 – k)]
case (i)
If k = 1
ρ(A) = 2, ρ(A | B) = 3.
ρ(A) ≠ ρ(A | B)
The system is inconsistent and it has no solution.

Case (ii)
If k ≠ 1, k ≠ -2
ρ(A) = 3, ρ(A | B) = 3 = n
The system is consistent and it has unique solution.

Case (iii)
If k = -2
ρ(A) = 2, ρ(A | B) = 2
The system is consistent and it has infinitely many solution.

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.6

Question 3.
Investigate the values of λ and µ the system of linear equations 2x + 3y + 5z = 9, 7x + 3y – 5z = 8, 2x + 3y + λz = µ, have
(i) no solution
(ii) a unique solution
(iii) an infinite number of solutions.
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.6 7
Case (i)
If λ = 5, µ ≠ 9
ρ(A) = 2, ρ(A | B) = 3
ρ(A) ≠ ρ(A | B)
The system is inconsistent. It has no solution.

Case (ii)
If λ = 5, µ ≠ 9
ρ(A) = 3, ρ(A | B) = 3
ρ(A) = ρ(A | B) = 3 = n
The system is consistent. It has unique solution.

Case (iii)
If λ = 5, µ = 9
ρ(A) = 2, ρ(A | B) =2
ρ(A) = ρ(A | B) = 2 < n
The system is consistent. It has infinitely many solution.

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.6

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.5

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 1 Applications of Matrices and Determinants Ex 1.5 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5

Question 1.
Solve the following systems of linear equations by Gaussian elimination method:
(i) 2x – 2y + 3z = 2, x + 2y – z = 3, 3x – y + 2z = 1
Solution:
Augmented matrix
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.5 1
Writing the equivalent equations from echelon from.
x – y + 2z = 3 …………. (1)
5y – 6z = -4 ………….. (2)
-z = -4
z = 4
(2) ⇒ 5y – 6z = -4
5y – 24 = -4
5y = -4 + 24
5y = 20
y = 4
(1) ⇒ x – y + 2z = 3
x – 4 + 8 = 3
x = 3 + 4 – 8
x = -1
∴ x = -1, y = 4, z = 4

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.5

(ii) 2x + 4y + 6z = 22, 3x + 8y + 5z = 27, -x + y + 2z = 2.
Solution:
Augmented matrix
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.5 2
Writing the equivalent equations from echelon from.
x + 2y + 3z = 11 …………. (1)
y – 2z = -3 ………….. (2)
11z = 22
z = 2
(2) ⇒ y – 2z = -3
y – 4 = -3
y = -3 + 4
y = 1
(1) ⇒ x + 2y + 3z = 11
x + 2(1) + 3(2) = 11
x + 2 + 6 = 11
x = 11 – 8 = 3
∴ x = 3, y = 1, z = 2

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.5

Question 2.
If ax² + bx + c is divided by x + 3, x – 5, and x – 1, the remainders are 21, 61 and 9 respectively. Find a, b and c. (Use Gaussian elimination method.)
Solution:
Given: f(x) = ax² + bx + c
In Remainder Theorem
f(-3) = 21
a(-3)² + b(-3) + c = 21
9a – 3b + c = 21 ……….. (1)
f(5) = 61
25a + 5b + c = 61 …………. (2)
f(1) = 9
a + b + c = 9 ………… (3)
Augmented matrix
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.5 3
Writing the equivalent equations from echelon from.
a + b + c = 9 …………. (1)
b + 2c = 5 ………….. (2)
-4c = -8
c = 2
(2) ⇒ b + 2c = 5
b + 4 = 5
b = 5 – 4
b = 1
(1) ⇒ a + b + c = 9
a + 1 + 2 = 9
a = 9 – 3
a = 6
a = 6, b = 1, c = 2

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.5

Question 3.
An amount of Rs 65,000 is invested in three bonds at the rates of 6%, 8% and 10% per annum respectively. The total annual income is Rs 5,000. The income from the third bond is Rs 800 more than that from the second bond. Determine the price of each bond. (Use Gaussian elimination method.)
Solution:
Let the amounts of 3 bounds be x, y, z
x + y + z = 65,000
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.5 4
Writing the equivalent equations from echelon from.
x + y + z = 65000 …………. (1)
2y + 3z = 90000 ………….. (2)
21z = 42000
z = 20000
(2) ⇒ 2y = 90000 – 3 × 20000
2y = 30000
y = 15000
(1) ⇒ x + 15000 + 20000 = 65000
x = 30000
∴ x = 30000, y = 15000, z = 20000

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.5

Question 4.
A boy is walking along the path y = ax² + bx + c through the points (-6, 8),(-2, -12), and (3, 8). He wants to meet his friend at P(7, 60). Will he meet his friend? (Use Gaussian elimination method.)
Solution:
y = ax² + bx + c
At(-6, 8) ⇒ 8 = 36a – 6b + c ………… (1)
At(-2, -12) ⇒ -12 = 4a – 2b + c ………… (2)
At(3, 8) ⇒ 8 = 9a + 3b + c ………… (3)
Augmented matrix
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.5 5
Writing the equivalent equations from the echelon.
36a – 6b + c = 8 …………. (1)
3b – 2c = 29 ………….. (2)
5c = -50
c = -10
(2) ⇒ 3b – 2c = 29
3b – 20 = 29
3b = 9
b = 3
(1) ⇒ 36a – 18 – 10 = 8
36a = 8 + 18 + 10
36a = 36
a = 1
At P (7, 60), y = ax² + bx + c
60 = 1(7²) + 3(7) – 10
60 = 49 – 21 – 10
60 = 60
He will meet his friend at P (7, 60)

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.5

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.4

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 1 Applications of Matrices and Determinants Ex 1.4 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.4

Question 1.
Solve the following systems of linear equations by Cramer’s rule:
(i) 5x – 2y + 16 = 0, x + 3y – 7 = 0
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.4 1

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.4

(ii) \(\frac{3}{x}\) + 2y =12, \(\frac{2}{x}\) + 3y = 13
Solution:
Let \(\frac{1}{x}\) = a
3a + 2b = 12
2a + 3b = 13
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.4 2

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.4

(iii) 3x + 3y – z = 1 1, 2x – y + 2z = 9, 4x + 3y + 2z = 25
Solution:
Δ = \(\left| \begin{matrix} 3 & 3 & -1 \\ 2 & -1 & 2 \\ 4 & 3 & 2 \end{matrix} \right| \)
= 3(-2 – 6) -3 (4 – 8) -1(6 + 4)
= 3(-8) -3(-4) -1(10)
= -24 + 12 – 10 = -22 ≠ 0
Δx = \(\left| \begin{matrix} 11 & 3 & -1 \\ 9 & -1 & 2 \\ 25 & 3 & 2 \end{matrix} \right| \)
= 11 (-2 – 6) – 3(18 – 50) – 1(27 + 25)
= 11(-8) -3(32) -1(52)
= -88 + 96 – 52 = -44
Δy = \(\left| \begin{matrix} 3 & 11 & -1 \\ 2 & 9 & 2 \\ 4 & 25 & 2 \end{matrix} \right| \)
= 3(18 – 50) – 11(4 – 8) – 1(50 – 36)
= 3(32) -11(4) -1(14)
= -96 + 44 – 14 = -66
Δx = \(\left| \begin{matrix} 3 & 3 & 11 \\ 2 & -1 & 9 \\ 4 & 3 & 25 \end{matrix} \right| \)
= 3(-25 – 27) – 3(50 – 36) + 11(6 + 4)
= 3(-52) -3(14) + 11(10)
= -156 – 42 + 110
= -88
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.4 3

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.4

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.4 4
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.4 5
3a – 4b – 2c = 1 ……….. (1)
a + 2b + c = 2 …………… (2)
2a – 5b – 4c = -1 ………….. (3)
Δ = \(\left| \begin{matrix} 3 & -4 & -2 \\ 1 & 2 & 1 \\ 2 & -5 & -4 \end{matrix} \right| \)
= 3(-8 + 5) + 4 (-4 – 2) – 2(-5 – 4)
= 3(-3) +4(-6) -2(-9)
= -9 – 24 + 18
= -15 ≠ 0
Δa = \(\left| \begin{matrix} 1 & -4 & -2 \\ 2 & 2 & 1 \\ -1 & -5 & -4 \end{matrix} \right| \)
= 1(-8 + 5) + 4(-8 + 1) – 2(-10 + 2)
= 1(-3) + 4(-7) – 2(-8)
= -3 – 28 + 16
= -15
Δb = \(\left| \begin{matrix} 3 & 1 & -2 \\ 1 & 2 & 1 \\ 2 & -1 & -4 \end{matrix} \right| \)
= 3(-8 + 1) – 1(-4 – 2) – 2(-1 – 4)
= 3(-7) -1(-6) – 2(-5)
= – 21 + 6 + 10 = -5
Δc = \(\left| \begin{matrix} 3 & -4 & 1 \\ 1 & 2 & 2 \\ 2 & -5 & -1 \end{matrix} \right| \)
= 3(-2 + 10) + 4(-1 – 4) + 1 (-5 – 4)
= 24 – 20 – 9 = -5
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.4 6

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.4

Question 2.
In a competitive examination, one mark is awarded for every correct answer while \(\frac{1}{4}\) mark is deducted for every wrong answer. A student answered 100 questions and got 80 marks. How many questions did he answer correctly? (Use Cramer’s rule to solve the problem).
solution:
No. of Questions answered = 100
Let the No. of questions answered correctly be x
and the No. of questions answered wrongly be y
Here, x + y = 100 and x – \(\frac { 1 }{ 4 }\) y = 80
(i.e) x + y = 100 and 4x – y = 320
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.4 7
correct questions = 84
wrong questions = 16.

Question 3.
A chemist has one solution which is 50% acid and another solution which is 25% acid. How much each should be mixed to make 10 litres of a 40% acid solution? (Use Cramer’s rule to solve the problem).
solution:
Let two solutions x and y
x + y = 10 …….. (1)
0.25 x + (0.50)y = (0.40) ……….. (2)
(2) × 100 ⇒ 25x + 50y = 400
(2) ÷ 5 ⇒ 5x + 10y = 80 …………. (3)
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.4 8

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.4

Question 4.
A fish tank can be filled in 10 minutes using both pumps A and B simultaneously. However, pump B can pump water in or out at the same rate. If pump B is inadvertently run in reverse, then the tank will be filled in 30 minutes. How long would it take each pump to fill the tank by itself? (Use Cramer’s rule to solve the problems).
solution:
pump A fills \((\frac {1}{x})^{th}\) of the tank in 1 hour.
pump B fills \((\frac {1}{y})^{th}\) of the tank in 1 hour.
Both can filled \((\frac {1}{10})^{th}\) of the tank in 1 hour.
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.4 9
using Cramer’s rule
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.4 10
Pump A takes 15 minutes
Pump B takes 30 minutes

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.4

Question 5.
A family of 3 people went out for dinner in a restaurant. The cost of two dosai, three idlies and two vadais is Rs 150. The cost of the two dosai, two idlies and four vadais is Rs 200. The cost of five dosai, four idlies and two vadais is T 250. The family has Rs 350 in hand and they ate 3 dosai and six idlies and six vadais. Will they be able to manage to pay the bill within the amount they had?
solution:
Let the Cost of dosai, Idlies and vadais be x, y, z
2x + 3y + 2z = 150
2x + 2y + 4z = 200
5x + 4y + 2z = 250
Δ = \(\left| \begin{matrix} 2 & 3 & 2 \\ 2 & 2 & 4 \\ 5 & 4 & 2 \end{matrix} \right| \)
= 2(4 – 16) – 3(4 – 20) + 2(8 – 10)
= 2(-12) – 3(-16) + 2(-2)
= -24 + 48 – 4
= 20 ≠ 0
Δx = \(\left| \begin{matrix} 150 & 3 & 2 \\ 200 & 2 & 4 \\ 250 & 4 & 2 \end{matrix} \right| \)
= 150(4 – 16) – 3(400 – 1000) + 2(800 – 500)
= 150(-12) – 3(-600) + 2(300)
= -1800 + 1800 + 600
= 600
Δy = \(\left| \begin{matrix} 2 & 150 & 2 \\ 2 & 200 & 4 \\ 5 & 250 & 2 \end{matrix} \right| \)
= 2(400 – 1000) – 150(4 – 20) + 2(500 – 1000)
= 2(-600) – 150(-16) + 2(-500)
= -1200 + 2400 – 1000
= 200
Δz = \(\left| \begin{matrix} 2 & 3 & 150 \\ 2 & 2 & 200 \\ 5 & 4 & 250 \end{matrix} \right| \)
= 2(500 – 800) – 3(500 – 1000) + 150(8 – 10)
= 2(-300) – 3(-500) + 150(-2)
= -600 + 1500 – 300
= 600
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.4 11
x = Rs 30, y = Rs 10, z = Rs 30
There are 3 dosai, 6 idlies and 6 vadais
= 3x + 6y + 6z
= 3(30) + 6(10) + 6 (30)
= 90 + 60 + 180
= Rs. 330
They can eat within the amount.

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.4

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.3

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 1 Applications of Matrices and Determinants Ex 1.3 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.3

Question 1.
Solve the following system of linear equations by matrix inversion method.
(i) 2x + 5y = -2, x + 2y = -3
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.3 1

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.3

(ii) 2x – y = 8, 3x + 2y = -2
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.3 2
x = 2, y = -4

(iii) 2x + 3y – z = 9, x + y + z = 9, 3x – y – z = -1
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.3 3
|A| = 2(-1+1)-3(-1-3)-1(-1-3)
= 0 + 12 + 4 =16 ≠ 0 A-1 exists.
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.3 4
∴ x = 2, y = 3, z = 4

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.3

(iv) x + y + z – 2 = 0, 6x – 4y + 5z – 31 = 0, 5x + 2y + 2z = 13
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.3 5
AX = B
X = A-1B
A = \(\left[\begin{array}{ccc}
1 & 1 & 1 \\
6 & -4 & 5 \\
5 & 2 & 2
\end{array}\right]\)
|A| = 1(-8-10)-1(12-25)+1(12+20)
= 18 + 13 +32 = 27
≠ 0
A-1 Exists
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.3 6
∴ x = 3, y = -2, z = 1

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.3

Question 2.
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.3 7
Find the products AB and BA and hence solve the system of equations x + y + 2z = 1, 3x + 2y + z = 7, 2x + y + 3z = 2
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.3 8
AB = BA = 4I3
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.3 9
∴ x = 2, y = 1, z = -1

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.3

Question 3.
A man is appointed in a job with a monthly salary of a certain amount and a fixed amount of annual increment. If his salary was Rs 19,800 per month at the end of the first month after 3 years of service and Rs 23,400 per month at the end of the first month after 9 years of service, find his starting salary and his annual increment. (Use the matrix inversion method to solve the problem.)
Solution:
Let the man starting the salary be Rs x and his annual increment be Rs y.
Given x + 3y = 19,800
x + 9y = 23,400
The equation can be written as
\(\left[\begin{array}{ll}
1 & 3 \\
1 & 9
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{c}
19800 \\
23400
\end{array}\right]\)
AX = B
X = A-1B
A = \(\left[\begin{array}{ll}
1 & 3 \\
1 & 9
\end{array}\right]\)
To find A-1
|A| = 9 – 3 = 6 ≠ 0 A-1 Exists.
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.3 10
Monthly salary = Rs 18000
Annual increment = Rs 1800

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.3

Question 4.
4 men and 4 women can finish a piece of work jointly in 3 days while 2 men and 5 women can finish the same work jointly in 4 days. Find the time taken by one man alone and that of one woman alone to finish the same work by Using the matrix inversion method.
Solution:
Let the time taken by one man alone be x days and one woman alone be y days.
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.3 11
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.3 12
∴ One man can do 18 days
One woman can do 36 days

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.3

Question 5.
The prices of three commodities A, B, and C are Rs x,y, and z per unit respectively. A person P purchases 4 units of B and sells two units of A and 5 units of C. Person Q purchases 2 units of C and sells 3 units of A and one unit of B. Person R purchases one unit of A and sells 3 unit of B and one unit of C. In the process, P, Q and R earn Rs 15,000, Rs 1,000 and 14,000 respectively. Find the prices per unit of A, B, and C. (Use the matrix inversion method to solve the problem.)
Solution:
Let x, y, z are commodities of A, B, C
2x – 4y + 5z = 15,000 ………… (1)
3x + y – 2z = 1000 ………… (2)
-x + 3y + z = 4000 ………… (3)
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.3 13
|A| = 2(1 + 6)+ 4(3 – 2) + 5(9 + 1)
= 2(7) + 4(1) + 5(10)
= 14 + 4 + 50 = 68
≠ 0
A-1 Exists.
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.3 14
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.3 15
x = Rs 2000, y = Rs 1000, z = Rs 3000
The prices per unit of A, B, and C are Rs 2000, Rs 1000, Rs 3000

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.3

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

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 1 Applications of Matrices and Determinants Ex 1.2 Textbook Questions and Answers, Notes.

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 minor method:
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.2 1
Solution:

(i) A = \(\begin{bmatrix} 2 & -4 \\ -1 & 2 \end{bmatrix}\)
A is a matrix of order 2 × 2 and p(A) ≤ 2
Second order minor
|A| = \(\begin{bmatrix} 2 & -4 \\ -1 & 2 \end{bmatrix}\)
= 4 – 4 = 0
∴p(A) ≠ 2
First order minor is non vanishing
p(A) = 1

(ii) A = \(\left[\begin{array}{rr}
-1 & 3 \\
4 & -7 \\
3 & -4
\end{array}\right]\)
A is a matrix of order 3 × 2 and p(A) ≤ 2
Second order minor
\(\begin{bmatrix} -1 & 3 \\ 4 & -7 \end{bmatrix}\)
= 7 – 12 = -5 ≠ 0
∴ p(A) = 2

(iii) A = \(\left[\begin{array}{rrrr}
1 & -2 & -1 & 0 \\
3 & -6 & -3 & 1
\end{array}\right]\)
A is a matrix of order 2 × 4 and p(A) ≤ 2
Second order minor

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.2 2
= 1(-4 + 6) + 2(-2 + 30) + 3(2 – 20)
= 2 + 56 – 54 = 4 ≠ 0
∴p(A) = 3
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.2 3

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

Question 2.
Find the rank of the following matrices by row reduction method:
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.2 4
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.2 5
The last equivalent matrix is in row echelon form. It has two non-zero rows.
∴ p(A) = 2
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.2 6
The last equivalent matrix is in row echelon form. It has three non-zero rows.
∴ p(A) = 3

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

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.2 7
The last equivalent matrix is in row echelon form. It has three non-zero rows.
∴ p(A) = 3

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

Question 3.
Find the inverse of each of the following by Gauss-Jordan method:
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.2 8
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.2 9

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

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.2 10

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

Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.2 11

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

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 1 Applications of Matrices and Determinants Ex 1.1 Textbook Questions and Answers, Notes.

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

Question 1.
Find the adjoint of the following:
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 1
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 2
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 3

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

Question 2.
Find the inverse (if it exists) of the following.
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 4
Solution:
\(\begin{bmatrix} -2 & 4 \\ 1 & -3 \end{bmatrix}\)
|A| = 6 – 4 = 2 ≠ 0
∴ A-1 exists. A is non singular.
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 5
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 6
|A| = 2(8-7)-3(6-3)+1(21-12)
= 2 – 9 + 9 = 2 ≠ 0. A-1 exists.
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 7

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

Question 3.
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 8
Solution:
\(\left[\begin{array}{ccc}
\cos \alpha & 0 & \sin \alpha \\
0 & 1 & 0 \\
-\sin \alpha & 0 & \cos \alpha
\end{array}\right]\)
|F(α)| = cos α(cos α – 0) – 0 + sin α(0 + sin α)
= cos²α + sin²α = 1
|f(α)| = 1 ≠ 0. [F(α)]-1 exists.
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 9
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 10
[∵ cos (-θ) = cos θ ; sin(-θ) = -sin θ]
from (1) and (2) we have
[F(α)]-1 = F(-α)

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

Question 4.
If A = \(\begin{bmatrix} 5 & 3 \\ -1 & -2 \end{bmatrix}\), show that A² – 3A – 7I2 = O2. hence find A-1
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 11
∴ A² -3A – 7I2 = O2
Post multiply this equation by A-1
A2A-1 – 3A A-1 – 7I2 A-1 = 0
A – 3I – 7A-1 = 0
A – 3I = 7 A-1
A-1 = \(\frac {1}{7}\) (A – 3I)
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 12

Question 5.
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 13
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 14
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 15
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 16

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

Question 6.
If A = \(\begin{bmatrix} 8 & -4 \\ -5 & 3 \end{bmatrix}\) verify that A(adj A) = (adj A) A = \(\left| A \right|\)I2.
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 17
(1), (2) and (3) ⇒ A (adj A) = (adj A)A = |A| I2.

Question 7.
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 18
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 19
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 20
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 21

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

Question 8.
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 22
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 23
|adj (A)| = 2 (24 – 0) + 4 (- 6 – 14) + 2(0 + 24)
= 48 – 80 + 48 = 16
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 24

Question 9.
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 25
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 26
|adj A| = 0 + 2(36 – 18) + 0 = 2(18) = 36
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 27

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

Question 10.
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 28
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 29

Question 11.
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 30
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 31
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 32
Hence proved

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

Question 12.
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 33
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 34
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 35

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

Question 13.
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 36
Solution:
Given A × B × C
⇒ A-1 A × B B-1 = A-1 C B-1
I × I = A-1 C B-1
⇒ X = A-1 CB-1
let us find A-1 and B-1
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 37
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 38

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

Question 14.
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 39
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 40
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 41
Hence proved.

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

Question 15.
Decrypt the received encoded message [2 -3] [20 4] with the encryption matrix \(\begin{bmatrix} -1 & -1 \\ 2 & 1 \end{bmatrix}\) 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:
Samacheer Kalvi 12th Maths Guide Chapter 1 Applications of Matrices and Determinants Ex 1.1 42
So the sequence of decoded row matrics is [8 5] [12 16]
The receiver reads the message as “HELP”.