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Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 2 Integral Calculus I Ex 2.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.4

Question 1.
Integrate the following with respect to x.
2 cos x – 3 sin x + 4 sec² x – 5 cosec² x
Solution:
∫2 cos x – 3 sin x + 4 sec² x – 5 cosec² x
= 2 ∫ cos x dx – 3 ∫ sin x dx + 4 ∫ sec² x – 5 ∫ cosec² x dx
= 2 sin x + 3 cos x + 4 tan x + 5 cot x + c

Question 2.
sin³x
Solution:
sin³x = 3 sin x – 4sin³x
4sin³x = 3 sin x – sin3x
⇒ sin³x = \(\frac { 1 }{4}\) [3 sin x – sin 3x]
∫sin³x dx = \(\frac { 1 }{4}\) ∫(3 sin x – sin3x) dx
= \(\frac { 1 }{4}\) [3(-cos x) – (\(\frac { -cos3x }{2}\)) + c
= \(\frac { -3 }{4}\) [cos x] + \(\frac { 1 }{12}\) (cos 3x) + c

Question 3.
\(\frac { cos 2x+2sin^2x }{cos^2x}\)
Solution:

Question 4.
\(\frac { 1} {sin^2x cos^2x}\)
Solution:

Question 5.
\(\sqrt { 1-sin 2x }\)
Solution:
\(\sqrt { 1-sin 2x }\) dx = ∫\(\sqrt { sin^2x+cos^2x-2sinx cos x dx }\)
∫\(\sqrt { (sin x-cos x)^2 }\) dx
∫(sin x – cos x) dx = ∫sin x dx – ∫cos x dx
= -cos x – sin x + c
= -[cos x + sin x] + c

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 2 Integral Calculus I Ex 2.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.3

Question 1.
Integrate the following with respect to x.
∫(e^{x log a} + e^{a log a} – e^{n log x}
Solution:
(e^{x log a} + e^{a log a} – e^{a log x}) dx
= ∫(e^{log ax} + e^{log aa} – e^{a log xn}) dx
= ∫(a¹ + a^a – xⁿ) dx
= [\(\frac { a^x }{log|a|}\) + a^a (x) – \(\frac { x^{n+1} }{(n+1)}\)] + c

Question 2.
\(\frac { a^x-e^{x log b} }{e^{x log a} b^x}\)
Solution:

Question 3.
(e^x + 1)² e^x
Solution:
∫(e^x + 1)² e^x dx = ∫[(e^x)² + 2e^x + 1]e^x dx
= ∫(e^2x. e^x + 2e^x. e^x + e^x) dx
= ∫(e^3x + 2e^2x + e^x) dx
= \(\frac { e^{3x} }{3}\) + \(\frac { 2e^{2x} }{2}\) + e^x + c
= \(\frac { e^{3x} }{3}\) + e^2x + e^x + c

Question 4.
\(\frac { e^{3x}-e^{-3x} }{e^{x}}\)
Solution:

Question 5.
\(\frac { e^{3x}+e^{5x} }{e^{x}+e^{-x}}\)
Solution:

Question 6.
[1 – \(\frac { 1 }{x^2}\)] e^{[x + \(\frac { 1 }{x}\)]}
Solution:

Question 7.
\(\frac { 1 }{x(log x)^2}\)
Solution:

Question 8.
Given f'(x) = e^x and f(0) = 2 then find f(x)
Solution:
f'(x) = e^x
Integrating both sides of the equation,
∫ f'(x) dx = ∫ e^x dx
⇒ f(x) = e^x + c
given f(0) = 2
2 = e⁰ + c
⇒ c = 1
Thus f(x) = e^x + 1

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 2 Integral Calculus I Ex 2.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.2

Question 1.
Integrate the following with respect to x.
[\(\sqrt { 2x}\) – \(\frac { 1 }{\sqrt { 2x}}\)]²
Solution:

Question 2.
\(\frac { x^4+x^2+2 }{(x-1)}\)
Solution:

Question 3.
\(\frac { x^3 }{x+2}\)
Solution:

Question 4.
\(\frac { x^3+3x^2-7x+11 }{x+5}\)
Solution:

Question 5.
\(\frac { 3x+2 }{(x-2)(x-3)}\)
Solution:
By partial fraction

Question 6.
\(\frac { 4x^2+2x+6 }{(x+1)^2(x-3)}\)
Solution:
By partial fraction

⇒ 4x² + 2x + 6 = A (x + 1)² + B (x – 3) (x + 1) + c (x – 3)
Put x = 3
4(9) + 2(3) + 6 = A(4)²
36 + 6 + 6 = 16A ⇒ A = \(\frac { 48 }{16}\)
A = 3
Put x = -1
4(1) + 2(-1) + 6 = C(-4)
4 – 2 + 6 = -4C ⇒ -4c = 8
C = -2
Put x = 0
6 = A(1) + B(-3) + C(-3)
6 = 3(1) – 3B + (-2) (-3)
6 = 3 – 3B + 6 ⇒ 3B = 3
B = 1

Question 7.
\(\frac { 3x^2-2x+5}{(x-1)(x^2+5)}\)
Solution:

Question 8.
Given f'(x) = \(\frac { 1 }{x}\) and f (1) = \(\frac { 1 }{π}\), then find f(x)
Solution:
f'(x) = \(\frac { 1 }{x}\)
f(x) = ∫f'(x) dx = ∫\(\frac { 1 }{x}\) dx
f(x) = log|x| + c
f(1) = π/4 ⇒ log|1| + c = π/4.
⇒ 0 + c = π/4
∴ c = π/4
∴ Required f (x) = log |x| + π/4

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 2 Integral Calculus I Ex 2.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.1

Question 1.
Integrate the following with respect to x.
\(\sqrt { 3x+5 }\)
Solution:

Question 2.
(9x² – \(\frac { 4 }{x^2}\))²
Solution:
A = \( \left[\begin{array}{cccc}
-2 & 1 & 3 & 4 \\
0 & 1 & 1 & 2 \\
1 & 3 & 4 & 7
\end{array}\right]\)
The order of A is 3 × 4
∴ P(A) < 3
Let us transform the matrix A to an echelon form

The number of non-zero rows = 3
∴ P(A) = 3

Question 3.
(3 + x) (2 – 5x)
Solution:
∫(3 + x) (2 – 5x) dx
= ∫(6 – 15x + 2x – 5x²) dx
= ∫(6 – 13x – 5x²) dx
= 6x – \(\frac { 13x^2 }{2}\) – \(\frac { 5x^3 }{3}\) + c

Question 4.
√x (x³ – 2x + 3)
Solution:

Question 5.
\(\frac { 8x+13 }{\sqrt{4x+7}}\)
Solution:

Question 6.
\(\frac { 1 }{\sqrt{x+1}+\sqrt{x-1}}\)
Solution:

Question 7.
Given f'(x) = x + b, f(1) = 5 and f(2) = 13, then find f(x)
Solution:
f(x) = ∫f'(x) dx
f(x) = ∫(x + b) dx = \(\frac { x^2 }{2}\) + bx + c
f (1) = 5 ⇒ \(\frac { (1)^2 }{2}\) + b(1) + c = 5
\(\frac { 1 }{2}\) + b + c = 5 ⇒ b + c = 5 – 1/2
b + c = \(\frac { 9 }{2}\) ⇒ 2b + 2c = 9 …….. (1)
f(2) = 13 ⇒ \(\frac { (2)^2 }{2}\) + b(2) + c = 13
2 + 2b + c = 13
2b + c = 11 ……… (2)
Solving eqn (1) & (2)

Substitute c = -2 in eqn (2)
2b – 2 = 11 ⇒ 2b = 11 + 2
b = 13/2
f(x) = \(\frac { x^2 }{2}\) + \(\frac { 13x }{2}\) – 2

Question 8.
Given f'(x) = 8x³ – 2x and f (2) = 8, then find f(x)
Solution:
f(x) = ∫f'(x)dx
= ∫(8x³ – 2x)dx = 8(\(\frac { x^4 }{4}\)) -2(\(\frac { x^2 }{2}\)) + c
f(x) = 2x⁴ – x² + c
f (2) = 8 ⇒ 2(2)⁴ – (2)² + c = 8
32 – 4 + c = 8 ⇒ c = -20
∴ f (x) = 2x⁴ – x² – 20

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 1 Applications of Matrices and Determinants Miscellaneous Problems Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 1 Applications of Matrices and Determinants Miscellaneous Problems

Question 1.
Find the rank of the matrix
A = \(\left(\begin{array}{cccc}
1 & -3 & 4 & 7 \\
9 & 1 & 2 & 0
\end{array}\right)\)
Solution:
\(\left(\begin{array}{cccc}
1 & -3 & 4 & 7 \\
9 & 1 & 2 & 0
\end{array}\right)\)
The order of A is 2 × 4
∴ P(A) ≤ 2
Let us transform the matrix A to an echelon form

The number of non-zero rows is 2
∴ P(A) = 2

Question 2.
Find the rank of the matrix
A = \( \left[\begin{array}{cccc}
-2 & 1 & 3 & 4 \\
0 & 1 & 1 & 2 \\
1 & 3 & 4 & 7
\end{array}\right]\)
Solution:
A = \( \left[\begin{array}{cccc}
-2 & 1 & 3 & 4 \\
0 & 1 & 1 & 2 \\
1 & 3 & 4 & 7
\end{array}\right]\)
The order of A is 3 × 4
∴ P(A) < 3
Let us transform the matrix A to an echelon form

The number of non-zero rows = 3
∴ P(A) = 3

Question 3.
Find the rank of the matrix
A = \(\left[\begin{array}{llll}
4 & 5 & 2 & 2 \\
3 & 2 & 1 & 6 \\
4 & 4 & 8 & 0
\end{array}\right]\)
Solution:
A = \(\left[\begin{array}{llll}
4 & 5 & 2 & 2 \\
3 & 2 & 1 & 6 \\
4 & 4 & 8 & 0
\end{array}\right]\)
The order of A is 3 × 4
∴ p(A) < 3
Let us transform the matrix A to an echelon form


The last equivalent matrix is in the echelon form.
Number of non-zero rows = 3
∴ P(A) = 3

Question 4.
Examine the consistency of the system of equations;
x + y + z = 7, x + 2y + 3z = 18, y + 2z = 6
Solution:
x + y + z = 7
x + 2y + 3z = 18
y + 2z = 6
The matrix form of the equations

Here p(A) ≠ p(A, B)
∴ The given system is inconsistent and has no solution.

Question 5.
Find k if the equations 2x + 3y – z = 5, 3x – y + 4z = 2, x + 7y – 6z = k are consistent.
Solution:
2x + 3y – z = 5
3x – y + 4z = 2
x + 7y – 6z = k
The matrix form of these equations

Obviously, the last equivalent matrix is in the ech-elon form.
Since the equations are consistent
P(-A) = p(A, B)
p(A) = 2 and p (A, B) = 2 then
k = 8

Question 6.
Find k if the equations x + y + z = 1, 3x – y – z = 4, x + 5y + 5z = k are inconsistent.
Solution:
x + y + z = 1
3x – y – z = 4
x + 5y + 5z = k
The matrix equation corresponding to the given system is

Obviously, the last equivalent matrix is in the ech-elon form.
Since the equations are inconsistent
p(A) ≠ p (A, B)
Here p(A) = 2 but p(A, B) should not equal to 2
∴ k ≠ 0
The equations are inconsistent when k assume any real value other than 0.

Question 7.
Solve the equations x + 2y + z = 7, 2x – y + 2z = 4, x + y – 2z = -1 by using Cramer’s rule.
Solution:
x + 2y + z = 7
2x – y + 2z = 4
x + y – 2z = -1
Here Δ = \(\left|\begin{array}{ccc}
1 & 2 & 1 \\
2 & -1 & 2 \\
1 & 1 & -2
\end{array}\right|\)
= 1(2 – 2)-2 (-4 – 2) + 1(2 + 1)
= 1(0) -2 (-6) + 1(3)
= 12 + 3 = 15 ≠ 0
∴ We can apply Cramer’s Rule and the system is consistent and it has unique solution.
Δ_x = \(\left|\begin{array}{ccc}
7 & 2 & 1 \\
4 & -1 & 2 \\
-1 & 1 & -2
\end{array}\right|\)
= 7 (2 – 2) -2 (-8 + 2) + 1(4 – 1)
= 7(0)-2 (-6) + 1(3)
= 12 + 3
= 15
Δ_y = \(\left|\begin{array}{ccc}
1 & 7 & 1 \\
2 & 4 & 2 \\
1 & -1 & -2
\end{array}\right|\)
= 1(-8 + 2) -7 (-4 – 2) + 1(-2 – 4)
= 1 (-6) – 7 (-6) + 1 (-6)
= -6 + 42 – 6 = 30
Δ_z = \(\left|\begin{array}{ccc}
1 & 2 & 7 \\
2 & -1 & 4 \\
1 & 1 & -1
\end{array}\right|\)
= 1(1 – 4) -2 (-2 – 4) + 7 (2 + 1)
= 1 (-3) -2 (-6) + 7 (3)
= -3 + 12 + 21 = 30
∴ By cramer’s rule

∴ The Solution is (x, y, z) = (1, 2, 2)

Question 8.
The cost of 2kg. of wheat and 1kg. of sugar is Rs 100. The cost of 3kg. of wheat, 2kg. of sugar and 1kg of rice is Rs 220. Find the cost of each per kg. using Cramer’s rule.
Solution:
Let cost of 1 kg of wheat be x; cost of 1kg of sugar be y and cost of 1 kg of rice be z.
2x + y = 100
x + z = 80
3x + 2y + z = 220
Here Δ = \(\left|\begin{array}{ccc}
2 & 1 & 0 \\
1 & 0 & 1 \\
3 & 2 & 1
\end{array}\right|\)
= 2 (0 – 2) -1 (1 – 3)
= – 4 + 2 = -2
≠ 0
∴ We can apply Cramer’s Rule and the system is consistent and it has unique solution.
Δ_x = \(\left|\begin{array}{ccc}
100 & 1 & 0 \\
80 & 0 & 1 \\
220 & 2 & 1
\end{array}\right|\)
= 100 (0 – 2) -1 (80 – 220)
= – 200 -1 (-140)
= – 200 +140
= -60
Δ_y = \(\left|\begin{array}{ccc}
2 & 100 & 0 \\
1 & 80 & 0 \\
3 & 220 & 1
\end{array}\right|\)
= 2 (80 – 220) -100 (1 – 3)
= 2 (-140) – 100 (-2)
= – 280 + 200
= -80
Δ_x = \(\left|\begin{array}{ccc}
2 & 1 & 100 \\
1 & 0 & 80 \\
3 & 2 & 220
\end{array}\right|\)
= 2 (0 -160) -1 (220 – 240) + 100 (2 – 0)
= 2 (-160) -1 (-20) + 100 (2)
= -320 + 20 + 200
= -100
∴ By cramer’s rule

The solution is (x, y, z) = (30, 40, 50)
∴ The cost of 1 kg of wheat is Rs 30; the cost of 1 kg of sugar is Rs 40 and the cost of 1 kg of rice is Rs 50.

Question 9.
A salesman has the following record of sales during three items A, B, and C, which has different rates of commission.

Find out the rate of commission on items A, B, and C by using Cramer’s rule.
Solution:
Let x, y, and z be the commission on items A, B, and C respectively.
90x + 100y + 20z = 800 ⇒ 9x + 10y + 2z = 80 ……… (1)
130x + 50y + 40z = 900 ⇒ 13x + 5y + 4z = 90 ………. (2)
60x + 100y + 30z = 850 ⇒ 6x + 10y + 3z = 85 ………. (3)
Here Δ = \(\left|\begin{array}{ccc}
9 & 10 & 2 \\
13 & 5 & 4 \\
6 & 10 & 3
\end{array}\right|\)
= 9 (15 – 40) -10 (39 – 24) + 2 (130 – 30)
= 9 (-25) – 10 (15) + 2 (100)
= -225 – 150 + 200
= -175
≠ 0
∴ We can apply Cramer’s Rule and the system is consistent and it has unique solution.
Δ_x = \(\left|\begin{array}{ccc}
80 & 10 & 2 \\
90 & 5 & 4 \\
85 & 10 & 3
\end{array}\right|\)
= 80 (15 – 40) -10 (270 – 340) + 2 (900 – 425)
= 80 (-25) -10 (-70) + 2 (475)
= -2000 + 700 + 950
= -350
Δ_y = \(\left|\begin{array}{ccc}
9 & 80 & 2 \\
13 & 90 & 4 \\
6 & 85 & 3
\end{array}\right|\)
= 9 (270 – 340) – 80 (39 – 24) + 2 (1105 – 540)
= 9 (-70)-80 (15) + 2 (565)
= -630 – 1200 + 1130
= -700
Δ_z = \(\left|\begin{array}{ccc}
9 & 10 & 80 \\
13 & 5 & 90 \\
6 & 10 & 85
\end{array}\right|\)
= 9 (425 – 900) -10 (1105 – 540) + 80 (130 – 30)
= 9 (-475) -10 (565) + 80 (100)
= – 4275 – 5650 + 8000
= -1925
∴ By cramer’s rule

The solution is (x, y, z) = (2, 4, 11)
∴ The rates of commission for A, B and C are Rs 2, Rs 4 and Rs 11 respectively.

Question 10.
The subscription department of a magazine sends out a letter to a large mailing list inviting subscriptions for the magazine. Some of the people receiving this letter already subscribe to the magazine while others do not. From this mailing list, 60% of those who already subscribe will subscribe again while 25% of those who do not now subscribe will subscribe. In the last letter, it was found that 40% of those receiving it ordered a subscription. What percent of those receiving the current letter can be expected to order a subscription?
Solution:
The transition probability matrix is S

∴ 39% of those receiving the current letter can be expected to order a subscription.

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

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

Choose the most suitable answer from the given four alternatives:

Question 1.
A = (1, 2, 3), then the rank of AA^T is
(a) 0
(b) 2
(c) 3
(d) 1
Solution:
(d) 1
Hint:
A = (1, 2, 3) then
A^T = \(\left(\begin{array}{l}
1 \\
2 \\
3
\end{array}\right)\)
AA^T = (1, 2, 3) \(\left(\begin{array}{l}
1 \\
2 \\
3
\end{array}\right)\) = (1 + 4 + 9) = (14)
Number of non-zero rows = 1
∴ P (AA^T) = 1

Question 2.
The rank of m × n matrix whose elements are unity is
(a) 0
(b) 1
(c) m
(d) n
Solution:
(b) 1

Question 3.
If T = is a transition probability matrix, then at equilibriuium A is equal to
(a) \(\frac { 1 }{ 4 }\)
(b) \(\frac { 1 }{ 5 }\)
(c) \(\frac { 1 }{ 6 }\)
(d) \(\frac { 1 }{ 8 }\)
Solution:
(a) \(\frac { 1 }{ 4 }\)
Hint:

At equilibrium (A B) \(\left(\begin{array}{ll}
0.4 & 0.6 \\
0.2 & 0.8
\end{array}\right)\)
0.4 A + 0.2 B = A
0.4 A + 0.2(1 – A) = A
0.4 A + 0.2 – 0.2 A = A
0.2 A + 0.2 = A
0.2 = A – 0.2 A
0.2 = 0.8 A
A = \(\frac { 0.2 }{ 0.8 }\) = \(\frac { 1 }{ 4 }\)

Question 4.
If A = \(\left(\begin{array}{ll}
2 & 0 \\
0 & 8
\end{array}\right)\), then p(A) is
(a) 0
(b) 1
(c) 2
(d) n
Solution:
(c) 2
Hint:
A = \(\left(\begin{array}{ll}
2 & 0 \\
0 & 8
\end{array}\right)\)
No. of Non-zero rows = 2
∴ p(A) = 2

Question 5.
The rank of the matrix \(\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 2 & 3 \\
1 & 4 & 9
\end{array}\right]\) is
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(d) 3
Hint:

No. of Non-zero rows = 3
∴ p(A) = 3

Question 6.
The rank of the unit matrix of order n is
(a) n – 1
(b) n
(c) n + 1
(d) n²
Solution:
(b) n

Question 7.
If p(A) = r then which of the following is correct?
(a) all the minors of order r which does not vanish
(b) A has at least one minor of order r which does not vanish
(c) A has at least one (r + 1) order minor which vanishes
(d) all (r + 1) and higher order minors should not vanish
Solution:
(b) A has at least one minor of order r which does not vanish.

Question 8.
If A = \(\left(\begin{array}{l}
1 \\
2 \\
3
\end{array}\right)\) then the rank of AA^T is
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(b) 1
Hint:

No. of non-zero rows = 1
p (AA^T) = 1

Question 9.
If the rank of the matrix \(\left[\begin{array}{ccc}
\lambda & -1 & 0 \\
0 & \lambda & -1 \\
-1 & 0 & \lambda
\end{array}\right]\) is 2. then λ is
(a) 1
(b) 2
(c) 3
(d) only real number
Solution:
(a) 1
Hint:
A = \(\left[\begin{array}{ccc}
\lambda & -1 & 0 \\
0 & \lambda & -1 \\
-1 & 0 & \lambda
\end{array}\right]\)
Since the Rank is 2, third order matrix vanishes
∴ |A| = 0
\(\left[\begin{array}{ccc}
\lambda & -1 & 0 \\
0 & \lambda & -1 \\
-1 & 0 & \lambda
\end{array}\right]\) = 0
λ(λ² – 0) + 1 (0 – 1) = 0
λ³ – 1 = 0 ⇒ λ³ = 1
∴ λ = 1

Question 10.

(a) 0
(b) 2
(c) 3
(d) 5
Solution:
(c) 3
Hint:
Number of non-zero rows = 3
∴ Rank = 3

Question 11.
If is a transition probability matrix, then the value of x is
(a) 0.2
(b) 0.3
(c) 0.4
(d) 0.7
Solution:
(c) 0.4
Hint:

is a transition probability matrix then 0.6 + x = 1
x = 1 – 0.6
x = 0.4

Question 12.
Which of the following is not an elementary transformation?
(a) R_i ↔ Rj
(b) R_i → 2R_i + 2Cj
(c) R_i → 2R_i + 4Rj
(d) C_i → C_i + 5Cj
Solution:
(b) R_i → 2R_i + 2Cj

Question 13.
If p(A) = p(A,B)= then the system is
(a) Consistent and has infinitely many solutions
(b) Consistent and has a unique solution
(c) inconsistent
(d) consistent
Solution:
(d) Consistent

Question 14.
If p(A) = p(A,B)= the number of unknowns, then the system is
(a) Consistent and has infinitely many solutions
(b) Consistent and has a unique solution
(c) inconsistent
(d) consistent
Solution:
(b) Consistent and has a unique solution

Question 15.
If p(A) ≠ p(A, B) =, then the system is
(a) Consistent and has infinitely many solutions
(b) Consistent and has a unique solution
(c) inconsistent
(d) consistent
Solution:
(c) inconsistent

Question 16.
In a transition probability matrix, all the entries are greater than or equal to
(a) 2
(b) 1
(c) 0
(d) 3
Solution:
(c) 0

Question 17.
If the number of variables in a non-homogeneous system AX = B is n, then the system possesses a unique solution only when
(a) p(A) = p(A, B) > n
(b) p(A) = p(A, B) = n
(c) p(A) = p(A, B) < n
(d) none of these
Solution:
(b) p(A) = p(A, B) = n

Question 18.
The system of equations 4x + 6y = 5, 6x + 9y = 7 has
(a) a unique solution
(b) no solution
(c) infinitely many solutions
(d) none of these
Solution:
(b) no solution
Hint:
The matrix form of the equations

p(A) ≠ p(A, B)
∴ The system is inconsistent.

Question 19.
For the system of equations x + 2y + 3z = 1, 2x + y – z = 3, 3x + 2y + k = 4
(a) there is only one solution
(b) there exists infinitely many solution
(c) there is no solution
(d) none of these
Solution:
(a) there is only one solution
Hint:
The matrix form of the equations

The last equivalent matrix in the echelon form
p(A) = p (A, B) = no. of unknowns
∴ The system is consistent.

Question 20.
If |A| ≠ 0, then A is
(a) non – singular matrix
(b) singular matrix
(c) zero matrix
(d) none of these
Solution:
(a) non-singular matrix

Question 21.
The system of linear equations x = y + z = 2, 2x + y – z = 3, 3x + 2y + k = 4 has unique solution, if k is not equal to
(a) 1
(b) 0
(c) 3
(d) 7
Solution:
(b) 0
Hint:
The matrix form of the equations

Since the system has unique solution.
P(A) = p(A, B) ≠ n
∴ K ≠ 0

Question 22.
Cramer’s rule is applicable only to get an unique solution when
(a) Δ_z ≠ 0
(b) Δ_x ≠ 0
(c) Δ ≠ 0
(d) Δ_y ≠ 0
Solution:
(c) Δ ≠ 0

Question 23.

Then (x, y) is

Solution:
(d) (\(\frac { -Δ_1 }{ Δ_2 }\), \(\frac { -Δ_1 }{ Δ_3 }\))

Question 24.
|A_{n × n}| = 3 |adj A| = 243 = (3)5 then the value n is
(a) 4
(b) 5
(c) 6
(d) 7
Solution:
(c) 6

Question 25.
Rank of a null matrix is
(a) 0
(b) -1
(c) ∞
(d) 1
Solution:
(a) 0

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

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

Question 1.
The subscription of a magazine sends out a letter to a large mailing list inviting subscriptions for the magazine. Some of the people receiving this letter already subscribe to the magazine while others do not. From this mailing list, 45% of those who already subscribe will subscribe again while 30% of those who do not now subscribe will subscribe. On the last letter, it was found that 40% of those receiving it ordered a subscription. What percent of those receiving the current letter can be expected to order a subscription?
Solution:
Transition Probability Matrix

S = 36%; F = 64%
∴ 36% of those receiving the current letter can be expected to order a subscription

Question 2.
Anew transit system has just gone into operation in Chennai. Of those who use the transit system this year, 30% will switch over to using metro train next year and 70% will continue to use he transit system. Of those who use metro train this year and 70% will continue to use metro train next year and 30% will switch over to transit system. Suppose the population of Chennai city remains constant and that 60% of the commuters use the transit system and 40% of the commuters use metro train this year.
(i) What percent of commuters will be using the transit system after one year?
(ii) What percent of commuters will be using the transit system in the one run?
Solution:
Transition Probability Matrix

S = 54%; C = 46%
Equilibrium will be reached in the long run
At equilibrium we must have
(S C) T = (S C) where S + C = 1
(S C) = \(\left(\begin{array}{ll}
0.7 & 0.3 \\
0.3 & 0.7
\end{array}\right)\) = (S C)
0.7S + 0.3C = S
0.7S + 0.3 (1 – S) = S
0.7S + 0.3 – 0.3S = S
0.4S + 0.3 = S
0.3 = S – 0.4S ⇒ 0.6S = 0.3
S = \(\frac { 0.3 }{ 0.6 }\) = 0.50
∴ 50% of the commuters will be transit system the long run.

Question 3.
Two types of soaps A and B are in the market. Their present market shares are 15% for A and 85% for B. Of those who bought A the previous year, 65% contionues to buy it again while 35% switch over to B. Of those who bought B the previous year, 55% buy it again and 45% switch over to A. Find their market shares after one year and when is the equilibrium reached?
Solution:
Transition Probability Matrix

Equilibrium will be reached in the long run
At equilibrium we must have
(A B) T = (A, B) where A + B = 1
(A B) = \(\left(\begin{array}{ll}
0.65 & 0.35 \\
0.45 & 0.55
\end{array}\right)\) = (A B)
0.65 A + 0.45 B = A
0.65 A + 0.45 (1 – A) = A
0.65 A + 0.45 – 0.45 A = A
0.20 A + 0.45 = A
A – 0.20 A = 0.45
0.80 A = 0.45
A = \(\frac { 0.45 }{ 0.80 }\) = 0.5625 and B = 04375
In the long sum {∴ B = 1 – A}
∴ A = 56.25 % and B = 43.75%

Question 4.
Two products A and B currently share the market with shares of 50% and 50% each respectively. Each week some brand switching takes place. Of those who bought A the previous week, 60% buy it again whereas 40% switch over to B. Of those who bought B the previous week, 80% buy it again whereas 20% switch over to A. Find their shares after one week and after two weeks. If the price war continues, when is the equilibrium reached?
Solution:
Transition Probability Matrix

Equilibrium will be reached in the long run.
At equilibrium, we must have
(A B) T = (A B) where A + B = 1
(A B) = \(\left(\begin{array}{ll}
0.60 & 0.40 \\
0.20 & 0.80
\end{array}\right)\) = (A B)
0.60 A + 0.20 B = A
0.60 A+ 0.20 (1 – A) = A
0.60 A + 0.20 – 0.20 A = A
0.40 A + 0.20 = A
0.20 = A – 0.40 A
0.60 A = 0.20
A = \(\frac { 0.20 }{ 0.60 }\) = 0.33 and B = 0.67
{∴ B = 1 – A}
In the long run
∴ A = 33% and B = 67%

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

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

Question 1.
Solve the following . equations by using Cramer’s rule
(i) 2x + 3y = 7; 3x + 5y = 9
(ii) 5x + 3v = 17; 3x + 7y = 31
(iii) 2x + y – z = 3; x + y + z – 1; x – 2y – 3z = 4
(iv) x + y + z = 6; 2x + 3y – z = 56; x – 2y – 3z = -7
Solution:
The equations are
2x + 3y = 7
3x + 5y = 9
Here Δ = \(\left|\begin{array}{ll}
2 & 3 \\
3 & 5
\end{array}\right|\) = 10 – 9 = 1
≠ 0
∴ We can apply Cramer’s Rule

∴ x = 8; y = -3

(ii) The equations are
5x + 3y = 17
3x + 7y = 31
Here Δ = \(\left|\begin{array}{ll}
5 & 3 \\
3 & 7
\end{array}\right|\) = 35 – 9 = 26
≠ 0
We can apply Cramer’s Rule

∴ x = 1; y = 4

(iii) The equations are
2x + y – z = 3
x + y + z = 1
x – 2y – 3z = 4
Here Δ = \(\left|\begin{array}{ccc}
2 & 1 & -1 \\
1 & 1 & 1 \\
1 & -2 & -3
\end{array}\right|\)
= 2 (-3 + 2) – 1 (-3 – 1) – 1 (-2 – 1)
= 2 (-1) -1 (-4) -1 (-3)
= -2 + 4 + 3
Δ = 5 ≠ 0
∴ We can apply Cramer’s Rule and the system is consistant and it has unique solution.
Δ_x = \(\left|\begin{array}{ccc}
3 & 1 & -1 \\
1 & 1 & 1 \\
4 & -2 & -3
\end{array}\right|\)
= 3 (-3 + 2) – 1 (-3 – 4) – 1 (-2 – 4)
= 3 (-1) -1 (-7) -1 ( 6)
= -3 + 7 + 6 = 10
Δ_y = \(\left|\begin{array}{ccc}
2 & 3 & -1 \\
1 & 1 & 1 \\
1 & 4 & -3
\end{array}\right|\)
= 2 (- 3 – 4) -3 (-3 – 1) -1 (4 – 1)
= 2 (-7) – 3 (-4) – 1 (3)
= -14 + 12 – 3
= -5
Δ_z = \(\left|\begin{array}{ccc}
2 & 1 & 3 \\
1 & 1 & 1 \\
1 & -2 & 4
\end{array}\right|\)
= 2 (4 + 2) – 1 (4 – 1) + 3 (- 2 – 1)
= 2 (6) – 1(3) + 3 (-3)
= 12 – 3 – 9 = 0
∴ By Cramer’s rule

∴ (x, y, z) = (2, -1, 0)

(iv) The equations are
x + y + z = 6
2x + 3y – z = 5
6x – 2y – 3z = -7
Here Δ = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
2 & 3 & -1 \\
6 & -2 & -3
\end{array}\right|\)
= 1 (-9 – 2) -1 (-6 + 6) + 1 (-4 – 18)
= 1 (-11) – 1 (0) +1 (-22)
= -11 – 22
= -33 ≠ 0
∴ We can apply Cramer’s Rule and the system is consistent and it has unique solution.
Now,
Δ_x = \(\left|\begin{array}{ccc}
6 & 1 & 1 \\
5 & 3 & -1 \\
-7 & -2 & -3
\end{array}\right|\)
= 6 (- 9 – 2) – 1 (- 15 – 7) + 1 (- 10 + 21)
= 6 (-11) – 1 (-22)+ 1 (11)
= -66 + 22 + 11
= -33
Δ_y = \(\left|\begin{array}{ccc}
1 & 6 & 1 \\
2 & 5 & -1 \\
6 & -7 & -3
\end{array}\right|\)
= 1 (-15 – 7) – 6 (- 6 + 6) + 1 (- 14 – 30)
= 1 (-22) – 6 (0) + 1 (-44)
= -66
Δ_z = \(\left|\begin{array}{ccc}
1 & 1 & 6 \\
2 & 3 & 5 \\
6 & -2 & -7
\end{array}\right|\)
= 1 (- 21 + 10) – 1 (- 14 – 30) + 6 (- 4 – 18)
= 1 (-11) – 1 (-44) + 6 (-22)
= -11 + 44 – 132
= -99
∴ By Cramer’s rule

(x, y, z) = (1, 2, 3)

(v) The equations are
x + 4y + 3z = 2
2x – 6y + 6z = -3
5x – 2y + 3z = -5
Here Δ = \(\left|\begin{array}{ccc}
1 & 4 & 3 \\
2 & -6 & 6 \\
5 & -2 & 3
\end{array}\right|\)
= 1 (-18 + 12) -4 (6 -30) + 3 (-4 + 30)
= 1 (-6) – 4 (-24) + 3 (26)
= -6 + 96 + 78
= 168 ≠ 0
We can apply Cramer’s Rule and the system is consistent and it has unique solution.
Δ_x = \(\left|\begin{array}{ccc}
2 & 4 & 3 \\
-3 & -6 & 6 \\
-5 & -2 & 3
\end{array}\right|\)
= 2 (-18 + 12) – 4 (-9 + 30) + 3 (6 – 30)
= 2 (-6) -4 (21) + 3 (-24)
= -12 – 84 – 72
= -168
Δ_y = \(\left|\begin{array}{ccc}
1 & 2 & 3 \\
2 & -3 & 6 \\
5 & -5 & 3
\end{array}\right|\)
= 1 (-9 + 30) – 2 (6 – 30) + 3 (-10 + 15)
= 1(21) – 2 (-24) + 3 (5)
= 21 + 48 + 15
= 84
Δ_z = \(\left|\begin{array}{ccc}
1 & 4 & 2 \\
2 & -6 & -3 \\
5 & -2 & -5
\end{array}\right|\)
= 1 (30 – 6) – 4 (-10 +15) + 2 (-4 + 30)
= 1 (24) – 4 (5) + 2 (26)
= 24 – 20 + 52
= 56
∴ By Cramer’s rule

(x, y, z) = (-1, \(\frac { 1 }{ 2 }\), \(\frac { 1 }{ 3 }\))

Question 2.
A commodity was produced by using 3 units of labour and 2 units of capital, the total cost is Rs 62. If the commodity had been produced by using 4 units of labour and one unit of capital, the cost is Rs 56. What is the cost per unit of labour and capital? (Use determinant method).
Solution:
Let ‘x’ be the cost per unit of labour
Let ‘y’ be the capital
∴ 3x + 2y = 62
4x + y = 56

∴ The cost per unit of labour is Rs 10 and Cost per unit of capital is Rs 16

Question 3.
A total of Rs 8,600 was invested in two accounts. One account earned 4\(\frac { 3 }{ 4 }\)% annual interest and the other earned 6\(\frac { 1 }{ 2 }\)% annual interest. If the total interest for one year was Rs 431.25, how much was invested in each accout? (Use determinant method)
Solution:
Let the two investments be x and y
Given that
x + y = 8600 ……. (1)

Multiply by 4 (both sides)
19x + 26y = 172500 ……….. (2)
Here Δ = \(\left|\begin{array}{ll}
1 & 1 \\
19 & 26
\end{array}\right|\) = 26 – 19 = 7 ≠ 0
∴ We can apply Cramer’s Rule
Δ_x = \(\left|\begin{array}{ll}
8600 & 1 \\
172500 & 26
\end{array}\right|\)
= 223600 – 172500
= 51100
Δ_y = \(\left|\begin{array}{ll}
1 & 8600 \\
19 & 172500
\end{array}\right|\)
= 172500 – 163400
= 9100
By Cramer’s Rule
x = \(\frac { Δ_x }{ Δ }\) = \(\frac { 51100 }{ 7 }\) = 7300
y = \(\frac { Δ_y }{ Δ }\) = \(\frac { 9100 }{ 7 }\) = 1300
∴ Amount invested at 4\(\frac { 3 }{ 4 }\)% is Rs 7,300 and
Amount invested at 6\(\frac { 1 }{2 }\)% is Rs 1,300

Question 4.
At marina two types of games viz., Horse riding and Quad Bikes riding are available on hourly rent. Keren and Bentia spent Rs 780 and Rs 560 during the month of May

Find the hourly charges for the two games (rides). (Use determinant method).
Solution:
let the hourly charges for horse riding be x and let the hourly charges for Quad Bikes riding be y.
For Keren
3x + 4y = 780 ……… (1)
For Benita
2x + 3y = 560 ……… (2)
Here Δ = \(\left|\begin{array}{ll}
3 & 4 \\
2 & 3
\end{array}\right|\) = 9 – 8 = 1 ≠ 0
∴ We can apply Cramer’s Rule

∴ The hourly charges for horse riding is Rs 100 and Quad Bikes riding is Rs 120

Question 5.
In a market survey three commodities A, B and C were considered. In finding out the index number some fixed weights were assigned to the three varieties in each of the commodities. The table below provides the information regarding the consumption of three commodities according to the three varieties and also the total weight received by the commodity.

Find the weights assigned to the three varieties by using Cramer’s Rule.
Solution:
Let x, y and z be are consumption of three commodities A, B and C respectively.
Given that
x + 2y + 3z = 11 ……… (1)
2x + 4y + 5z = 21 ……… (2)
3x + 5y + 6z = 27 ……… (2)
Here Δ = \(\left|\begin{array}{ccc}
1 & 2 & 3 \\
2 & 4 & 5 \\
3 & 5 & 6
\end{array}\right|\)
= 1 (24 – 25) – 2 (12 – 15) + 3 (10- 12)
= 1 ( 1) 2 ( 3) + 3 (- 2)
= -1 + 6 – 6 = -1 ≠ 0
∴ We can apply Cramer’s Rule
Now Δ_x = \(\left|\begin{array}{ccc}
11 & 2 & 3 \\
21 & 4 & 5 \\
27 & 5 & 6
\end{array}\right|\)
= 11 (24 – 25) -2 (126 – 135) + 3 (105 – 108)
= 11(-1) -2 (-9)+ 3 (-3)
= -11 + 18 – 9
= -2
Δ_y = \(\left|\begin{array}{ccc}
1 & 11 & 3 \\
2 & 21 & 5 \\
3 & 27 & 6
\end{array}\right|\)
= 1 (126 – 135) -11 (12 – 15) + 3 (54 – 63)
= 1 (-9) -11 (-3) + 3 (-9)
= -9 + 33 – 27
= -3
Δ_z = \(\left|\begin{array}{ccc}
1 & 2 & 11 \\
2 & 4 & 21 \\
3 & 5 & 27
\end{array}\right|\)
= 1 (108 – 105) – 2 (54 – 63) + 11 (10 – 12)
= 1(3) – 2 (-9) + 11 (-2)
= 3 + 18 – 22 = -1
By Cramer’s Rule

∴ (x, y, z) = (2, 3, 1)

Question 6.
A total of Rs 8,500 was invested in three interest-earning accounts. The interest rates were 2%, 3%, and 6%, if the total simple interest for one year was Rs 380 and the amount invested at 6% was equal to the sum of the amounts in the other two accounts, then how much was invested in each account? (use Cramer’s rule)
Solution:
Let the amount invested at 2%, 3%, and 6% are x, y, and z respectively.
x + y + z = 8500 ………. (1)

2x + 3y + 6z = 38000 ………. (2)
x + y = z
x + y – z = 0 ……… (3)
Here Δ = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
2 & 3 & 6 \\
1 & 1 & -1
\end{array}\right|\)
= 1 (-3 – 6) -1 (-2 – 6) + 1 (2 – 3)
= 1 (-9) -1 (-8) + 1 (-1)
= -9 + 8 – 1 = -2
≠ 0
∴ We can apply Cramer’s Rule
Δ_x = \(\left|\begin{array}{ccc}
8500 & 1 & 1 \\
38000 & 3 & 6 \\
0 & 1 & -1
\end{array}\right|\)
= 8500 (-3 – 6) -1 (-38000 – 0) + 1 (38000 – 0)
= 8500 (-9) + 38000 + 38000
= -76500 + 76000
= -500
Δ_y = \(\left|\begin{array}{ccc}
1 & 8500 & 1 \\
2 & 38000 & 6 \\
1 & 0 & -1
\end{array}\right|\)
= 1(-38000 – 0) – 8500 (-2 – 6) + 1 (0 – 38000)
= -38000 + 68000-38000
= 68000 – 76000
= -8000
Δ_z = \(\left|\begin{array}{ccc}
1 & 1 & 8500 \\
2 & 3 & 38000 \\
1 & 1 & 0
\end{array}\right|\)
= (0 – 38000) -1 (0 – 38000) + 8500 (2 – 3)
= -38000 + 38000 – 8500
= -8500
By Cramer’s Rule

∴ Amount invested at 2% is Rs 250; Amount invested at 3% is Rs 4,000 and Amount invested at 6% is Rs 4,250

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

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

Question 1.
Find the rank of each of the following matrices
(i) \(\left(\begin{array}{ll}
5 & 6 \\
7 & 8
\end{array}\right)\)
Solution:
Let A = \(\left|\begin{array}{ll}
5 & 6 \\
7 & 8
\end{array}\right|\)
Order of A is 2 × 2 [∴ P(A) ≤ 2]
Consider the second order minor = \(\left|\begin{array}{ll}
5 & 6 \\
7 & 8
\end{array}\right|\)
= 40 – 42
|A| = -2 ≠ 0
There is a minor of order 2, which is not zero
ρ(A) = 2

(ii) \(\left(\begin{array}{ll}
1 & -1 \\
3 & -6
\end{array}\right)\)
Solution:
Let A = \(\left(\begin{array}{ll}
1 & -1 \\
3 & -6
\end{array}\right)\)
Order of A is 2 × 2 [∴ ρ(A) ≤ 2]
Consider the second order minor = \(\left|\begin{array}{ll}
1 & -1 \\
3 & -6
\end{array}\right|\)
= -6 – (-3)
= -6 + 3 = -3
|A| ≠ 0
There is a minor of order 2, which is not zero
∴ ρ(A) = 2

(iii) \(\left(\begin{array}{ll}
1 & 4 \\
2 & 8
\end{array}\right)\)
Solution:
Let A = \(\left(\begin{array}{ll}
1 & 4 \\
2 & 8
\end{array}\right)\)
Order of A is 2 × 2 [∴ ρ(A) ≤ 2]
Consider the second order minor = \(\left|\begin{array}{ll}
1 & 4 \\
2 & 8
\end{array}\right|\)
= 8 – 8 = 0
Since the second are minor vanishes, ρ(A) ≠ 2
Consider a first order minor |1| ≠ 0
There is a minor of order 1, which is not zero
∴ ρ(A) = 1

(iv) \(\left[\begin{array}{ccc}
2 & -1 & 1 \\
3 & 1 & -5 \\
1 & 1 & 1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
2 & -1 & 1 \\
3 & 1 & -5 \\
1 & 1 & 1
\end{array}\right]\)
Order of A is 3 × 3
∴ ρ(A) ≤ 2
Consider the third order minor \(\left|\begin{array}{ccc}
2 & -1 & 1 \\
3 & 1 & -5 \\
1 & 1 & 1
\end{array}\right|\)
= 2(1 + 5) – (-1) (3 + 5) + 1 (3 – 1)
= 2 (6) + 1(8) + 1(2)
= 12 + 8 + 2
= 22 ≠ 0
There is a minor of order 3, which is not zero
∴ ρ(A) = 3

(v) \(\left[\begin{array}{ccc}
-1 & 2 & -2 \\
4 & -3 & 4 \\
-2 & 4 & -4
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
-1 & 2 & -2 \\
4 & -3 & 4 \\
-2 & 4 & -4
\end{array}\right]\)
Order of A is 3 × 3
∴ ρ(A) ≤ 3
Consider the third order minor \(\left|\begin{array}{ccc}
-1 & 2 & -2 \\
4 & -3 & 4 \\
-2 & 4 & -4
\end{array}\right|\)
= -1(12 – 16) – 2 (-16 + 8) – 2 (16 – 6)
= -1(-4) – 2 (-8) – 2 (10)
= 4 + 16 – 20
= 0
Since the third order minor vanishes, therefore
∴ ρ(A) ≠ 3
Consider a second order minor \(\left|\begin{array}{ll}
-1 & 2 \\
4 & -3
\end{array}\right|\)
= 3 – 8
= -5 ≠ 0
There is a minor of order 2, which is not zero.
∴ ρ(A) = 2

The order of A is 3 × 4
∴ ρ(A) ≤ 3

The number of non zero rows is 2
∴ ρ(A) = 2

The order of A is 3 × 4
∴ ρ(A) ≤ 3
Consider the third order minor \(\left|\begin{array}{ccc}
3 & 1 & -5 \\
1 & -2 & 1 \\
1 & 5 & -7
\end{array}\right|\)
= 3 (14 – 5) – 1 (- 7 – 1) – 5 (5 + 2)
= 3 (9) – 1 (-8) – 5 (7)
= 27 + 8 – 35
= 0
\(\left|\begin{array}{ccc}
1 & -5 & 1 \\
-2 & 1 & -5 \\
5 & -7 & 2
\end{array}\right|\)
= 1 (2 – 35) + 5 (-4 + 25) – 1 (14 – 5)
= 1 (-33) + 5(21) – 1 (9)
= -33 + 105 – 9
= 63 ≠ 0
There is a minor of order 3, which is not zero.
∴ ρ(A) = 3

Order of A is 3 × 4
∴ ρ(A) ≤ 3
Consider the third order minor \(\left|\begin{array}{ccc}
1 & -2 & 3 \\
-2 & 4 & -1 \\
-1 & 2 & 7
\end{array}\right|\)
= 1 (28 + 2)+ 2 (-14 – 1) + 3 (-4 + 4)
= 1 (30) + 2 (- 15) + 3 (0)
= 0
\(\left|\begin{array}{ccc}
-2 & 3 & 4 \\
4 & -1 & -3 \\
2 & 7 & 6
\end{array}\right|\)
= -2 (-6 + 21) – 3 (24 + 6) + 4 (28 + 2)
= -2(15) – 3 (30)+ 4 (30)
= 0
\(\left|\begin{array}{ccc}
1 & 3 & 4 \\
-2 & -1 & -3 \\
-1 & 7 & 6
\end{array}\right|\)
= 1 (-6 + 21) – 3 (-12 – 3) + 4 (-14 – 1)
= 1 (15) -3 (-15) + 4 (-15)
= 15 + 45 – 60
= 0
\(\left|\begin{array}{ccc}
1 & -2 & 4 \\
-2 & 4 & -3 \\
-1 & 2 & 6
\end{array}\right|\)
= 1 (24 + 6) + 2 (-12 – 3) + 4 (-4 + 4)
= 1 (30) + 2 (-15) + 4(0)
= 30 – 30
= 0
Since all third order minors vanishes, ρ(A) ≠ 3
Now, let us consider the second order minors.
Consider one of the second order minors
\(\left|\begin{array}{ll}
-2 & 3 \\
4 & -1
\end{array}\right|\) = 2 – 12 = -10 ≠ 0
There is a minor of order 2 which is not zero
∴ ρ(A) = 2

Question 2.
If A = \(\left(\begin{array}{ccc}
1 & 1 & -1 \\
2 & -3 & 4 \\
3 & -2 & 3
\end{array}\right)\) and B = \(\left(\begin{array}{ccc}
1 & -2 & 3 \\
-2 & 4 & -6 \\
5 & 1 & -1
\end{array}\right)\)
find the rank of A B and the rank of B A.
Solution:

To find the rank of AB
Order of AB is 3 × 3
∴ ρ(AB) ≤ 3
Consider the third order minor |AB| = \(\left|\begin{array}{ccc}
-6 & 3 & -2 \\
28 & -12 & 20 \\
22 & -11 & 18
\end{array}\right|\)
= -6(-216 + 220) -3(504 – 440) – 2(-308 + 264)
= – 6(4) – 3(64) – 2(-44)
= -24 – 192 + 88
= 128 ≠ 0
There is a minor of order 3, which is not zero.
∴ ρ(AB) = 3
To find the rank of BA
Order of BA is 3 × 3
∴ ρ(BA) ≤ 3
Consider the third order minor |BA| = \(\left|\begin{array}{ccc}
6 & 1 & 0 \\
-12 & -2 & 0 \\
4 & 4 & -4
\end{array}\right|\)
= 6(8 – 0) – 1(48 – 0) + 0(-48 + 8)
= 6(8) – 1(48) + 0(-40)
= 48 – 48 + 0
= 0
Since the third order minor vanishes, therefore P(BA) ≠ 3
Consider a second order minor \(\left|\begin{array}{ll}
-12 & -2 \\
4 & 4
\end{array}\right|\) = -48 + 8 = -40 ≠ 0
There is a minor of order 2 which is not zero
∴ ρ(BA) = 2

Question 3.
Solve the following system of equations by rank method
x + y + z = 9, 2x + 5y + 7z = 52, 2x – y – z = 0
Solution:
x + y + z = 9
2x + 5y + 7z = 52
2x – y – z = 0
The matrix equation corresponding to the given system is

Obviously the last equivalent matrix is in the echelon form. It has three non-zero rows.
P(A) = P (A, B) = 3 = Number of unknowns
The given system is consistent and has unique solution.
To find the solution, let us rewrite the above ech-elon form into the matrix form.
\(\left[\begin{array}{lll}
1 & 1 & 1 \\
0 & 3 & 5 \\
0 & 0 & 2
\end{array}\right]\) \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right] =\left[\begin{array}{c}
9 \\
34 \\
16
\end{array}\right]\)
x + y + z = 9 ……..(1)
3y + 5z = 34 ……… (2)
2z = 16 ……… (3)
z = \(\frac { 16 }{ 2 }\) = 8
z = 8
Substitute z = 8 in eqn (2)
3y + 5 (8) = 34
3y + 40 = 34
3y = 34 – 40
3y = -6
y = -2
Substitute y = -2 and z = 8 in eqn (1)
x = (-2) + 8 = 9
x + 6 = 9
x = 9 – 6
x = 3
∴ x = 3, y = -2, z = 8

Question 4.
Show that the equations 5x + 3y + 7z = 4, 3x + 26y + 2z = 9, 7x + 2y + 10z = 5 are consistent and solve them by rank method.
Solution:
5x + 3y + 7z = 4
3x + 26y + 2z = 9
7x + 2y + 10z = 5
The matrix equation corresponding to the given systematic


Obviously, the last equivalent matrix is in the echelon form. It has two non-zero rows.
P([A, B]) = 2 ; P(A) = 2
P(A) = P ([A, B]) = 2 < Number of unknowns
The given system is consistent and has infinitely many solutions.
The given system is equivalent to the matrix equation
\(\left[\begin{array}{lll}
1 & 5 & 1 \\
0 & -11 & 1 \\
0 & 0 & 0
\end{array}\right]\) \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right] =\left[\begin{array}{c}
2 \\
-3 \\
0
\end{array}\right]\)
x + 5y + z = 2 ……. (1)
-11y + z = -3 ……… (2)
let z = k

Where K ε R

Question 5.
Show that the following system of equations have unique solutions:
x + y + z = 3, x + 2y + 3z = 4, x + 4y + 9z = 6 by rank method.
Solution:
x + y + z = 3
x + 2y + 3z = 4
x + 4y + 9z = 6
The matrix equation corresponding to the given system is

The last equivalent matrix is in the echelon form [A, B] has 3 non-zero rows and [A] has 3 non-zero rows.
p([A,B]) = 3; ρ(A) = 3
ρ([A, B]) = ρ(A) = No. of unknowns
∴ The system of equations have unique solution.

Question 6.
For what values of the parameter λ, will the following equations fail to have unique solution:
3x – y + λz = 1, 2x + y + z = 2, x + 2y – λz = -1
Solution:
3x – y + λz = 1
2x + y + z = 2
x + 2y – λz = -1
The matrix equation corresponding to the given system is

If the equations fail to have unique solution.
ρ(A) ≠ ρ(A, B)
ρ(A, B) = 3
ρ(A) ≠ 3
Therefore 7 + 2λ = 0
2λ = -7 and λ = -7/2

Question 7.
The price of three commodities. X, Y, and Z are x,y, and z respectively Mr. Anand purchases 6 units of Z and sells 2 units of Y. Mr. Amar purchases a unit of Y and sells 3 units of X and 2 units of Z. Mr. Amit purchases a unit of X and sells 3 units of Y and a unit of Z. In the process they earn Rs 5,000/-, Rs 2,000/- and Rs 5,500/- respectively. Find the prices per unit of three commodities by the rank method.
Solution:
Let the equations for Mr. Anand, Mr. Amar, and Mr. Amit are
2x + 3y – 6z = 5000
3x – y + 2z = 2000
-x + 3y + z = 5500 respectively
The matrix equation corresponding to the given system is

∴ The given system is equivalent to the matrix equation
\(\left[\begin{array}{lll}
-1 & 4 & -8 \\
0 & 1 & -2 \\
0 & 0 & 7
\end{array}\right]\) \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right] =\left[\begin{array}{c}
3000 \\
1000 \\
3500
\end{array}\right]\)
-x + 4y – 8z = 3000 …….. (1)
y – 2z = 1000 ………. (2)
7z = 3500 ……….. (3)
Eqn (3) ⇒ z = \(\frac { 3500 }{ 7 }\)
∴ z = 500
Eqn (2) ⇒ y – 2(500) = 1000
y = 1000 + 1000
∴ y = 2000
Eqn (1) ⇒ -x + 4 (2000) – 8 (500) = 3000
-x + 8000 – 4000 = 3000
-x + 4000 = 3000
-x = 3000 – 4000
-x = – 1000
∴ x = 1000
The price of three commodities, x, y and z are Rs. 1000; Rs. 2000 and Rs. 500 respectively.

Question 8.
An amount of Rs 5,000/- is to be deposited in three different bonds bearing 6%, 7%, and 8% per year respectively. Total annual income is Rs 358/-, If the income from the first two investments is Rs 70/- more than the income from the third, then find the amount of investment in each bond by the rank method.
Solution:
Let the three different bonds be x, y, and z
x + y + z = 5000 …….. (1)

6x + 7y = 8z + 7000
6x + 7y – 8z = 7000 ……… (3)
The matrix equation corresponding to the given system is

∴ The given system is equivalent to the matrix equation
\(\left[\begin{array}{lll}
1 & 1 & 1 \\
0 & 1 & 2 \\
0 & 0 & -16
\end{array}\right]\) \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right] =\left[\begin{array}{c}
5000 \\
5800 \\
-28800
\end{array}\right]\)
x + y + z = 5000 ……… (1)
y + 2z = 5800 …….. (2)
-16z = -28800 ……… (3)
eqn (3) ⇒ z = \(\frac { -28800 }{ -16 }\)
∴ z = 1800
eqn (2) ⇒ y + 2(1800) = 5800
y + 3600 = 5800
y = 5800 – 3600
∴ y = 2200
eqn (1) ⇒ x + 2200 + 1800 = 5000
x + 4000 = 5000
x = 5000 – 4000
∴ x = 1000
The amount of investment in each bond is Rs 1000, Rs 2200 and Rs 1800.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 12 Discrete Mathematics Ex 12.3 Text Book Back Questions and Answers, Notes.

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

Choose the most suitable answer from the given four alternatives:

Question 1.
A binary operation on a set Sis 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:
Let 1, 2 ∈ N
1 * 2 = 1 – 2 = -1 ∉ N

Question 3.
Which one of the following is a binary operation on N?
(a) Subtraction
(b) Multiplication
(c) Division
(d) 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:
Since -2, 1/2 ∈ R , but (-2)^1/2 ∉ R.

Question 5.
The operation * defined by a * b = \(\frac { ab }{ 7 }\) is not a binary operation on
(a) Q⁺
(b) Z
(c) R
(d) C
Solution:
(b) Z
Hint:
a * b = \(\frac { ab }{ 7 }\) ∉ Z
as Z is set of all integers.

Question 6.
In the set Q define a \(\bigodot\) b = a + b + ab. For what value of y, 3 \(\bigodot\) (y \(\bigodot\) 5) = 7?
(a) y = \(\frac { 2 }{ 3 }\)
(b) y = \(\frac { -2 }{ 3 }\)
(c) y = \(\frac { -3 }{ 2 }\)
(d) y = 4
Solution:
(b) y = \(\frac { -2 }{ 3 }\)
Hint:
a \(\bigodot\) b = a + b + ab
Given 3 \(\bigodot\) (y \(\bigodot\) 5) = 7
3 \(\bigodot\) (y + 5 + 5y) = 7
3 \(\bigodot\) (6y + 5) = 7
3 + 6y + 5 + 3 (6y + 5) = 7
8 + 6y + 18y + 15 = 7
24 y = 7 – 23 = -16
y = \(\frac { -16 }{ 24 }\) = \(\frac { -2 }{ 3 }\)

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
Solution:
(c) Both commutative and associative
Hint:

∴ (a * b) * c = a * (b * c)
* is 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 complex number and its conjugate is purely imaginary
(d) \(\sqrt{5}\) is an irrational number
Solution:
(d) \(\sqrt{5}\) is an irrational number

Question 9.
Which one of the following statements has truth value F?
(a) Chennai is in India or √2 is an integer.
(b) Chennai is in India or √2 is an irrational number.
(c) Chennai is in China or √2 is an integer.
(d) Chennai is in China or √2 is an irrational number.
Solution:
(c) Chennai is in China or √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:
(i.e.) 2³ = 8

Question 11.
Which one is the inverse of the statement (p v q) → (p ∧ q)?
(a) (p ∧ q) → (p v q)
(b) ¬(p v q) → (p ∧ q)
(c) (¬P v ¬q) → (¬p ∧ ¬q)
(d) (¬p ∧ ¬q) → (¬p v ¬q)
Solution:
(d) (¬p ∧ ¬q) → (¬p v ¬q)
Hint:
(p v q) → (p ∧ q)
¬(p v q) → ¬(p ∧ q)
(¬P ∧ ¬q) → (¬p v ¬q)

Question 12.
Which one is the contrapositive of the statement (p v q) → r?
(a) ¬r → (¬p ∧ ¬q)
(b) ¬r → (p v q)
(c) r → (p ∧ q)
(d) p → (q v r)
Solution:
(a) ¬r → (¬p ∧ ¬q)
Hint:
(p v q) → r
Contrapositive is ¬r → ¬(p v q)
¬r → (¬p ∧ ¬q)

Question 13.
The truth table for (p ∧ q) v ¬q is given below

Solution:
(c) (1) T, (2) T, (3) F, (4) T
Hint:

1 = T, 2 = T, 3 = F, 4 = T

Question 14.
In the last column of the truth table for ¬(p v ¬q) the number of final outcomes of the truth value ‘F’ is
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(c) 3
Hint:

Number of ‘F’ in last column is 3

Question 15.
Which one of the following is incorrect? For any two propositions p and q, we have
(a) ¬(p v q) ≡ ¬p ∧ ¬q
(b) ¬(p ∧ q) ≡ ¬p v ¬q
(c) ¬(p v q) ≡ ¬p v ¬q
(d) ¬(¬p) ≡ p
Solution:
(c) ¬(p v q) ≡ ¬p v ¬q

Question 16.

Which one of the following is correct for the truth value of (p ∧ q) → ¬p

Solution:
(b) (1) F, (2) T, (3) T, (4) T
Hint:

Question 17.
The dual of ¬(p v q) v [p v(p ∧ ¬r)] is
(a) ¬(p ∧ q) ∧ [p v(p ∧ ¬r)]
(b) (p ∧ q) ∧ [p v(p v ¬r)]
(c) ¬(p ∧ q) ∧ [p ∧ (p ∧ r)]
(d) ¬(p ∧ q) ∧ [p ∧ (p v ¬r)]
Solution:
(d) ¬(p ∧ q) ∧ [p ∧ (p v ¬r)]
Hint:
Dual is obtained by changing ∧ into v and vice versa.

Question 18.
The proposition \(p \wedge(\neg p \vee q)]\) is ……..
(a) a tautology
(b) a contradiction
(c) logically equivalent to \(p \wedge q\)
(d) logically equivalent to \(p \vee q\)
Solution:
(c) logically equivalent to \(p \wedge q\)

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:
(a) (1) F, (2) T, (3) F, (4) T
Hint:
(1) F and T = F
(2) T and T =T
(3) T and F = F
(4) T and T = 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.