Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 1 Applications of Matrices and Determinants Ex 1.7 Textbook Questions and Answers, Notes.

## 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

Solution:

ρ(A) = 2 ρ[A | B] = 2

ρ(A) ρ[A | B] = 2 < n

The system is consistent. It has non trivial solution.

Writing the equivalent equations from echelon form

3x + 2y + 7z = 0 ………… (1)

-17y – 34z = 0 ……….. (2)

Put z = t

(2) ⇒ -17y = 34t

y = \(\frac {34t}{-17}\) = -2t

(1) ⇒ 3x + 2(-2t) + 7t = 0

3x – 4t + 7t = 0

3x + 3t = 0

3x = -3t

x = -t

(x, y, z) (-t, -2t, t) ∀ t ∈ R

(ii) 2x + 3y – z = 0, x – y – 2z = 0, 3x + y + 3z = 0

Solution:

ρ(A) = 3 ρ[A | B] = 3

ρ(A) = ρ[A | B] = 3

The system is consistent. It has trivial solution.

x = 0, y = 0, z = 0

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:

Case (i):

if λ ≠ 8

ρ(A) = 3 ρ(A | B) = 3

ρ(A) = ρ(A | B) = 3 = n

The system is consistent. It has unique (trivial) solution.

∴ Solution x = 0, y = 0, z = 0

Case (ii):

if λ = 8

ρ(A)= ρ(A | B) = 2

ρ(A) = ρ(A | B) = 2 < n

The system is consistent. It has non trivial solution.

Question 3.

By using Gaussian elimination method, balance the chemical -reaction equation:

C_{2}H_{6} + O_{2} → H_{2}O + CO_{2}.

Solution:

We are searching for positive integers x_{1}, x_{2}, x_{3} and x_{4}

x_{1 }C_{2}H_{6} + x_{2 }O_{2} → x_{3 }H_{2}O + x_{4 }CO_{2} ……….(1)

The number of carbon atoms on the LHS of (1) should be equal to the number of carbon atoms on the RHS of (1) so we get a linear

homogeneous equation.

2x_{1} x_{4} = 2x_{1} – x_{4} = 0 ……..(2)

6x_{1} = 2x_{3} = 6x_{1} – 2x_{3} = 0

÷ 2 ⇒ 3x_{1} – x_{3} = 0 ………(3)

2x_{2} = x_{3} + 2x_{4} ⇒ 2x_{2} – x_{3} – 2x_{4} = 0 ……… (4)

Equation (2), (3) and (4) constitute a homogeneous system of linear equations in four unknowns.

Augmented matrix

The system is consistent and has an infinite number of solutions.

Writing the equations using the echelon form we get

-2x_{3} + 3x_{4} = 0 ……….. (1)

2x_{2} – x_{3} – 2x_{4} = 0 ………… (2)

2x_{1} – x_{4} = 0 …………. (3)

Put x_{4} = t

(3) ⇒ 2x_{1} – t = 0

x_{1} = \(\frac {t}{2}\)

(1) ⇒ -2x_{3} + 3x_{4} = 0

-2x_{3} = -3t

x_{3} = \(\frac {3}{2}\) t

(2) ⇒ 2x_{2} – x_{3} – 2x_{4} = 0

2x_{2} – \(\frac {3}{2}\) t – 2t = 0

2x_{2} = \(\frac {3}{2}\) t + 2t = \(\frac {7t}{2}\)

x_{2} = \(\frac {7t}{4}\)

(x_{1}, x_{2}, x_{3}, x_{4}) = (\(\frac {t}{2}\), \(\frac {7t}{4}\), \(\frac {3}{2}\)t, t) ∀ t ∈ R

since x_{1}, x_{2}, x_{3} and x_{4} are positive integers.

Let us choose t = 4

So the balanced equation is

2C_{2}H_{6} + 7O_{2} → 6H_{2}O + 4CO_{2}.