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

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:
C₂H₆ + O₂ → H₂O + CO₂.
Solution:
We are searching for positive integers x₁, x₂, x₃ and x₄
x₁ C₂H₆ + x₂ O₂ → x₃ H₂O + x₄ CO₂ ……….(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.
2x₁ x₄ = 2x₁ – x₄ = 0 ……..(2)
6x₁ = 2x₃ = 6x₁ – 2x₃ = 0
÷ 2 ⇒ 3x₁ – x₃ = 0 ………(3)
2x₂ = x₃ + 2x₄ ⇒ 2x₂ – x₃ – 2x₄ = 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
-2x₃ + 3x₄ = 0 ……….. (1)
2x₂ – x₃ – 2x₄ = 0 ………… (2)
2x₁ – x₄ = 0 …………. (3)
Put x₄ = t
(3) ⇒ 2x₁ – t = 0
x₁ = \(\frac {t}{2}\)
(1) ⇒ -2x₃ + 3x₄ = 0
-2x₃ = -3t
x₃ = \(\frac {3}{2}\) t
(2) ⇒ 2x₂ – x₃ – 2x₄ = 0
2x₂ – \(\frac {3}{2}\) t – 2t = 0
2x₂ = \(\frac {3}{2}\) t + 2t = \(\frac {7t}{2}\)
x₂ = \(\frac {7t}{4}\)
(x₁, x₂, x₃, x₄) = (\(\frac {t}{2}\), \(\frac {7t}{4}\), \(\frac {3}{2}\)t, t) ∀ t ∈ R
since x₁, x₂, x₃ and x₄ are positive integers.
Let us choose t = 4

So the balanced equation is
2C₂H₆ + 7O₂ → 6H₂O + 4CO₂.