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

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

(ii) 3x + y + z = 2, x – 3y + 2z = 1, 7x – y + 4z = 5

Solution:

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

(iii) 2x + 2y + z = 5, x – y + z = 1, 3x + y + 2z = 4

Solution:

ρ(A) = 2 ρ[A | B] = 3

ρ(A) ≠ ρ[A | B] = 2 < n

∴ The system is inconsistent. It has no solution.

(iv) 2x – y + z = 2, 6x – 3y + 3z = 6, 4x – 2y + 2z = 4

Solution:

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

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:

[∵ 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.

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:

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.