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## 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

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

Solution:

(i) The matrix form of the above equations is

The above matrix is in echelon form.

Here ρ(A, B) = ρ(A) < number of unknowns.

The system is consistent with infinite number of solutions. To find the solutions.

Writing the equivalent equations.

We get 3x + 2y + 7z = 0 ……. (1)

-17y – 34z = 0 ……. (2)

Taking z = t in (2) we get -17y – 34t = 0

⇒ -17y = 34t

⇒ y = -2t

Taking z = t, y = -2t in (1) we get

3x + 2(-2t) + 7t = 0

⇒ 3x – 4t + 7t = 0

⇒ 3x = -3t

⇒ x = -t

So the solution is x = -t; y = -2t; and z = t, t ∈ R

(ii) The matrix form of the equations is

\(\left(\begin{array}{ccc}{2} & {3} & {-1} \\ {1} & {-1} & {-2} \\ {3} & {1} & {3}\end{array}\right)\left(\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right)=\left(\begin{array}{l}{0} \\ {0} \\ {0}\end{array}\right)\)

(i.e) AX = B

The augmented matrix [A, B] is

The above matrix is in echelon form also ρ(A, B) = ρ(A) = 3 = number of unknowns

The system is consistent with unique solution, x = y = z = 0

(i.e) The system has trivial solution only.

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:

The matrix form of the equation is

\(\left(\begin{array}{lll}{1} & {1} & {3} \\ {4} & {3} & {\lambda} \\ {2} & {1} & {2}\end{array}\right)\left(\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right)=\left(\begin{array}{l}{0} \\ {0} \\ {0}\end{array}\right)\)

(i.e) AX = B

The augmented matrix [A, B] is

The above matrix is in echelon form

Case 1: When λ ≠ 8, ρ(A, B) = ρ(A) = 3 = number of unknowns

The system is consistent with a unique solution.

Case 2: When λ = 8, ρ(A, B) = ρ(A) = 2 < number of unknowns.

The system is consistent with non-trivial solutions.

Question 3.

By using the 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} such that

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 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)

Similarly considering hydrogen and oxygen atoms we get respectively.

2x_{2} = x_{3} + 2x_{4}

⇒ 2x_{2} – x_{3} – 2x_{4} = 0 …… (3)

and -2x_{3}+ 3x_{4} = 0 …… (4)

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

The augmented matrix (A, B) is

Now ρ(A, B) = ρ(A) = 3 < number of unknowns.

So the system is consistent and has an infinite number of solutions.

Writing the equations using the echelon form we get

2x_{1} – x_{4} = 0 …… (5)

2x_{2} – x_{3} – 2x_{4} = 0 ……. (6)

-2x_{3} + 3x_{4} = 0 ……. (7)

Since x_{1}, x_{2}, x_{3} and x_{4} are positive integers. Let us choose t = 4t.

Then we get x_{1} = 2, x_{2} = 7, x_{3} = 6, and x_{4} = 4

So the balanced equation is 2C_{2}H_{6} + 7O_{2} → 6H_{2}O + 4CO_{2}.

### Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.7 Additional Problems

Question 1.

Solve the following homogeneous linear equations.

x + 2y – 5z = 0,

3x + 4y + 6z = 0,

x + y + z = 0

Solution:

The given system of equations can be written in the form of matrix equation

AX = B

The augmented matrix is

This is in the echelon form.

Clearly ρ[A, B] = 3 and ρ(A) = 3

∴ ρ(A) = ρ[A, B] = 3 = number of unknowns

∴ The given system of equations is consistent and has a unique solution. i.e., trivial solution

∴ x = 0, y = 0 and z = 0

Note: Since ρ(A) = 3, | A | ≠ 0 i.e. A is non-singular;

∴ The given system has only trivial solution x = 0, y = 0, z = 0

Question 2.

For what value of n the equations.

x + y + 3z = 0,

4x + 3y + µz = 0,

2x + y + 2z = 0 have a

(i) trivial solution,

(ii) non-trivial solution.

Solution:

The system of equations can be written as AX = B

Case (i): If µ ≠ 8 then 8 – µ ≠ 0 and hence there are three non-zero rows.

∴ ρ[A] = ρ[A, B] = 3 = the number of unknowns.

∴ The system has the trivial solution x = 0, y = 0, z = 0

Case (ii): If µ = 8 then.

ρ[A, B] = 2 and ρ(A) = 2

∴ ρ(A) = ρ[A, B] = 2 < number of unknowns.

The given system is equivalent to

x + y + 3z = 0; y + 4z = 0

∴ y= – 4z ; x = z

Taking z = k, we get x = k,y = – 4k, z = k [k ∈ R – {0}] which are non-trivial solutions.

Thus the system is consistent and has infinitely many non-trivial solutions.

Note: In case (ii) the system also has trivial solution. For only non-trivial solutions we removed k = 0.