Cramer's Rule MCQ Quiz in मल्याळम - Objective Question with Answer for Cramer's Rule - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

The system of equations A X = 0 is said to be homogenous system of equations, then

If |A| ≠ 0, then its solution X = 0, is called trivial solution.

If |A| = 0. Then A X = 0 has a non-trivial solution which means the system will have infinitely many solutions.

Calculation:

Given:  2x – 3y = 0 and 2x + αy = 0

These equations can be written as: A X = B where 

As we know that, the given system is a homogenous system of equation. So, in order to say that the system has infinitely many solutions: |A| = 0.

⇒ |A| = 2α + 6 = 0 ⇒ α = -3.

Calculation

⇒ λ = 17

⇒ μ ≠ 18

Hence option 1 is correct

Concept:

Let us consider a system of equations in three variables:

a₁ × x + b₁ × y + c₁ × z = d₁

a₂ × x + b₂ × y + c₂ × z = d₂

a₃ × x + b₃ × y + c₃ × z = d₃

Then, 

By cramer’s rule:

I. If Δ ≠ 0, then the system of equation has unique solution and it is given by: 

II. If Δ = 0 and atleast one of the determinants Δ, Δ₁, Δ₂ and Δ₃ is non-zero, then the given system is inconsistent.

III. If Δ = 0 and Δ₁ =  Δ₂ =  Δ₃ = 0, then the system is consistent and has infinitely many solutions.

Calculation:

Given: x + y + z = 3, x – y + 2z = 6 and x + y + α z = β.

As we know that,

⇒ Δ = 2 – 2α, Δ₁ = 3β – 9α, Δ₂ =  3α - β and Δ₃ = 6 – 2β.

As we know that, for the given system of equation to have infinitely many solutions according to cramer’s rule: Δ = 0 and Δ₁ =  Δ₂ =  Δ₃ = 0

⇒ Δ = 2 – 2α = 0 ⇒ α = 1.

For α = 1 , we have: Δ₁ = 3β – 9, Δ₂ =  3 – β and Δ₃ =  6 – 2β

So, in order to have infinitely many solutions, Δ₁ =  Δ₂ =  Δ₃ = 0.

⇒ β = 3.

Concept:

Let us consider a system of equations in two variables:

a₁ × x + b₁ × y = c₁

a₂ × x + b₂ × y = c₂

Then, 

By Cramer's rule, the solution of a system of the equation has a unique solution if Δ ≠ 0 and the solution is given by: 

By cramer’s rule:

I. If Δ ≠ 0, then the system of the equation has a unique solution and it is given by: 

II. If Δ = 0 and at least one of the determinants Δ, Δ1, Δ2, and Δ3 is non-zero, then the given system is inconsistent.

III. If Δ = 0 and Δ1 =  Δ2 =  Δ3 = 0, then the system is consistent and has infinitely many solutions.

Calculation:

Given: 4x + y = α and βx + 2y = 3

As we know that, the given system of equation will be inconsistent if if Δ = 0 and at least one of the determinants Δ₁ and Δ₂ is non-zero.

∵ The given system has no solution

⇒ Δ = 8 - β = 0 ⇒ β = 8.

So, for α = 3 / 2, we can see that both Δ₁ and Δ₂ are zero.

for α = 3 / 2 and β = 8 the given system is consistent and has infinitely many solutions.

Hence, option 4 is correct.

Calculation

Given:

The system of simultaneous linear equations:

⇒

⇒

Since , the system has a unique solution.

⇒

⇒

Thus, we have a unique solution with .

Hence option 4 is correct

Calculation

For non-zero solution,

Since (x−y)(y−z)(z−x)(x+y+z)

Hence option 3 is correct
$\left[\because \begin{vmatrix} 1 & 1 & 1 \ x&y&z \x^3&y^3&z^3 \end{vmatrix}=(x-y)(y-z)(z-x)(x+y+z)\right]$