Solution of Linear Equations MCQ Quiz - Objective Question with Answer for Solution of Linear Equations - Download Free PDF

Calculation:

Given,

Also,

Putting in previous equation:

Now,

∴ The correct answer is Option 2.

Calculation:

Given:

System of linear equations:

 and 

The coefficient matrix A is:

The augmented matrix [A|B] is:

Calculate the determinant of A, |A|:

⇒ For unique solution, :

For no solution or infinite solutions, , so .

⇒ If , the system becomes:

⇒ From the second and third equations, for a solution to exist, .

⇒ If and , infinite solutions exist.

⇒ If and , no solution exists.

(I) Unique solution: . can take 9 values (1 to 10 except 4). can take 10 values. Total pairs: 9 × 10 = 90.

(II) No solution: and . 9 values for . 1 pair for . Total pairs: 1 × 9 = 9.

(III) Infinite solutions: and . 1 pair only.

(IV) At least one solution: Total pairs - No solution pairs = 100 - 9 = 91.

∴ (I) - (S), (II) - (Q), (III) - (P), (IV) - (R)

Calculation

⇒ 

D₁ = 2λµ + 99λ – 10µ + 255

D₂ = 13 – µ

D₃ = 5λ + 5

D₂ = 0 ⇒ µ = 13 & D₃ = 0 ⇒ λ = –1

For these values D & D₁ = 0

 λ + μ = 13 - 1 = 12

Hence option 4 is correct

Calculation

⇒ 2(λμ + 141) + (5μ – 300) – 235 – 100λ = 0 …(1)

⇒ 6λ = –12 ⇒ λ = –2

Put λ = - 2 in (1)

⇒ 2(–2μ + 141) + 5μ – 300 – 235 + 200 = 0

⇒ μ = 53

µ – 2λ = 57

Hence option 4 is correct

Calculation

As the given system of equations has a non-trivial solution.

Applying C₂ → C₂ - C₁ and C₃ → C₃ - C₁

Expanding along C₃, we get

⇒ (a - p)(-c)(q - b) + (r - c){p(q - b) - b(a - p)} = 0

⇒ (p - a)(q - b)c + p(r − c)(q - b) + b(r − c)(p - a) = 0

Dividing by (p - a)(q - b)(r - c), we get

⇒ 

⇒ 

⇒ 

Hence option 2 is correct

Concept

Let the system of equations be,

a₁x + b₁y + c₁z = d₁

a₂x + b₂y + c₂z = d₂

a₃x + b₃y + c₃z = d₃

⇒ AX = B

⇒ X = A^-1 B = 

⇒ If det (A) ≠ 0, the system is consistent having unique solution.

⇒ If det (A) = 0 and (adj A). B = 0, system is consistent, with infinitely many solutions.

⇒ If det (A) = 0 and (adj A). B ≠ 0, system is inconsistent (no solution)

Calculation:

Given: 

kx + y + z = 1, x + ky + z = k and x + y + kz = k²

⇒ For the given equations to have no solution, |A| = 0

⇒ k(k² – 1) - 1(k – 1) + 1(1 – k) = 0

⇒ k³ – k – k + 1 + 1 – k = 0

⇒ k³ -3k +2 = 0

⇒ (k – 1) (k – 1) (k + 2) = 0

⇒ k = 1, -2

If we put k = 1 in the above given equations, then all the equations will become the same.

Hence, the given equations have no solution if k = - 2. 

Concept:

Consider three linear eqaution in two variable:

a₁x + b₁y + c_{1 = 0}

a₂x + b₂y + c₂ = 0

a₃x + b₃y + c₃ = 0

Condition for the consistency of three simultaneous linear equations in 2 variables:

​​​​

Formula for Quadratic equation:

ax² + bx + c = 0

x = 

Calculation:

2k2x + 3y - 1 = 0      ....(1)

7x - 2y + 3 = 0      ....(2)

6kx + y + 1 = 0      ....(3)

For consistency of given simultaneous equation,

⇒ 2k²(-2- 3) - 3(7 - 18k) - 1(7 + 12k) = 0

⇒ -10k² - 21 + 54k - 7 - 12k = 0

⇒ -10k² + 42k - 28 =  0

⇒ 5k2 - 21k + 14 =  0

By using the formula,

Concept:

A = [aij]m × n = coefficient of matrix, X = Column matrix of the variables

B = column matrix of constants

The system AX = B has

1) A unique solution If Rank of A = Rank [A|B] and is equal to the number of variables.

2) Infinitely many solutions, If Rank of A = Rank of [A|B]

3) no solution, If Rank of A ≠ Rank of [A|B], i.e. Rank of A

Calculation:

Let 

We can see that, at λ = 3, R₂ = R₃ and hence, |A| = 0.

For λ = 3 either infinite solutions exist or no solution exists.

The coefficient matrix of the given linear equation is

R3 → R3 - R2

⇒ 

Let the augmented matrix be

For no solution,  μ = 10 

Hence, λ = 3, μ = 10 is the correct answer.

Shortcut Trickx + y + z = 6

x + 2y + 3z = 10

x + 2y + λz = μ 

Hence, if we put λ = 3, μ = 10, two-equation will get coincide, which result in an infinite number of solution.

Concept

Let the system of equations be,

a₁x + b₁y + c₁z = d₁

a₂x + b₂y + c₂z = d₂

a₃x + b₃y + c₃z = d₃

⇒ AX = B

⇒ X = A^-1 B = 

⇒ If det (A) ≠ 0, system is consistent having unique solution.

Calculation:

Given system of linear equation are kx + y + z = 1, x + ky + z = 1 and x + y + kz = 1

Let A = 

det (A) = |A| = k (k² – 1) – 1(k -1) + 1 (1 – k)

⇒ |A| = k³ – k – k + 1 + 1 – k = k³ – 3k + 2

For unique solution,

det (A) ≠ 0

⇒ k³ – 3k + 2 ≠ 0

⇒ (k – 1) (k² + k - 2) ≠ 0

⇒ (k – 1) (k – 1) (k + 2) ≠ 0

∴ k ≠ 1 and k ≠ -2

Concept

Let the system of equations be,

a₁x + b₁y + c₁z = d₁

a₂x + b₂y + c₂z = d₂

a₃x + b₃y + c₃z = d₃

⇒ AX = B

⇒ X = A^-1 B = 

⇒ If det (A) ≠ 0, system is consistent having unique solution.

⇒ If det (A) = 0 and (adj A). B = 0, system is consistent, with infinitely many solutions.

⇒ If det (A) = 0 and (adj A). B ≠ 0, system is inconsistent (no solution)

Calculation:

Given system of equation x + y + z = 8, x – y + 2z = 6 and 3x – y + 5z = k

⇒ AX = B

Determinant of A = |A| = 1 (-5 + 2) – 1 (5 – 6) + 1 (-1 + 3) = -3 + 1 + 2 = 0

So we can say that equations have either an infinite solution or no solution.

A unique solution is not possible.

 ∴ Statement 3 is wrong.

We have adj A = 

If k = 15,

B = 

Now (adj A). B will be 

⇒ no solution

If K = 20,

B = 

Now (adj A). B will be 

⇒ infinitely many solutions

Hence Option 1 is correct.

The given system of linear equations:

x + y + z = 5;

x + 2y + 2z = 6;

and x + 3y + λz = μ have infinite solution.

∴ Δ = 0, Δx = Δy = Δz = 0

Now, forming determinant from the given equations,

⇒ 1(2λ – 6)-1(λ – 2)+1(3 – 2) = 0

⇒ 2λ – 6 – λ + 2 + 3 – 2 = 0

⇒ λ – 8 + 5 = 0

⇒ λ – 3 = 0

∴ λ = 3

Now, the determinant of y is:

R₂ → R₂ – R₁

R₃ → R₃ – R₁

⇒ 1(2 – (μ – 5)) – 5(0 – 0) + 1(0 – 0) = 0

⇒ 1(2 – (μ – 5)) = 0

⇒ 2 – μ + 5 = 0

⇒ 7 – μ = 0

∴ μ = 7

Now,

∴ λ + μ = 3 + 7 = 10

Ax = B

X = (adj(A).B)/(|A|)

|A|  = - 12 - (- 12) = - 12 + 12 = 0

|A| = 0

Adj (A) =

0 =

Hence, no solution.

Concept:

Let us consider the system of linear equations:

a₁₁ × x + a₁₂ × y = b₁

a₂₁ × x + a₂₂ × y = b₂

We can write these equations in matrix form as: A X = B, where 

Case -1: If A is a non-singular matrix then |A| ≠ 0.

Then, X = A^-1 B where A^-1 will exist if and only if |A| ≠ 0 and it is given by: 

Case – 2: If A is a singular matrix then |A| = 0.

In this case, we have to calculate (adj (A)) × B.

If (adj (A)) × B ≠ O, where O is the null matrix then the system of equations is inconsistent and has no solution.

If (adj (A)) × B = O, where O is the null matrix then the system of equations is consistent and has infinitely many solutions.

Calculation:

Given: x + 3y = 5 and 2x + ky = 8

We can write these equations in matrix form as: A X = B, where 

In order to say that the given system of linear equations is consistent and has unique solution, |A| ≠ 0.

⇒ |A| = k – 6 ≠ 0 ⇒ k ≠ 6

Concept: 

If two linear equations

a1​x + b1y ​= c1​ and a2​x + b2y ​= c2​. Then,

(a) If a1​​/a2 = b1​​/b2 ​​= c1​​/c2 ​​, the system is consistent and has infinitely many solutions.

(b) If a1​​/a2 = b1​​/b2 ≠ c1​​/c2 the system has no solution and is inconsistent

Calculation:

The equations 2x - ky + 7 = 0 and 6x - 12y + 15 = 0

2/6 = - k/-12 ≠ 7/15

Using, 2/6 = - k/-12

⇒ k =  24/6 =  4

∴ At k =  4, The equations 2x - ky + 7 = 0 and 6x - 12y + 15 = 0 have no solution 

Explanation:

b must be linearly dependent on the columns of A

Consider A_3x3 then [A : b]_3x4, i.e., Augmented matrix have 4 vectors.

But by property of rank, we have 

Which is definitely less than 4 

So [A : b] has L.D. set of vectors so we can conclude that matrix b must be linearly dependent of column of matrix A.

The correct option is (3).