JEE Main System of Linear Equations Previous Year Questions With Solutions

JEE Previous year’s set of questions on System of Linear Equations gives a clear understanding of the topic. These solutions help break down all the complex problems by the simple step-by-step method of solving. BYJU’S provides accurate solutions, which are prepared by subject experts. This section deals with the solution of linear systems using different ways. 

How To Solve a Linear Equation System? 

A linear system may appear in any one of 3 possible ways: 

(i) The system has no solution, 

(ii) A single unique solution or 

(iii) Infinitely many solutions. 

Linear equation systems can be solved using various methods such as Graphical Method, Elimination Method, Cross Multiplication Method, Substitution Method, Matrix Method and Determinants Method. The set of all possible solutions is called the solution set.

Students can easily access and download chapter-wise solutions in PDF format for free. Download System of Linear Equations solved questions PDF from the link mentioned below.

Download System of Linear Equations Previous Year Solved Questions PDF

JEE Main Past Year Questions With Solutions on System of Linear Equations

Question 1: Consider the system of equations x + y + z = 1, 2x + 3y + 2z = 1, 2x + 3y + (a² – 1)z = a + 1 then

(a) System has a unique solution for |a| = √3

(b) System is inconsistence for |a| = √3

(c) System is inconsistence for a = 4

(d) System is inconsistence for a = 3

Answer: (b)

Solution:

Given system of linear equations:

x + y + z = 1 ….(1)

2x + 3y + 2z = 1 ….(2)

2x + 3y + (a² – 1)z = a + 1 …..(3)

= 1[3(a² – 1) – (3)(2)] – 1[2(a² – 1) – (2)(2)] + 1[2(3) – 2(3)]

= 3a² – 3 – 6 – 2a² + 2 + 4 + 0

= a² – 3

Consider D ≠ 0

So, a² – 3 ≠ 0

⇒ a² ≠ 3

⇒ |a| ≠ ±√3

That means the given linear equation system is inconsistent for |a| = ±√3.

So option (b) is correct.

Question 2: If the system of linear equations are

2x + 2ay + az = 0

2x + 3by + bz = 0 and

2x + 4cy + cz = 0,

where a, b, c ∈ R are non-zero and distinct; has a non-zero solution, then

(a) a + b + c = 0

(b) 1/a, 1/b, 1/c are in A.P.

(c) a, b, c are in A.P.

(d) a, b, c are in G.P.

Answer: (b)

Solution:

Given system of linear equations

2x + 2ay + az = 0

2x + 3by + bz = 0 and

2x + 4cy + cz = 0,

As the given system of equations has the non-zero solution, we have D = 0.

⇒ (3b − 2a)(c − a) − (4c − 2a)(b − a) = 0

⇒ 3bc − 2ac − 3ab + 2a² − [4bc − 4ac − 2ab + 2a²] = 0

⇒ −bc + 2ac − ab = 0

⇒ ab + bc = 2ac

⇒ 1/c + 1/a = 2/b

Which shows that 1/a, 1/b, and 1/c are in A.P.

Question 3: If the system of linear equations

x + y + z = 6

x + 2y + 3z = 10 and

3x + 2y + λz = μ

has more than two solutions, then μ − λ² is equal to ________.

Solution:

The system of equations has more than 2 solutions.

Find for D = D₃ = 0

Question 4: For which of the following ordered pairs (μ, δ), the system of linear equations

x + 2y + 3z = 1

3x + 4y + 5z = μ

4x + 4y + 4z = δ

is inconsistent?

(a) (4, 6)

(b) (3, 4)

(c) (1, 0)

(d) (4, 3)

Answer: (d)

Solution:

Given,

x + 2y + 3z = 1

3x + 4y + 5z = μ

4x + 4y + 4z = δ

For an inconsistent system, D = 0, and anyone of D_x, D_y, and D_z should not be equal to 0.

Now,

= 1(16 – 20) – 2(12 – 20) + 3(12 – 16)

= -4 – 2(-8) + 3(-4)

= -4 + 16 – 12

= 0

Now, consider D_x.

= 1(16 – 20) – 2(4μ – 5δ) + 3(4μ – 4δ)

= -4 – 8μ + 10δ + 12μ – 12δ

= -4 + 4μ – 2δ

Since, Dx ≠ =0

4μ – 2δ ≠ 4

2μ – δ ≠ 2

For an inconsistent system, 2μ ≠ δ + 2

Therefore, the system will be inconsistent for μ = 4, δ = 3.

Question 5: The system of linear equations

λx + 2y + 2z = 5

2λx + 3y + 5z = 8

4x + λy + 6z = 10 has:

(a) no solution when λ = 2

(b) infinitely many solutions when λ = 2

(c) no solution when λ = 8

(d) a unique solution when λ = -8

Answer: (a)

Solution:

Given,

λx + 2y + 2z = 5

2λx + 3y + 5z = 8

4x + λy + 6z = 10

Therefore, the given system of equations has no solution for λ= 2.

Question 6: The following system of linear equations

7x + 6y − 2z = 0 ,

3x + 4y + 2z = 0

x − 2y − 6z = 0, has

(a) infinitely many solutions, (x, y, z) satisfying y = 2z

(b) infinitely many solutions, (x, y, z) satisfying x = 2z

(c) no solution

(d) only the trivial solution

Answer: (b)

Solution:

Given system of linear equations are:

7x + 6y − 2z = 0

3x + 4y + 2z = 0

x − 2y − 6z = 0

Here, the system of equations is Homogeneous.

⇒ The system is consistent.

⇒ Infinite solutions exist (both trivial and non-trivial solutions)

When y = 2z

Let’s take y = 2 and z = 1

When (x, 2, 1)is substituted in the system of equations

⇒ 7x + 10 = 0,

3x + 10 = 0 and

x − 10 = 0 (which is not possible)

Therefore, for y = 2z, infinitely many solutions do not exist.

Consider, x = 2z.

Let’s take x = 2, z = 1, y = y.

Substitute (2, y, 1) in the system of equations.

⇒ y = −2

So, for each pair of (x, z), we get a value of y.

Therefore, for x = 2z, infinitely many solutions exist.

Question 7: If the system of linear equations

x + ky + 3z = 0

3x + ky – 2z = 0 and

2x + 4y – 3z = 0

has a non-zero solution (x, y, z), then xz/y² is equal to

(a) -10

(b) 10

(c) -30

(d) 30

Answer: (b)

Solution:

Given system of linear equations are:

x + ky + 3z = 0

3x + ky – 2z = 0 and

2x + 4y – 3z = 0

The system has a non-zero solution, so D = 0.

1(-3k + 8) – k(-9 + 4) + 3(12 – 2k) = 0

Solving the above equation, we have -4k = -44

or k = 11

x + 11y + 3z = 0 …(i)

3x + 11y – 2z = 0 …..(ii)

2x + 4y – 3z = 0 ….(iii)

Solving (i) and (iii)

x = -5y

Using x = -5y in (iii), -10y + 4y – 3z = 0

-6y – 3z = 0

or z = -2y

Now, xz/y² = (-5y)(-2y)/y² = 10

Question 8: The number of real values of λ for which the system of linear equations

2x + 4y − λz = 0

4x + λy + 2z = 0

λx + 2y + 2z = 0

has infinitely many solutions, is :

(a) 0

(b) 1

(c) 2

(d) 3

Answer: (b)

Solution:

For infinitely many solutions, D = 0, D_x = 0, D_y = 0 and D_z = 0

Similarly, D_y = 0 and D_z = 0

The equation λ³ + 4λ – 40 has only one solution.

Since λ(-∞) = -∞ and λ(∞) = ∞

Here α is only one solution, so λ(α) = 0

Now, differentiating λ³ + 4λ – 40 w.r.t. λ we get

3λ² + 4 > 0

The equation can not have y, m, so

λ(m) = 0 and λ(y) = 0

Thus, the number of real values of λ is 1.

Question 9: If x = a, y = b, z = c is a solution of the system of linear equations

x + 8y + 7z = 0

9x + 2y + 3z = 0

x + y + z = 0

such that the point (a, b, c) lies on the plane x + 2y + z = 6, then 2a + b + c equals :

(a) −1

(b) 0

(c) 1

(d) 2

Answer: (c)

Solution:

Given system of linear equations

x + 8y + 7z = 0 …(i)

9x + 2y + 3z = 0 ….(ii)

x + y + z = 0 ….(iii)

Operate: (ii) – 3 × (iii)

6x – y = 0

or

y = 6x …..(iv)

Substituting (iv) in (i), we get;

x + 8(6x) + 7z = 0

z = -7x ……(v)

Since x = a, y = b, z = c (Given)

b = 6a and c = -7a

Also, (a, b, c) lies on the plane x + 2y + z = 6.

Therefore, a + 2b + c = 6 …..(vi)

Putting the values of b and c in (vi),

a + 2(6a) – 7a = 6

⇒ a = 1

Also, we get b = 6 and c = -7

Now, 2a + b + c = 2(1) + 6 – 7 = 1

Question 10: It S is the set of distinct values of ‘b’ for which the following system of linear equations

x + y + z = 1

x + ay + z = 1

ax + by + z = 0

has no solution, then S is :

(a) an empty set

(b) an infinite set

(c) a finite set containing two or more elements

(d) a singleton

Answer: (d)

Solution:

= -(a – 1)²

For no solution, D = 0

So, -(a – 1)² = 0

⇒ a = 1

We get the first two planes co-incident for a = 1.

x + y + z = 1

x + y + z = 1

x + by + z = 0

If b = 1, the system will be inconsistent, hence no solution.

If b ≠ 1, the system will produce infinite solutions.

Hence, S has to be a singleton set {1} for no solution.

Also, Read:

JEE Advanced Maths Chapter-wise Previous Year Questions With Solutions