JEE Previous year set of questions on topic System of Linear Equations gives a clear understanding of the topic. These solutions help in breaking down all the difficult problems by the simple step-by-step method of solving. BYJUâ€™S provides accurate solutions, which are prepared by subject experts. This section deals with the solution of linear systems using different methods. How To Solve a Linear Equation System? Linear equations system can be solved using different 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. A linear system may behave in any one of 3 possible ways: The system has no solution, a single unique solution or infinitely many solutions.

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**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 + (a^{2} – 1)z = a + 1 then

(a) System has a unique solution for |a| = âˆš3

(b) System is inconsistence for |a| = âˆš3

(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 + (a^{2} – 1)z = a + 1 …..(3)

Consider a^{2} – 1 = 2

then LHS of (2) and (3) are same but RHS are not.

Hence a^{2} = 3 => |a| = âˆš3

For |a| = âˆš3, system is inconsistence.

So option (b) is correct.

**Question 2:** If the system of linear equations

2x + 2ay + az = 0

2x + 3by + bz = 0 and

2x + 4cy + cz = 0,

where a, b, c Ð„ R are non-zero and distinct; has 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,

Now,

=> (3b âˆ’ 2a)(c âˆ’ a) âˆ’ (4c âˆ’ 2a)(b âˆ’ a) = 0

=> 3bc âˆ’ 2ac âˆ’ 3ab + 2a^{2} âˆ’ [4bc âˆ’ 4ac âˆ’ 2ab + 2a^{2}] = 0

=> âˆ’bc + 2ac âˆ’ ab = 0

=> ab + bc = 2ac

=> 1/c + 1/a = 2/b

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

**Question 3:** If system of linear equations

x + y + z = 6

x + 2y + 3z = 10 and

3x + 2y + Î»z = Î¼

has more than two solutions, then Î¼ âˆ’ Î»^{2} is equal to ________.

**Solution: **

The system of equations has more than 2 solutions.

Find for D = D_{3} = 0

**Question 4:** For which of the following ordered pairs (Î¼, Î´), the system of linear equations

x + 2y + 3z = 1

3x + 4y + 5z = Î¼

4x + 4y + 4z = Î´

is inconsistent?

(a) (4, 6) (b) (3, 4) (c) (1, 0) (d) (4, 3)

**Answer: (d) **

**Solution:**

For inconsistent system, one of D_{x}, D_{y}, D_{z} should not be equal to 0.

Now,

For inconsistent system, 2Î¼ â‰ Î´ + 2

Therefore, the system will be inconsistent for Î¼ = 4, Î´ = 3.

**Question 5:** The system of linear equations

Î»x + 2y + 2z = 5

2Î»x + 3y + 5z = 8

4x + Î»y + 6z = 10 has:

(a) no solution when Î» = 2

(b) infinitely many solutions when Î» = 2

(c) no solution when Î» = 8

(d) a unique solution when Î» = -8

**Answer: (a)**

**Solution: **

Therefore, Equations have no solution for Î»= 2.

**Question 6:** The following system of linear equations

7x + 6y âˆ’ 2z = 0 ,

3x + 4y + 2z = 0

x âˆ’ 2y âˆ’ 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

7x + 6y âˆ’ 2z = 0 ,

3x + 4y + 2z = 0

x âˆ’ 2y âˆ’ 6z = 0,

As the system of equations are Homogeneous

=> The system is consistent.

=> Infinite solutions exist (both trivial and non-trivial solutions)

When, y = 2z

Letâ€™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 âˆ’ 10 = 0 (which is not possible)

Therefore, y = 2z

=> Infinitely many solutions not exist.

For x = 2z, letâ€™s take x = 2, z = 1, y = y

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

=> y = âˆ’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/y^{2} is equal to

(a) -10 (b) 10 (c) -30 (d) 30

**Answer: (b)**

**Solution: **

Given system of linear equations

x + ky + 3z = 0

3x + ky – 2z = 0 and

2x + 4y – 3z = 0

System has non zero solution, so

D = 0

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

Solving 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/y^{2} = (-5y)(-2y)/y^{2} = 10

**Question 8: **The number of real values of Î» for which the system of linear equations

2x + 4y âˆ’ Î»z = 0

4x + Î»y + 2z = 0

Î»x + 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, Î»^{3} + 4Î» – 40 has only one solution.

Since Î»(-âˆž) = -âˆž and Î»(âˆž) = âˆž

Here Î± is only one solution, so Î»(Î±) = 0

[Using intermediate value property]Now, differentiating Î»^{3} + 4Î» – 40 w.r.t. Î» we get

3Î»^{2} + 4 > 0

The equation can not have y, m, so

Î»(m) = 0 and Î»(y) = 0

Thus, the number of real values of Î» 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) âˆ’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 x (iii)

6x – y = 0 or y = 6x …..(iv)

Using (iv) in (i)

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) â€“ 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 â€˜bâ€™ 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} = 0

=> a = 1

We get 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 and hence no solution.

If b â‰ 1, the system will produce infinite solutions.

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